Definite Integrals
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Inequalities
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Riemann Sum and Integrability
0/22 completed

- Motivation to Riemann Sum
- Riemann Sum and Integrability
- Exercise 1
- Exercise 2
- Exercise 3
- Exercise 4
- Exercise 5
- Exercise 6
- Exercise 7
- Exercise 8
- Exercise 9
- Exercise 10
- Exercise 11
- Exercise 12
- Exercise 13
- Exercise 14 part 1
- Exercise 14 part 2
- Exercise 15 part 1
- Exercise 15 part 2
- Exercise 16
- Exercise 17
- Exercise 18

Fundamental Theorm of Calculus
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- Fundamental Theorem of Calculus
- First Fundamental Theorem I
- First Fundamental Theorem II
- Second Fundamental Theorem
- Exercise 1
- Exercise 2
- Exercise 3
- Exercise 4
- Exercise 5
- Exercise 6
- Exercise 7
- Exercise 8
- Exercise 9 part 1
- Exercise 9 part 2
- Exercise 9 part 3
- Exercise 9 part 4
- Exercise 10 part 1
- Exercise 10 part 2
- Exercise 10 part 3
- Exercise 11

Riemann Integration and Integrability
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- Introcution, Partition and Upper,Lower Riemann sums
- Examples - Upper,Lower Riemann sums
- Upper,Lower Riemann integrals and Integrability
- Examples - Riemann Integrability
- Exercise 1a
- Exercise 1b
- Exercise 2
- Partition and Refinements
- Riemann s criterion for Integrability
- Applications of Riemann s criterion for Integrability
- Exercise 3
- Exercise 4
- Exercise 5
- Exercise 6 - Properties of Definite Integrals
- Exercise 7 - Properties of Definite Integrals
- Exercise 8 - Properties of Definite Integrals
- Exercise 9
- Exercise 10
- Exercise 11

Further Exercises - Criterion for Integrability
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Advanced exercises - Riemann Sum and FTC
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- MVT for Integrals
- Exercise 1
- Exercise 2
- Exercise 3
- Exercise 4
- Exercise 5
- Exercise 6
- Exercise 7
- Exercise 8
- Exercise 9
- Exercise 10
- Exercise 11
- Exercise 12
- Exercise 13
- Exercise 14
- Exercise 15
- Exercise 16
- Exercise 17
- Exercise 18
- Exercise 19
- Exercise 20
- Exercise 21
- Exercise 22
- Exercise 23
- Exercise 24
- Exercise 25
- Exercise 26
- Exercise 27

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[{"Name":"Definite Integrals","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Introduction","Duration":"6m 40s","ChapterTopicVideoID":6358,"CourseChapterTopicPlaylistID":84700,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/6358.jpeg","UploadDate":"2019-12-11T21:04:45.8200000","DurationForVideoObject":"PT6M40S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.700","Text":"In this clip, I\u0027m going to talk about definite integral,"},{"Start":"00:02.700 ","End":"00:04.530","Text":"first of all how to write it."},{"Start":"00:04.530 ","End":"00:06.990","Text":"We also use the integral sign,"},{"Start":"00:06.990 ","End":"00:10.230","Text":"but we put 2 numbers, 1 here,"},{"Start":"00:10.230 ","End":"00:12.810","Text":"let\u0027s call it a, and 1 here b in practice,"},{"Start":"00:12.810 ","End":"00:14.430","Text":"they could be actual numbers."},{"Start":"00:14.430 ","End":"00:17.085","Text":"We have a function f. Let\u0027s assume f is continuous,"},{"Start":"00:17.085 ","End":"00:19.320","Text":"not always necessary, but just to be safe,"},{"Start":"00:19.320 ","End":"00:21.300","Text":"we\u0027ll deal with continuous functions."},{"Start":"00:21.300 ","End":"00:23.190","Text":"I have the integral,"},{"Start":"00:23.190 ","End":"00:25.860","Text":"let me say from a to b of f of x dx,"},{"Start":"00:25.860 ","End":"00:27.645","Text":"and this is how it\u0027s written."},{"Start":"00:27.645 ","End":"00:29.730","Text":"Now how do we define it as"},{"Start":"00:29.730 ","End":"00:32.160","Text":"actually 2 definitions under"},{"Start":"00:32.160 ","End":"00:35.295","Text":"the fundamental theorem of calculus that says they\u0027re the same."},{"Start":"00:35.295 ","End":"00:36.900","Text":"1 of them is geometric,"},{"Start":"00:36.900 ","End":"00:39.150","Text":"1 relates to the indefinite integral."},{"Start":"00:39.150 ","End":"00:42.000","Text":"I\u0027ll go with the geometric definition first."},{"Start":"00:42.000 ","End":"00:46.145","Text":"I brought in a picture I found on the Internet to illustrate this."},{"Start":"00:46.145 ","End":"00:49.430","Text":"Here we have a pair of axis, x and y."},{"Start":"00:49.430 ","End":"00:54.675","Text":"Here is the graph of the function f of x here with the points a and b here."},{"Start":"00:54.675 ","End":"00:58.050","Text":"Usually, a is on the left b is on the right, but not necessarily."},{"Start":"00:58.050 ","End":"01:00.800","Text":"For the moment we\u0027ll assume a is on the left b is on the right."},{"Start":"01:00.800 ","End":"01:03.650","Text":"If not, you\u0027ll have to reverse the answer."},{"Start":"01:03.650 ","End":"01:08.240","Text":"The integral from a to b of f of x dx as written here,"},{"Start":"01:08.240 ","End":"01:13.265","Text":"is just the area that\u0027s between the curve and the x-axis."},{"Start":"01:13.265 ","End":"01:16.985","Text":"Also going to assume that here we have a positive function."},{"Start":"01:16.985 ","End":"01:19.040","Text":"If a is bigger than b,"},{"Start":"01:19.040 ","End":"01:21.455","Text":"if a is on the right and b is on the left,"},{"Start":"01:21.455 ","End":"01:24.920","Text":"then we just take minus of the answer."},{"Start":"01:24.920 ","End":"01:31.340","Text":"I would say that the integral from b to a of f of"},{"Start":"01:31.340 ","End":"01:39.305","Text":"x dx is minus the integral from a to b of f of x dx."},{"Start":"01:39.305 ","End":"01:40.985","Text":"That\u0027s 1 thing."},{"Start":"01:40.985 ","End":"01:44.390","Text":"We can always assume that b is on the right and if not,"},{"Start":"01:44.390 ","End":"01:46.955","Text":"we just reverse and take the negative answer."},{"Start":"01:46.955 ","End":"01:48.829","Text":"The other thing is in this picture,"},{"Start":"01:48.829 ","End":"01:51.050","Text":"I assumed that the function was above."},{"Start":"01:51.050 ","End":"01:53.300","Text":"In general, that may not happen,"},{"Start":"01:53.300 ","End":"01:57.050","Text":"so I\u0027ll bring another picture where f goes from a to b,"},{"Start":"01:57.050 ","End":"02:00.320","Text":"but it\u0027s sometimes positive and sometimes negative,"},{"Start":"02:00.320 ","End":"02:01.535","Text":"and in this case,"},{"Start":"02:01.535 ","End":"02:05.180","Text":"it\u0027s not exactly the area under the curve."},{"Start":"02:05.180 ","End":"02:09.369","Text":"It is in essence if the curve is above the x-axis,"},{"Start":"02:09.369 ","End":"02:14.960","Text":"we take this area with a plus sign and the bits that the curve is below the x-axis,"},{"Start":"02:14.960 ","End":"02:17.380","Text":"we take those areas with a negative sign."},{"Start":"02:17.380 ","End":"02:21.230","Text":"This is like a negative area and then a bit of positive as well."},{"Start":"02:21.230 ","End":"02:23.705","Text":"We have this minus this plus this,"},{"Start":"02:23.705 ","End":"02:26.900","Text":"that will be what we call the definite integral from a to"},{"Start":"02:26.900 ","End":"02:30.800","Text":"b. I\u0027m not going to go into much more detail than that."},{"Start":"02:30.800 ","End":"02:34.070","Text":"There are exercises later on areas and things."},{"Start":"02:34.070 ","End":"02:37.850","Text":"But what I do want is the other definition of the definite integral,"},{"Start":"02:37.850 ","End":"02:39.890","Text":"which ties in with the indefinite."},{"Start":"02:39.890 ","End":"02:45.770","Text":"Suppose I have a function capital F of x,"},{"Start":"02:45.770 ","End":"02:48.470","Text":"which is the indefinite integral,"},{"Start":"02:48.470 ","End":"02:51.050","Text":"which is written also with this symbol,"},{"Start":"02:51.050 ","End":"02:55.415","Text":"but without anything written here and here of f of x dx."},{"Start":"02:55.415 ","End":"02:59.675","Text":"That means that F is an antiderivative of f,"},{"Start":"02:59.675 ","End":"03:02.300","Text":"usually become determinant precisely,"},{"Start":"03:02.300 ","End":"03:05.240","Text":"so we put this constant of integration."},{"Start":"03:05.240 ","End":"03:06.710","Text":"But in any event,"},{"Start":"03:06.710 ","End":"03:10.640","Text":"it means that F prime of x equals f."},{"Start":"03:10.640 ","End":"03:15.470","Text":"Let me just say that F will be 1 specific anti-derivative."},{"Start":"03:15.470 ","End":"03:16.610","Text":"Choose a particular C,"},{"Start":"03:16.610 ","End":"03:22.715","Text":"just fix some F. Then the other way of defining the integral from a to"},{"Start":"03:22.715 ","End":"03:29.150","Text":"b of f of x dx is just the antiderivative,"},{"Start":"03:29.150 ","End":"03:37.085","Text":"the indefinite integral applied to the point b minus f applied to the point a."},{"Start":"03:37.085 ","End":"03:40.220","Text":"There is a mathematical notation when you take a function,"},{"Start":"03:40.220 ","End":"03:42.395","Text":"substitute 2 values and subtract,"},{"Start":"03:42.395 ","End":"03:45.725","Text":"that\u0027s just f of x,"},{"Start":"03:45.725 ","End":"03:51.000","Text":"then we put a vertical line and we put here a and b,"},{"Start":"03:51.000 ","End":"03:54.750","Text":"which means substitute bx equals a and subtract,"},{"Start":"03:54.750 ","End":"03:57.395","Text":"sometimes we emphasize it by putting x equals,"},{"Start":"03:57.395 ","End":"04:01.400","Text":"there\u0027s a more old-fashioned notation which puts it in"},{"Start":"04:01.400 ","End":"04:06.050","Text":"square brackets and then you put the a here and the b here."},{"Start":"04:06.050 ","End":"04:08.900","Text":"Basically, if you see an expression not an integral with"},{"Start":"04:08.900 ","End":"04:11.990","Text":"a number on top and bottom usually means substitute this,"},{"Start":"04:11.990 ","End":"04:14.015","Text":"substitute this and subtract."},{"Start":"04:14.015 ","End":"04:18.185","Text":"I want to point out that the C would not make any difference."},{"Start":"04:18.185 ","End":"04:23.270","Text":"Suppose I took g of x is equal to f of x plus 5,"},{"Start":"04:23.270 ","End":"04:27.725","Text":"just some constant, and then if instead of F I use g,"},{"Start":"04:27.725 ","End":"04:33.260","Text":"I mean g of b minus g of a is going to come out the same because g"},{"Start":"04:33.260 ","End":"04:42.740","Text":"of b is going to be f of b plus 5 minus f of a plus 5."},{"Start":"04:42.740 ","End":"04:45.830","Text":"The 5s are going to cancel and it\u0027s just going to be"},{"Start":"04:45.830 ","End":"04:48.740","Text":"the same as this wherever constant I put,"},{"Start":"04:48.740 ","End":"04:50.660","Text":"it\u0027s going to appear plus and minus,"},{"Start":"04:50.660 ","End":"04:53.420","Text":"so it really doesn\u0027t depend on which"},{"Start":"04:53.420 ","End":"04:56.720","Text":"primitive or anti-derivative you choose here, though,"},{"Start":"04:56.720 ","End":"04:59.240","Text":"we don\u0027t really need the C. Let me just show you in"},{"Start":"04:59.240 ","End":"05:03.335","Text":"practice using this thing how we go about writing it."},{"Start":"05:03.335 ","End":"05:05.630","Text":"I\u0027m not going to relate to the area at the moment."},{"Start":"05:05.630 ","End":"05:10.160","Text":"But if I gave you an example let say I want the"},{"Start":"05:10.160 ","End":"05:17.715","Text":"integral from 1-2 of 3x squared dx,"},{"Start":"05:17.715 ","End":"05:20.020","Text":"then using this method,"},{"Start":"05:20.020 ","End":"05:23.470","Text":"we take F, which is an antiderivative."},{"Start":"05:23.470 ","End":"05:27.955","Text":"So what we do is we take the integral and irregular in the old sense,"},{"Start":"05:27.955 ","End":"05:32.230","Text":"which is the function x cubed right at the side."},{"Start":"05:32.230 ","End":"05:36.835","Text":"If I just say the integral of 3x squared dx,"},{"Start":"05:36.835 ","End":"05:42.850","Text":"then it\u0027s x cubed plus C. But here we just take any particular primitive."},{"Start":"05:42.850 ","End":"05:44.560","Text":"I\u0027ll take C equals 0 here,"},{"Start":"05:44.560 ","End":"05:48.730","Text":"and then we just put the numbers 1 and 2 here,"},{"Start":"05:48.730 ","End":"05:51.745","Text":"and then it means substitute 2,"},{"Start":"05:51.745 ","End":"05:56.080","Text":"so I\u0027ve got 2 cubed and then subtract substitute 1,"},{"Start":"05:56.080 ","End":"06:01.155","Text":"I get 1 cubed and that\u0027s 8 minus 1 is 7."},{"Start":"06:01.155 ","End":"06:03.650","Text":"Here\u0027s an example of an actual computation,"},{"Start":"06:03.650 ","End":"06:05.435","Text":"and this is how it typically goes."},{"Start":"06:05.435 ","End":"06:06.830","Text":"You\u0027re given numbers here,"},{"Start":"06:06.830 ","End":"06:10.520","Text":"you do the indefinite integral and you plug in the top limit,"},{"Start":"06:10.520 ","End":"06:13.595","Text":"plug in the bottom limit and subtract and you get the answer."},{"Start":"06:13.595 ","End":"06:17.570","Text":"But we also know that if you drew the graph of 3x squared,"},{"Start":"06:17.570 ","End":"06:20.905","Text":"and then I took 1 and I took 2,"},{"Start":"06:20.905 ","End":"06:26.775","Text":"then this area between 1 and 2 of the function 3x squared will also come out 7,"},{"Start":"06:26.775 ","End":"06:28.980","Text":"so that there are these 2 definitions,"},{"Start":"06:28.980 ","End":"06:34.265","Text":"1 involving area, 1 involving antiderivatives or indefinite integrals."},{"Start":"06:34.265 ","End":"06:35.630","Text":"That\u0027s all I have to say."},{"Start":"06:35.630 ","End":"06:41.250","Text":"The rest will be covered in the solved exercises. That\u0027s all for now."}],"ID":6370},{"Watched":false,"Name":"Exercise 1","Duration":"2m 10s","ChapterTopicVideoID":4509,"CourseChapterTopicPlaylistID":84700,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/4509.jpeg","UploadDate":"2017-01-26T13:14:24.9030000","DurationForVideoObject":"PT2M10S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.155","Text":"In this exercise, we have to compute a definite integral."},{"Start":"00:04.155 ","End":"00:05.370","Text":"How do I know it\u0027s definite?"},{"Start":"00:05.370 ","End":"00:07.260","Text":"Because it\u0027s got numbers here and here."},{"Start":"00:07.260 ","End":"00:09.840","Text":"The way we approach this, is,"},{"Start":"00:09.840 ","End":"00:16.260","Text":"we start off with the indefinite integral 2x squared gives us raise the power,"},{"Start":"00:16.260 ","End":"00:18.465","Text":"it\u0027s x cubed and divide by 3."},{"Start":"00:18.465 ","End":"00:21.480","Text":"It\u0027s 2/3 x cubed."},{"Start":"00:21.480 ","End":"00:24.360","Text":"Then the 4x raise the power by 1,"},{"Start":"00:24.360 ","End":"00:26.940","Text":"I get x squared and divide by 2."},{"Start":"00:26.940 ","End":"00:30.794","Text":"That gives me minus 2x squared."},{"Start":"00:30.794 ","End":"00:36.299","Text":"The 1 is just x. I don\u0027t need a constant in a definite integral."},{"Start":"00:36.299 ","End":"00:43.505","Text":"All I have to do is indicate that this is going to be from 1 to 4."},{"Start":"00:43.505 ","End":"00:49.290","Text":"Now, the 1 to 4 I colored for me like in the red means minus,"},{"Start":"00:49.290 ","End":"00:50.980","Text":"and then the black means plus,"},{"Start":"00:50.980 ","End":"00:53.690","Text":"is I substitute x equals 4 first,"},{"Start":"00:53.690 ","End":"00:58.340","Text":"and then I subtract from it what I get when I substitute x equals 1."},{"Start":"00:58.340 ","End":"01:09.275","Text":"This is going to equal 2-thirds times 4 cubed minus 2 times 4 squared plus 4."},{"Start":"01:09.275 ","End":"01:12.209","Text":"That\u0027s the 4 part."},{"Start":"01:12.209 ","End":"01:15.795","Text":"Then I\u0027m going to get minus."},{"Start":"01:15.795 ","End":"01:18.600","Text":"We\u0027re going to do the red part."},{"Start":"01:18.600 ","End":"01:21.435","Text":"2/3 times 1 cubed,"},{"Start":"01:21.435 ","End":"01:25.100","Text":"minus 2 times 1 squared, plus 1."},{"Start":"01:25.100 ","End":"01:27.980","Text":"In each case, I just copied this and instead of x,"},{"Start":"01:27.980 ","End":"01:29.450","Text":"I put 4, instead of x,"},{"Start":"01:29.450 ","End":"01:31.895","Text":"I took 1 and I subtract the 2."},{"Start":"01:31.895 ","End":"01:34.595","Text":"Now it\u0027s just computations."},{"Start":"01:34.595 ","End":"01:38.270","Text":"4 cubed is 64,"},{"Start":"01:38.270 ","End":"01:43.950","Text":"2/3 of 64 is 42 and 2/3."},{"Start":"01:43.950 ","End":"01:45.905","Text":"Here we have at C,"},{"Start":"01:45.905 ","End":"01:48.890","Text":"4 squared is 16 minus 32,"},{"Start":"01:48.890 ","End":"01:53.090","Text":"plus 4, is like minus 28."},{"Start":"01:53.090 ","End":"01:55.850","Text":"Then we\u0027re going to have here"},{"Start":"01:55.850 ","End":"02:04.935","Text":"minus 2/3 powers of 1 or just 1 minus 2, and plus 1."},{"Start":"02:04.935 ","End":"02:06.555","Text":"What do we get?"},{"Start":"02:06.555 ","End":"02:11.170","Text":"13? That\u0027s the answer. Done."}],"ID":4518},{"Watched":false,"Name":"Exercise 2","Duration":"1m 37s","ChapterTopicVideoID":4510,"CourseChapterTopicPlaylistID":84700,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/4510.jpeg","UploadDate":"2017-01-26T13:15:08.9070000","DurationForVideoObject":"PT1M37S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.930","Text":"In this exercise, we have to compute the definite end to"},{"Start":"00:03.930 ","End":"00:08.085","Text":"grow it\u0027s definite because it has numbers here of this function."},{"Start":"00:08.085 ","End":"00:11.130","Text":"We happen to be lucky here because I noticed"},{"Start":"00:11.130 ","End":"00:14.400","Text":"immediately and perhaps if you have sharp eyes,"},{"Start":"00:14.400 ","End":"00:15.840","Text":"you would notice it too,"},{"Start":"00:15.840 ","End":"00:19.410","Text":"that the derivative of x squared plus x plus"},{"Start":"00:19.410 ","End":"00:24.240","Text":"1 is exactly 2x plus 1 because from here I get 2x, from here I get 1."},{"Start":"00:24.240 ","End":"00:27.180","Text":"It\u0027s worth noting this thing because then we\u0027re lucky,"},{"Start":"00:27.180 ","End":"00:33.855","Text":"we can use the formula that the integral of f prime over f,"},{"Start":"00:33.855 ","End":"00:35.790","Text":"where f is a function of x,"},{"Start":"00:35.790 ","End":"00:38.845","Text":"is exactly the natural log of"},{"Start":"00:38.845 ","End":"00:43.250","Text":"f. Here this is exactly the case where f is the denominator,"},{"Start":"00:43.250 ","End":"00:50.510","Text":"f prime is the numerator so we get the natural log of x squared"},{"Start":"00:50.510 ","End":"00:58.710","Text":"plus x plus 1 and this we have to take between 0 and 2,"},{"Start":"00:58.710 ","End":"01:00.520","Text":"which means that we plugin 2,"},{"Start":"01:00.520 ","End":"01:02.855","Text":"we plugin 0 and we subtract."},{"Start":"01:02.855 ","End":"01:04.700","Text":"What we get if we put in 2?"},{"Start":"01:04.700 ","End":"01:11.760","Text":"2 squared is 4 plus 2 plus 1 is 7."},{"Start":"01:13.760 ","End":"01:16.520","Text":"Don\u0027t need the absolute value of the 7,"},{"Start":"01:16.520 ","End":"01:20.180","Text":"that\u0027s also 7 less plugin 0,"},{"Start":"01:20.180 ","End":"01:25.835","Text":"we get natural log of 1 because 0 squared plus 0 plus 1 is 1."},{"Start":"01:25.835 ","End":"01:33.080","Text":"Now, natural log of 1 is 0 so we\u0027re just left with the natural log of 7."},{"Start":"01:33.080 ","End":"01:38.010","Text":"No need to compute it this is precise. We\u0027re done."}],"ID":4519},{"Watched":false,"Name":"Exercise 3","Duration":"4m 2s","ChapterTopicVideoID":4511,"CourseChapterTopicPlaylistID":84700,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/4511.jpeg","UploadDate":"2017-01-26T13:16:18.8270000","DurationForVideoObject":"PT4M2S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.770","Text":"Here we have a definite integral to compute and in this case,"},{"Start":"00:04.770 ","End":"00:10.800","Text":"it looks like integration by parts and I\u0027m going to remind you of the formula that\u0027s the"},{"Start":"00:10.800 ","End":"00:19.455","Text":"integral of udv is equal to uv minus the integral of vdu."},{"Start":"00:19.455 ","End":"00:22.665","Text":"Now, when we have integration by parts,"},{"Start":"00:22.665 ","End":"00:26.985","Text":"there\u0027s 2 ways of solving it and I\u0027m going to do it both ways."},{"Start":"00:26.985 ","End":"00:33.030","Text":"One way, which I\u0027m going to do over here is just computing the indefinite integral"},{"Start":"00:33.030 ","End":"00:39.555","Text":"first so actually what I\u0027m going to do is just erase this and this and at the end,"},{"Start":"00:39.555 ","End":"00:42.195","Text":"I\u0027m going to substitute the 2 and the 3."},{"Start":"00:42.195 ","End":"00:45.890","Text":"This 1 is better to take as u and this 1 as"},{"Start":"00:45.890 ","End":"00:51.130","Text":"dv because we don\u0027t want to integrate the x we want to differentiate it."},{"Start":"00:51.130 ","End":"00:54.660","Text":"We get the u times v,"},{"Start":"00:54.660 ","End":"01:00.200","Text":"so here\u0027s u, we need to know du and v because we have u and dv,"},{"Start":"01:00.200 ","End":"01:03.410","Text":"du is just 1 dx,"},{"Start":"01:03.410 ","End":"01:06.890","Text":"which means dx and v is the integral of this,"},{"Start":"01:06.890 ","End":"01:09.830","Text":"so it\u0027s minus e to the minus x."},{"Start":"01:09.830 ","End":"01:13.160","Text":"Now if I substitute here, I get u,"},{"Start":"01:13.160 ","End":"01:14.930","Text":"which is x, v,"},{"Start":"01:14.930 ","End":"01:17.510","Text":"which is minus e to the minus x,"},{"Start":"01:17.510 ","End":"01:23.675","Text":"minus e to the minus x minus the integral of v,"},{"Start":"01:23.675 ","End":"01:32.040","Text":"which is minus e to the minus x and du, which is dx."},{"Start":"01:32.040 ","End":"01:35.345","Text":"This becomes minus x,"},{"Start":"01:35.345 ","End":"01:39.350","Text":"e to the minus x and then the integral of minus e to"},{"Start":"01:39.350 ","End":"01:43.625","Text":"the minus x is just plus e to the minus x."},{"Start":"01:43.625 ","End":"01:47.585","Text":"We still end up with a minus plus constant,"},{"Start":"01:47.585 ","End":"01:52.670","Text":"that\u0027s the indefinite and I want to go now to the definite,"},{"Start":"01:52.670 ","End":"01:55.235","Text":"so I just plug in the 2 and the 3."},{"Start":"01:55.235 ","End":"01:59.030","Text":"What I need to do here is to take minus x,"},{"Start":"01:59.030 ","End":"02:00.770","Text":"e to the minus x,"},{"Start":"02:00.770 ","End":"02:06.270","Text":"minus e to the minus x between 2 and 3."},{"Start":"02:06.270 ","End":"02:10.460","Text":"Then I\u0027ll take minus e to the minus x outside the brackets,"},{"Start":"02:10.460 ","End":"02:12.815","Text":"and I\u0027m left with x plus 1."},{"Start":"02:12.815 ","End":"02:16.530","Text":"That might be easier to substitute and then I\u0027ll take it from 2-3,"},{"Start":"02:16.530 ","End":"02:21.900","Text":"the whole thing and so we will get, if put in 3,"},{"Start":"02:21.900 ","End":"02:24.539","Text":"I\u0027ll get x plus 1 is 4,"},{"Start":"02:24.539 ","End":"02:26.130","Text":"so it\u0027s minus 4,"},{"Start":"02:26.130 ","End":"02:27.690","Text":"e to the minus 4,"},{"Start":"02:27.690 ","End":"02:31.559","Text":"and if I put in 2, that becomes 3,"},{"Start":"02:31.559 ","End":"02:33.950","Text":"so it\u0027s minus, minus,"},{"Start":"02:33.950 ","End":"02:35.500","Text":"that\u0027s a plus here,"},{"Start":"02:35.500 ","End":"02:39.240","Text":"3, e to the minus 2,"},{"Start":"02:39.240 ","End":"02:41.655","Text":"this should be a minus 3, sorry."},{"Start":"02:41.655 ","End":"02:46.445","Text":"In summary, this method is to find the indefinite integral"},{"Start":"02:46.445 ","End":"02:51.350","Text":"and when we found it then to plug in the 3 and the 2 and make it definite."},{"Start":"02:51.350 ","End":"02:57.200","Text":"The other method is to constantly stay with the definite with the 2 and the 3."},{"Start":"02:57.200 ","End":"03:00.950","Text":"Copying from here minus x,"},{"Start":"03:00.950 ","End":"03:06.680","Text":"e to the minus x and all this from 2-3, so not an integral,"},{"Start":"03:06.680 ","End":"03:09.290","Text":"just substitute because this is not an integral already,"},{"Start":"03:09.290 ","End":"03:17.875","Text":"minus the integral from 2-3 of minus e to the minus x, dx."},{"Start":"03:17.875 ","End":"03:19.670","Text":"I get minus x,"},{"Start":"03:19.670 ","End":"03:23.945","Text":"e to the minus x from 2-3,"},{"Start":"03:23.945 ","End":"03:32.945","Text":"minus this integral becomes e to the minus x also from 2-3."},{"Start":"03:32.945 ","End":"03:38.810","Text":"I can just combine them and have 1 thing going from 2-3, so let\u0027s see what we have."},{"Start":"03:38.810 ","End":"03:41.090","Text":"We have minus x,"},{"Start":"03:41.090 ","End":"03:43.384","Text":"e to the minus x,"},{"Start":"03:43.384 ","End":"03:51.620","Text":"minus e to the minus x from 2-3 and at this point,"},{"Start":"03:51.620 ","End":"03:54.935","Text":"we just finish it the same way as here."},{"Start":"03:54.935 ","End":"03:59.105","Text":"I mean, you can go back to here because this is the same as this."},{"Start":"03:59.105 ","End":"04:02.550","Text":"We\u0027re done, I mean, this is the answer."}],"ID":4520},{"Watched":false,"Name":"Exercise 4","Duration":"3m 36s","ChapterTopicVideoID":4512,"CourseChapterTopicPlaylistID":84700,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/4512.jpeg","UploadDate":"2017-01-26T13:17:27.9800000","DurationForVideoObject":"PT3M36S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.570","Text":"Here, we have to compute this definite integral,"},{"Start":"00:03.570 ","End":"00:05.970","Text":"has numbers here, so it\u0027s a definite integral,"},{"Start":"00:05.970 ","End":"00:09.960","Text":"and it looks like a case for integration by substitution,"},{"Start":"00:09.960 ","End":"00:12.360","Text":"where I\u0027m going to substitute the natural log of x."},{"Start":"00:12.360 ","End":"00:13.800","Text":"Now when we have a substitution,"},{"Start":"00:13.800 ","End":"00:16.500","Text":"there are 2 methods, and I\u0027m going to demonstrate both method."},{"Start":"00:16.500 ","End":"00:18.540","Text":"I\u0027ll demonstrate 1 method over here."},{"Start":"00:18.540 ","End":"00:21.690","Text":"Just copy the exercise from 1 to 4,"},{"Start":"00:21.690 ","End":"00:28.545","Text":"the integral of natural log^4 of x over x dx."},{"Start":"00:28.545 ","End":"00:33.800","Text":"I\u0027m going to make the substitution t equals natural log of x."},{"Start":"00:33.800 ","End":"00:37.340","Text":"Now in the first method, by substitution,"},{"Start":"00:37.340 ","End":"00:42.500","Text":"I\u0027m going to go into the world of t and I\u0027m never going to come back to the world of x."},{"Start":"00:42.500 ","End":"00:46.820","Text":"In the other method, we just do the indefinite integral first,"},{"Start":"00:46.820 ","End":"00:50.555","Text":"and then at the end we substitute the limits 1 and 4 and subtract."},{"Start":"00:50.555 ","End":"00:54.500","Text":"Let\u0027s go with this way where we just go totally over to t. What"},{"Start":"00:54.500 ","End":"00:59.970","Text":"we get is dt is 1 over x dx."},{"Start":"00:59.970 ","End":"01:04.045","Text":"I see here that I already have 1 over x dx,"},{"Start":"01:04.045 ","End":"01:07.010","Text":"so I don\u0027t need to play with this and"},{"Start":"01:07.010 ","End":"01:10.130","Text":"get what dx\u0027s separately because I only need dx over x,"},{"Start":"01:10.130 ","End":"01:11.780","Text":"so I get the integral."},{"Start":"01:11.780 ","End":"01:13.415","Text":"Before I write the limits,"},{"Start":"01:13.415 ","End":"01:16.175","Text":"these are limits for x, or let me just leave it blank for a moment."},{"Start":"01:16.175 ","End":"01:21.175","Text":"What I have is t^4 and dx over x is dt."},{"Start":"01:21.175 ","End":"01:26.130","Text":"But the thing is, that the limits have to be substituted also."},{"Start":"01:26.130 ","End":"01:28.460","Text":"If t is natural log of x,"},{"Start":"01:28.460 ","End":"01:30.665","Text":"when x is 1,"},{"Start":"01:30.665 ","End":"01:34.130","Text":"then t is natural log of 1."},{"Start":"01:34.130 ","End":"01:39.810","Text":"When x is 4, then t is natural log of 4."},{"Start":"01:39.810 ","End":"01:43.830","Text":"What I get is natural log of 1 is 0."},{"Start":"01:43.830 ","End":"01:49.170","Text":"Natural log of 4 is just natural log of t^4 dt."},{"Start":"01:49.170 ","End":"01:54.750","Text":"I\u0027m continuing here, so we get t^5 over"},{"Start":"01:54.750 ","End":"02:01.525","Text":"5 between the limits of 0 and natural log of 4."},{"Start":"02:01.525 ","End":"02:04.110","Text":"When t is 0,"},{"Start":"02:04.110 ","End":"02:05.520","Text":"this thing is 0."},{"Start":"02:05.520 ","End":"02:13.680","Text":"Basically, what I get is 1/5 natural log of 4^5 like the 5 here."},{"Start":"02:13.680 ","End":"02:15.570","Text":"That would be my answer."},{"Start":"02:15.570 ","End":"02:17.390","Text":"That\u0027s 1 method."},{"Start":"02:17.390 ","End":"02:23.375","Text":"Now, the other method is to just do the indefinite integral."},{"Start":"02:23.375 ","End":"02:26.360","Text":"Actually, I\u0027ve already done that in a way,"},{"Start":"02:26.360 ","End":"02:29.390","Text":"the indefinite integral of this."},{"Start":"02:29.390 ","End":"02:36.550","Text":"Then I make the substitution where t is natural log of x, the same substitution."},{"Start":"02:36.550 ","End":"02:38.240","Text":"I\u0027ll use it over there,"},{"Start":"02:38.240 ","End":"02:43.015","Text":"and then I get the integral of t^4 dt,"},{"Start":"02:43.015 ","End":"02:44.950","Text":"but indefinite, there\u0027s no limit."},{"Start":"02:44.950 ","End":"02:52.040","Text":"Then I get the same thing which is t^5 over 5 plus a constant,"},{"Start":"02:52.040 ","End":"02:54.110","Text":"and then I substitute back."},{"Start":"02:54.110 ","End":"02:55.730","Text":"Now here\u0027s the difference; here,"},{"Start":"02:55.730 ","End":"02:58.190","Text":"we didn\u0027t go back from t to x."},{"Start":"02:58.190 ","End":"03:02.720","Text":"Here, we go back and we say that t is natural log of x,"},{"Start":"03:02.720 ","End":"03:07.875","Text":"so I get natural log^5 of x over 5,"},{"Start":"03:07.875 ","End":"03:13.500","Text":"and then I put in the original limits 4 and 1."},{"Start":"03:13.500 ","End":"03:16.615","Text":"Then when x is equal to 4,"},{"Start":"03:16.615 ","End":"03:22.880","Text":"I get natural log^5 of 4 over 5."},{"Start":"03:22.880 ","End":"03:26.240","Text":"When x is 1, the natural log of x is 0,"},{"Start":"03:26.240 ","End":"03:31.035","Text":"so minus 0, and we basically got the same answer,"},{"Start":"03:31.035 ","End":"03:33.380","Text":"this, just erase the 0."},{"Start":"03:33.380 ","End":"03:36.570","Text":"These are the 2 ways and we are done."}],"ID":4521},{"Watched":false,"Name":"Exercise 5","Duration":"2m 18s","ChapterTopicVideoID":4513,"CourseChapterTopicPlaylistID":84700,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/4513.jpeg","UploadDate":"2017-01-26T13:18:09.3230000","DurationForVideoObject":"PT2M18S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.075","Text":"Here we have to compute this definite integral,"},{"Start":"00:03.075 ","End":"00:04.650","Text":"just copied it over here."},{"Start":"00:04.650 ","End":"00:07.700","Text":"I\u0027m going to use a trigonometric identity to help us,"},{"Start":"00:07.700 ","End":"00:10.950","Text":"and that is that the cosine squared"},{"Start":"00:10.950 ","End":"00:15.660","Text":"of an angle Alpha is just 1/2 of 1"},{"Start":"00:15.660 ","End":"00:19.950","Text":"plus cosine of twice the angle, 2 Alpha."},{"Start":"00:19.950 ","End":"00:23.295","Text":"In our case, the Alpha is 4x."},{"Start":"00:23.295 ","End":"00:25.620","Text":"What I\u0027m going to get"},{"Start":"00:25.620 ","End":"00:29.370","Text":"is the integral from 1 to Pi."},{"Start":"00:29.370 ","End":"00:31.800","Text":"Now, the half I can take in front,"},{"Start":"00:31.800 ","End":"00:33.360","Text":"that\u0027s the 1/2,"},{"Start":"00:33.360 ","End":"00:35.180","Text":"and I\u0027ve got 1 plus cosine,"},{"Start":"00:35.180 ","End":"00:39.810","Text":"2 Alpha is just 8x dx."},{"Start":"00:39.810 ","End":"00:42.940","Text":"This gives me 1/2."},{"Start":"00:42.940 ","End":"00:44.645","Text":"Now, the integral of this,"},{"Start":"00:44.645 ","End":"00:46.370","Text":"the integral of 1 is x"},{"Start":"00:46.370 ","End":"00:49.040","Text":"and the integral of cosine 8x."},{"Start":"00:49.040 ","End":"00:50.824","Text":"Well, I\u0027ll give you another formula."},{"Start":"00:50.824 ","End":"00:57.095","Text":"The integral of cosine of ax dx in general,"},{"Start":"00:57.095 ","End":"00:59.645","Text":"is equal to 1 over a,"},{"Start":"00:59.645 ","End":"01:03.245","Text":"the sine of ax plus constant."},{"Start":"01:03.245 ","End":"01:07.849","Text":"What I get here is 1/2 x plus,"},{"Start":"01:07.849 ","End":"01:13.655","Text":"1 over a is 1 over 8 sine 8x."},{"Start":"01:13.655 ","End":"01:16.715","Text":"Now, I have to also put in the limits,"},{"Start":"01:16.715 ","End":"01:19.850","Text":"so that\u0027s from 1 to Pi,"},{"Start":"01:19.850 ","End":"01:21.635","Text":"and let\u0027s see what we get."},{"Start":"01:21.635 ","End":"01:23.450","Text":"We get 1/2."},{"Start":"01:23.450 ","End":"01:26.130","Text":"Now, if I put in Pi,"},{"Start":"01:26.130 ","End":"01:30.795","Text":"I get Pi, and what is sine of 8 Pi?"},{"Start":"01:30.795 ","End":"01:34.815","Text":"Sine of 8 Pi is like 4 times 2 Pi."},{"Start":"01:34.815 ","End":"01:36.860","Text":"It\u0027s 4 complete circles,"},{"Start":"01:36.860 ","End":"01:38.780","Text":"it\u0027s like the sine of 0."},{"Start":"01:38.780 ","End":"01:41.209","Text":"So the sine of 0 is 0."},{"Start":"01:41.209 ","End":"01:44.800","Text":"I\u0027m saying that this thing here is 0,"},{"Start":"01:44.800 ","End":"01:46.655","Text":"sine of 8 Pi,"},{"Start":"01:46.655 ","End":"01:49.580","Text":"and that just gives me nothing."},{"Start":"01:49.580 ","End":"01:51.410","Text":"Now, we subtract the lower bit."},{"Start":"01:51.410 ","End":"01:53.750","Text":"So it\u0027s minus 1,"},{"Start":"01:53.750 ","End":"01:58.815","Text":"and minus 1/8 of sine of 8x,"},{"Start":"01:58.815 ","End":"02:00.850","Text":"which is just 8."},{"Start":"02:00.850 ","End":"02:03.290","Text":"If I combine this, what do I get?"},{"Start":"02:03.290 ","End":"02:08.690","Text":"This disappears and I get 1/2 of Pi"},{"Start":"02:08.690 ","End":"02:13.690","Text":"minus 1 minus 1/8 sine 8,"},{"Start":"02:13.690 ","End":"02:16.350","Text":"and that\u0027s the answer."},{"Start":"02:16.350 ","End":"02:18.820","Text":"We\u0027re done."}],"ID":4522},{"Watched":false,"Name":"Exercise 6","Duration":"3m 10s","ChapterTopicVideoID":4514,"CourseChapterTopicPlaylistID":84700,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/4514.jpeg","UploadDate":"2017-01-26T13:19:12.0270000","DurationForVideoObject":"PT3M10S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.810","Text":"In this exercise, we have to compute a definite integral which is defined piece-wise."},{"Start":"00:06.810 ","End":"00:12.390","Text":"I mean the function is defined piece-wise or to split function from 0 to 1, it\u0027s 1 way,"},{"Start":"00:12.390 ","End":"00:16.050","Text":"and from bigger than or equal to 1 it\u0027s another way."},{"Start":"00:16.050 ","End":"00:18.120","Text":"I\u0027ve copied the exercise."},{"Start":"00:18.120 ","End":"00:22.935","Text":"The way we handle this is we just split up the range from 0 to 4,"},{"Start":"00:22.935 ","End":"00:25.230","Text":"2 sub ranges from 0 to 1,"},{"Start":"00:25.230 ","End":"00:26.880","Text":"and from 1 to 4."},{"Start":"00:26.880 ","End":"00:32.100","Text":"What we get is the integral from 0 to 1 of f of"},{"Start":"00:32.100 ","End":"00:40.835","Text":"x dx plus the integral from 1 to 4 of f of x dx."},{"Start":"00:40.835 ","End":"00:42.515","Text":"By which I mean,"},{"Start":"00:42.515 ","End":"00:47.390","Text":"we can just rewrite this because we know what f of x is between 0 and 1."},{"Start":"00:47.390 ","End":"00:53.970","Text":"I\u0027ve got the integral of the square roots of x from 0 to 1 dx."},{"Start":"00:53.970 ","End":"00:55.265","Text":"Then the second bit,"},{"Start":"00:55.265 ","End":"00:57.650","Text":"f of x here is 1 over x squared."},{"Start":"00:57.650 ","End":"01:02.720","Text":"I have the integral from 1 to 4 of 1 over x squared dx."},{"Start":"01:02.720 ","End":"01:09.920","Text":"Now I\u0027d like to write this in exponential notation because we going to use the formula,"},{"Start":"01:09.920 ","End":"01:15.530","Text":"standard formula that the integral of x^n dx"},{"Start":"01:15.530 ","End":"01:22.925","Text":"is x^n plus 1 over n plus 1 plus the constant in the indefinite case."},{"Start":"01:22.925 ","End":"01:25.040","Text":"Coming back here, what we get,"},{"Start":"01:25.040 ","End":"01:27.020","Text":"first of all put into an exponential."},{"Start":"01:27.020 ","End":"01:29.815","Text":"This is x to the power of 1 half."},{"Start":"01:29.815 ","End":"01:35.925","Text":"This 1 is from 1 to 4 x to the minus 2 dx."},{"Start":"01:35.925 ","End":"01:38.030","Text":"At this point I apply this formula,"},{"Start":"01:38.030 ","End":"01:42.720","Text":"x to the 1 half plus 1 is x to the power of 3"},{"Start":"01:42.720 ","End":"01:48.020","Text":"over 2 and divide it by 3 over 2, like this."},{"Start":"01:48.020 ","End":"01:55.295","Text":"Then plus x to the minus 1 over minus 1."},{"Start":"01:55.295 ","End":"02:00.390","Text":"This 1 is taken from 0 to 1,"},{"Start":"02:00.390 ","End":"02:05.205","Text":"and this 1 I\u0027m taking from 1 to 4."},{"Start":"02:05.205 ","End":"02:07.425","Text":"Let\u0027s see, what do we get for the first 1?"},{"Start":"02:07.425 ","End":"02:12.275","Text":"When x is 1, it\u0027s 1 to the power of 3 over 2,"},{"Start":"02:12.275 ","End":"02:14.090","Text":"which is just 1,"},{"Start":"02:14.090 ","End":"02:17.735","Text":"1 over 3 over 2 is 2/3."},{"Start":"02:17.735 ","End":"02:21.560","Text":"When x is 0, it\u0027s just 0."},{"Start":"02:21.560 ","End":"02:24.980","Text":"This is 2/3 minus 0 plus."},{"Start":"02:24.980 ","End":"02:30.215","Text":"Now here I have to subtract what I get when x is 1 from what I get when x is 4."},{"Start":"02:30.215 ","End":"02:35.420","Text":"When x is 4, x to the minus 1 is 1 over 4,"},{"Start":"02:35.420 ","End":"02:40.795","Text":"so it\u0027s 1/4 over minus 1."},{"Start":"02:40.795 ","End":"02:45.075","Text":"When x is 1, it\u0027s 1 over 1,"},{"Start":"02:45.075 ","End":"02:48.915","Text":"which is 1 over minus 1."},{"Start":"02:48.915 ","End":"02:50.385","Text":"In other words, let\u0027s see,"},{"Start":"02:50.385 ","End":"02:57.965","Text":"this becomes minus 1/4 and this bit becomes plus 1,"},{"Start":"02:57.965 ","End":"03:00.395","Text":"so altogether what do I get;"},{"Start":"03:00.395 ","End":"03:04.160","Text":"2/3 minus 1/4 plus 1."},{"Start":"03:04.160 ","End":"03:08.010","Text":"I think it\u0027s 1 and 5/12 if I haven\u0027t made a mistake."},{"Start":"03:08.010 ","End":"03:10.990","Text":"Anyway, this should be our answer."}],"ID":4523},{"Watched":false,"Name":"Exercise 7","Duration":"6m 27s","ChapterTopicVideoID":4515,"CourseChapterTopicPlaylistID":84700,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/4515.jpeg","UploadDate":"2017-03-12T06:28:52.4330000","DurationForVideoObject":"PT6M27S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.160","Text":"In this exercise, we have to compute the definite integral of this function,"},{"Start":"00:05.160 ","End":"00:07.635","Text":"which contains an absolute value."},{"Start":"00:07.635 ","End":"00:09.720","Text":"I don\u0027t like the absolute value,"},{"Start":"00:09.720 ","End":"00:11.850","Text":"so I\u0027m going to rewrite it as a split function,"},{"Start":"00:11.850 ","End":"00:13.575","Text":"or piece-wise defined function."},{"Start":"00:13.575 ","End":"00:18.060","Text":"So what we\u0027re integrating is some function f of x,"},{"Start":"00:18.060 ","End":"00:26.680","Text":"which is equal to the square root of 4 plus x minus 1 in absolute value."},{"Start":"00:26.960 ","End":"00:30.400","Text":"Well, let\u0027s remember what the absolute value is."},{"Start":"00:30.400 ","End":"00:38.000","Text":"The absolute value is defined as absolute value of sum number a is equal to piece-wise,"},{"Start":"00:38.000 ","End":"00:39.695","Text":"is either equal to a itself,"},{"Start":"00:39.695 ","End":"00:41.990","Text":"if a is already non-negative,"},{"Start":"00:41.990 ","End":"00:43.550","Text":"but if a is negative,"},{"Start":"00:43.550 ","End":"00:46.700","Text":"then we have to make it minus a to make it positive,"},{"Start":"00:46.700 ","End":"00:49.115","Text":"so this is how it\u0027s defined."},{"Start":"00:49.115 ","End":"00:50.860","Text":"What we get here,"},{"Start":"00:50.860 ","End":"00:54.200","Text":"if x minus 1 is bigger, or equal to 0,"},{"Start":"00:54.200 ","End":"00:57.110","Text":"then we can just drop the absolute value and we say it\u0027s"},{"Start":"00:57.110 ","End":"01:01.550","Text":"4 plus just x minus 1 under the square root,"},{"Start":"01:01.550 ","End":"01:05.695","Text":"provided that x minus 1 is bigger or equal to 0,"},{"Start":"01:05.695 ","End":"01:09.170","Text":"and if x minus 1 is less than 0,"},{"Start":"01:09.170 ","End":"01:15.425","Text":"then we get the square root of 4 minus from this minus here, x minus 1."},{"Start":"01:15.425 ","End":"01:19.165","Text":"Let me simplify this a bit, this will equal,"},{"Start":"01:19.165 ","End":"01:24.450","Text":"let\u0027s see, 4 plus x minus 1 is 3 plus x,"},{"Start":"01:24.450 ","End":"01:30.440","Text":"so let\u0027s write this as the square root of 3 plus x or x plus 3,"},{"Start":"01:30.440 ","End":"01:33.320","Text":"this will be when x minus 1 is bigger equal to 0,"},{"Start":"01:33.320 ","End":"01:35.630","Text":"which means x is bigger or equal to 1,"},{"Start":"01:35.630 ","End":"01:41.845","Text":"and it will equal 4 minus x minus 1 is 5 minus x,"},{"Start":"01:41.845 ","End":"01:44.700","Text":"that\u0027s when x minus 1 less than 0,"},{"Start":"01:44.700 ","End":"01:47.565","Text":"or x less than 1."},{"Start":"01:47.565 ","End":"01:55.260","Text":"What I have here is actually the integral of f of x dx,"},{"Start":"01:55.260 ","End":"01:57.075","Text":"because this is what my f be,"},{"Start":"01:57.075 ","End":"01:58.735","Text":"this was from minus 1-4."},{"Start":"01:58.735 ","End":"02:01.235","Text":"Now, what\u0027s something happens at x equals 1?"},{"Start":"02:01.235 ","End":"02:05.430","Text":"So I have to break the integral up before and after x equals 1,"},{"Start":"02:05.430 ","End":"02:12.180","Text":"so a first bit is from minus 1 to 1 of f of x dx,"},{"Start":"02:12.180 ","End":"02:18.780","Text":"and the second bit is from 1 up to 4 of f of x dx,"},{"Start":"02:18.780 ","End":"02:21.275","Text":"but I need to replace f with what it is,"},{"Start":"02:21.275 ","End":"02:31.080","Text":"so what it is is the integral from minus 1 to 1 of f of x. X is on the less than 1 bit,"},{"Start":"02:31.080 ","End":"02:35.880","Text":"so it\u0027s square root of 5 minus x dx,"},{"Start":"02:35.880 ","End":"02:40.440","Text":"and from 1-4, it\u0027s only bigger than 1 bit,"},{"Start":"02:40.440 ","End":"02:45.060","Text":"so it\u0027s the square root of x plus 3 dx."},{"Start":"02:45.060 ","End":"02:47.510","Text":"Now, we have to compute these 2 integrals separately,"},{"Start":"02:47.510 ","End":"02:49.430","Text":"and then add the results."},{"Start":"02:49.430 ","End":"02:51.740","Text":"I\u0027m going to write a formula here,"},{"Start":"02:51.740 ","End":"02:53.410","Text":"which will help me here,"},{"Start":"02:53.410 ","End":"02:55.325","Text":"and the formula is this,"},{"Start":"02:55.325 ","End":"03:00.155","Text":"which basically tells me how to integrate an exponent when it\u0027s not x,"},{"Start":"03:00.155 ","End":"03:01.820","Text":"but a linear function of x,"},{"Start":"03:01.820 ","End":"03:03.860","Text":"and it\u0027s very similar to what we do for x,"},{"Start":"03:03.860 ","End":"03:05.540","Text":"just raise the power by 1,"},{"Start":"03:05.540 ","End":"03:06.980","Text":"and divide by the new power,"},{"Start":"03:06.980 ","End":"03:12.065","Text":"except that we have to also divide by the coefficient of x."},{"Start":"03:12.065 ","End":"03:13.910","Text":"Now in our case, we don\u0027t have an exponent,"},{"Start":"03:13.910 ","End":"03:15.410","Text":"but yes we do really,"},{"Start":"03:15.410 ","End":"03:18.050","Text":"because the square root is to the power of 1/2,"},{"Start":"03:18.050 ","End":"03:22.820","Text":"so this is equal to the integral of minus 1 to 1,"},{"Start":"03:22.820 ","End":"03:28.230","Text":"of 5 minus x to the power of 1/2 dx,"},{"Start":"03:28.230 ","End":"03:34.370","Text":"plus the integral from 1 to 4 x plus 3 to the power of 1/2 dx."},{"Start":"03:34.370 ","End":"03:39.755","Text":"Now, if I interpret this formula for the case where n is 1/2,"},{"Start":"03:39.755 ","End":"03:43.310","Text":"what I get is that the integral of the square root of ax,"},{"Start":"03:43.310 ","End":"03:49.370","Text":"plus b dx is going to equal 1 over a,"},{"Start":"03:49.370 ","End":"03:57.980","Text":"ax plus b to the power of 1/2 plus 1 is 3 over 2 divided by 3 over 2,"},{"Start":"03:57.980 ","End":"04:00.125","Text":"so in this case,"},{"Start":"04:00.125 ","End":"04:07.980","Text":"we get 5 minus x to the power of 3 over 2 over 3 over 2,"},{"Start":"04:07.980 ","End":"04:11.270","Text":"but the a here is minus 1 is the coefficient of x."},{"Start":"04:11.270 ","End":"04:13.730","Text":"So I have to put a minus here,"},{"Start":"04:13.730 ","End":"04:18.490","Text":"and this will be taken between minus 1 and 1,"},{"Start":"04:18.490 ","End":"04:25.310","Text":"and the next bit will be x plus 3 to the power of 3 over 2,"},{"Start":"04:25.310 ","End":"04:27.995","Text":"also over 3 over 2,"},{"Start":"04:27.995 ","End":"04:32.305","Text":"this time between 1 and 4."},{"Start":"04:32.305 ","End":"04:36.230","Text":"This divided by 3 over 2 is a bit of a nuisance."},{"Start":"04:36.230 ","End":"04:40.645","Text":"I\u0027m going to take that divided by 3 over 2 is like multiplying by 2/3,"},{"Start":"04:40.645 ","End":"04:45.135","Text":"so I\u0027m just going to write this as 2/3 of"},{"Start":"04:45.135 ","End":"04:52.730","Text":"minus 5 minus x to the 3 over 2 from minus 1 to 1,"},{"Start":"04:52.730 ","End":"04:57.725","Text":"plus x plus 3 to the power of 3 over 2,"},{"Start":"04:57.725 ","End":"05:01.550","Text":"from 1 to 4, so let\u0027s see what we get, 2/3."},{"Start":"05:01.550 ","End":"05:05.265","Text":"Now, here we have to plug in 1 and minus 1 and subtract."},{"Start":"05:05.265 ","End":"05:07.395","Text":"When x is 1,"},{"Start":"05:07.395 ","End":"05:10.994","Text":"we get 5 minus 1 is 4,"},{"Start":"05:10.994 ","End":"05:14.290","Text":"4 to the power of 3 over 2 is 8,"},{"Start":"05:14.290 ","End":"05:17.045","Text":"and this would give me minus 8."},{"Start":"05:17.045 ","End":"05:19.550","Text":"When x is minus 1,"},{"Start":"05:19.550 ","End":"05:23.240","Text":"I\u0027ve got 5 minus minus 1, which is 6,"},{"Start":"05:23.240 ","End":"05:26.275","Text":"so it\u0027s 6 to the power of 3 over 2,"},{"Start":"05:26.275 ","End":"05:28.650","Text":"minus 6 to the power of 3 over 2,"},{"Start":"05:28.650 ","End":"05:30.690","Text":"which makes it plus, because I\u0027m subtracting,"},{"Start":"05:30.690 ","End":"05:35.010","Text":"it\u0027s minus minus 6 to the power of 3 over 2."},{"Start":"05:35.010 ","End":"05:40.265","Text":"Then the next bit is where I put x equals 4,"},{"Start":"05:40.265 ","End":"05:46.370","Text":"so I\u0027ve got 4 plus 3 is 7 to the power of 3 over 2,"},{"Start":"05:46.370 ","End":"05:48.620","Text":"and when x is 1,"},{"Start":"05:48.620 ","End":"05:52.235","Text":"I get 4 to the power of 3 over 2,"},{"Start":"05:52.235 ","End":"05:57.399","Text":"which is 8, but that\u0027s in a minus sign."},{"Start":"05:57.399 ","End":"06:00.060","Text":"We could leave it like this."},{"Start":"06:00.060 ","End":"06:04.410","Text":"I just like to make 1 simplification or maybe 2,"},{"Start":"06:04.410 ","End":"06:10.845","Text":"and what I want to say is that the minus 8 with the minus 8 become minus 16,"},{"Start":"06:10.845 ","End":"06:13.380","Text":"and if you\u0027re going to do it on the calculator,"},{"Start":"06:13.380 ","End":"06:19.460","Text":"it maybe easier to write it as decimal 6 to the 1.5,"},{"Start":"06:19.460 ","End":"06:25.205","Text":"and here 7 to the 1.5, and still with the 2/3 in front,"},{"Start":"06:25.205 ","End":"06:28.050","Text":"and that\u0027s the answer."}],"ID":4524},{"Watched":false,"Name":"Exercise 8","Duration":"7m 2s","ChapterTopicVideoID":4516,"CourseChapterTopicPlaylistID":84700,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/4516.jpeg","UploadDate":"2017-03-13T05:23:45.5130000","DurationForVideoObject":"PT7M2S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.430","Text":"In this exercise, you have to compute the"},{"Start":"00:02.430 ","End":"00:05.160","Text":"following definite integral which I have copied over here."},{"Start":"00:05.160 ","End":"00:07.830","Text":"I want to warn you this exercise is going to be a bit"},{"Start":"00:07.830 ","End":"00:11.685","Text":"tedious and also we\u0027re going to use a trick that is not intuitive."},{"Start":"00:11.685 ","End":"00:13.019","Text":"Let\u0027s get started."},{"Start":"00:13.019 ","End":"00:19.740","Text":"The trick is to let x equals pi minus t. This trip won\u0027t work in general,"},{"Start":"00:19.740 ","End":"00:21.540","Text":"specifically with 0 and pi,"},{"Start":"00:21.540 ","End":"00:23.310","Text":"it\u0027s going to work and maybe with other numbers,"},{"Start":"00:23.310 ","End":"00:24.765","Text":"but in general not."},{"Start":"00:24.765 ","End":"00:31.485","Text":"Dx is equal to just minus 1 dt. It\u0027s minus dt."},{"Start":"00:31.485 ","End":"00:41.480","Text":"After the substitution, what we get is the integral x is pi minus t. Sine x is"},{"Start":"00:41.480 ","End":"00:45.005","Text":"sine of pi minus t"},{"Start":"00:45.005 ","End":"00:51.890","Text":"over 1 plus cosine x is cosine of pi minus t,"},{"Start":"00:51.890 ","End":"00:55.715","Text":"but it\u0027s squared, and dx is minus dt."},{"Start":"00:55.715 ","End":"01:00.065","Text":"I can put the minus here and the dt here."},{"Start":"01:00.065 ","End":"01:02.915","Text":"Finally, we have to also switch the limits."},{"Start":"01:02.915 ","End":"01:06.035","Text":"You see when x equals 0,"},{"Start":"01:06.035 ","End":"01:11.060","Text":"then t is equal to pi minus x."},{"Start":"01:11.060 ","End":"01:15.335","Text":"When x is 0, t is pi and when x is pi,"},{"Start":"01:15.335 ","End":"01:17.285","Text":"then t is 0."},{"Start":"01:17.285 ","End":"01:24.160","Text":"What we end up getting if we continue down here is minus the integral."},{"Start":"01:24.160 ","End":"01:27.739","Text":"Now, this was from pi to 0."},{"Start":"01:27.739 ","End":"01:32.260","Text":"Now, we don\u0027t usually like the upper limit to be less than the lower limit,"},{"Start":"01:32.260 ","End":"01:36.770","Text":"but there is a well-known rule that if you switch the top and the bottom,"},{"Start":"01:36.770 ","End":"01:38.875","Text":"then you can get rid of the minus."},{"Start":"01:38.875 ","End":"01:42.740","Text":"In fact, I\u0027m just going to erase this minus and write"},{"Start":"01:42.740 ","End":"01:47.400","Text":"it as the integral from 0 to pi of the same thing."},{"Start":"01:47.400 ","End":"01:51.695","Text":"Now I\u0027d like to remind you of some trigonometrical identities."},{"Start":"01:51.695 ","End":"01:57.880","Text":"1 of them is that the sine of pi minus t,"},{"Start":"01:57.880 ","End":"02:00.190","Text":"pi is 180 degrees, remember,"},{"Start":"02:00.190 ","End":"02:02.840","Text":"is the same as the sine of the angle."},{"Start":"02:02.840 ","End":"02:05.970","Text":"But with the cosine, it\u0027s a minus."},{"Start":"02:05.970 ","End":"02:13.880","Text":"Cosine of pi minus t is minus cosine t. We have the integral from 0 to pi of"},{"Start":"02:13.880 ","End":"02:22.145","Text":"pi minus t and sine of pi minus t is just sine t. Now cosine of pi t is minus cosine t,"},{"Start":"02:22.145 ","End":"02:28.235","Text":"but it\u0027s squared, so it\u0027s still cosine squared t because the minus squared is a plus."},{"Start":"02:28.235 ","End":"02:31.115","Text":"I\u0027m going to break it up into 2 bits."},{"Start":"02:31.115 ","End":"02:36.830","Text":"I\u0027m going to break up according to the pi minus t. There\u0027s a minus here,"},{"Start":"02:36.830 ","End":"02:40.760","Text":"and this minus enables me to break it up into"},{"Start":"02:40.760 ","End":"02:44.915","Text":"2 integrals pi with this thing and t with this thing."},{"Start":"02:44.915 ","End":"02:49.565","Text":"I\u0027m going to write it now as the integral from 0"},{"Start":"02:49.565 ","End":"02:54.350","Text":"to pi of pi times sine t. Now even write the pi"},{"Start":"02:54.350 ","End":"03:03.620","Text":"outside of sine t over 1 plus cosine squared t dt minus the"},{"Start":"03:03.620 ","End":"03:08.420","Text":"integral from 0 to pi t sine t"},{"Start":"03:08.420 ","End":"03:15.050","Text":"over 1 plus cosine squared of t. Now let\u0027s see where this gets us."},{"Start":"03:15.050 ","End":"03:17.530","Text":"Now here\u0027s the other trick that we\u0027re going to use."},{"Start":"03:17.530 ","End":"03:20.570","Text":"The original integral from 0 to pi."},{"Start":"03:20.570 ","End":"03:23.120","Text":"Let\u0027s give it a letter I for integrals,"},{"Start":"03:23.120 ","End":"03:32.095","Text":"so I\u0027ll call it letter I. I claim that this integral here is also equal to i."},{"Start":"03:32.095 ","End":"03:33.695","Text":"This is exactly the same as this."},{"Start":"03:33.695 ","End":"03:36.215","Text":"The t or the x is a dummy variable."},{"Start":"03:36.215 ","End":"03:39.290","Text":"I get 2i is equal to this thing,"},{"Start":"03:39.290 ","End":"03:41.690","Text":"so i is 1.5 of this thing."},{"Start":"03:41.690 ","End":"03:48.080","Text":"Basically what I get is i equals pi over 2 times the"},{"Start":"03:48.080 ","End":"03:55.300","Text":"integral of sine t over 1 plus cosine squared t dt."},{"Start":"03:55.300 ","End":"03:57.335","Text":"I\u0027ll explain that again."},{"Start":"03:57.335 ","End":"03:58.520","Text":"If I have that,"},{"Start":"03:58.520 ","End":"04:04.559","Text":"let\u0027s say i is equal to something minus i,"},{"Start":"04:04.559 ","End":"04:12.015","Text":"then 2i is equal to that something and i is equal to 1.5 of that something."},{"Start":"04:12.015 ","End":"04:14.550","Text":"That\u0027s just what I did here basically."},{"Start":"04:14.550 ","End":"04:18.420","Text":"I forgot the limits from 0 to pi."},{"Start":"04:18.420 ","End":"04:22.710","Text":"What remains is to compute this integral."},{"Start":"04:22.710 ","End":"04:26.445","Text":"This is equal to pi over 2."},{"Start":"04:26.445 ","End":"04:29.720","Text":"This yellow thing is actually equal to pi over 2,"},{"Start":"04:29.720 ","End":"04:31.370","Text":"and that\u0027s what we\u0027re going to leave to the end."},{"Start":"04:31.370 ","End":"04:33.470","Text":"I\u0027ll also highlight that."},{"Start":"04:33.470 ","End":"04:39.040","Text":"The final answer is pi squared over 4,"},{"Start":"04:39.040 ","End":"04:44.515","Text":"but I have the debt of showing you that this thing is equal to this,"},{"Start":"04:44.515 ","End":"04:46.660","Text":"and that\u0027s what I\u0027ll do now."},{"Start":"04:46.660 ","End":"04:52.325","Text":"Now I\u0027m going to show you that the integral of sine t"},{"Start":"04:52.325 ","End":"04:59.820","Text":"over 1 plus cosine squared t dt from 0 to pi is equal to this."},{"Start":"04:59.820 ","End":"05:01.685","Text":"I\u0027m going to do it by substitution."},{"Start":"05:01.685 ","End":"05:08.585","Text":"I\u0027m going to say that let z equal cosine t,"},{"Start":"05:08.585 ","End":"05:15.145","Text":"and then dz is equal to minus sine t dt."},{"Start":"05:15.145 ","End":"05:20.725","Text":"What I want to do is put a minus here and a minus here,"},{"Start":"05:20.725 ","End":"05:24.200","Text":"and then I will get minus sine t dt."},{"Start":"05:24.200 ","End":"05:32.385","Text":"I\u0027ll get the integral of 1 over 1 plus z squared dz."},{"Start":"05:32.385 ","End":"05:37.900","Text":"We have to substitute the limit when t is 0,"},{"Start":"05:37.900 ","End":"05:43.395","Text":"that means that z is cosine of 0 which is 1,"},{"Start":"05:43.395 ","End":"05:51.295","Text":"and when t is pi then z is cosine pi which is minus 1."},{"Start":"05:51.295 ","End":"06:01.115","Text":"This is equal to minus the integral from minus 1 to 1 of 1 over 1 plus z squared dz."},{"Start":"06:01.115 ","End":"06:05.775","Text":"This is an immediate integral and it\u0027s the arctangent."},{"Start":"06:05.775 ","End":"06:14.490","Text":"We get minus the arctangent of z from minus 1 to 1."},{"Start":"06:14.490 ","End":"06:17.240","Text":"I made a small mistake."},{"Start":"06:17.240 ","End":"06:20.240","Text":"I forgot to carry the minus over."},{"Start":"06:20.240 ","End":"06:22.310","Text":"If this is a minus,"},{"Start":"06:22.310 ","End":"06:24.505","Text":"then this becomes a plus."},{"Start":"06:24.505 ","End":"06:29.675","Text":"What we get is the arctangent of"},{"Start":"06:29.675 ","End":"06:35.510","Text":"1 minus the arctangent of minus 1."},{"Start":"06:35.510 ","End":"06:37.490","Text":"Now I know that the angle whose tangent is 1 is"},{"Start":"06:37.490 ","End":"06:42.365","Text":"45 degrees and here it\u0027s minus 45 degrees, but it\u0027s pi over 4."},{"Start":"06:42.365 ","End":"06:45.170","Text":"Basically, this is pi over 4 minus,"},{"Start":"06:45.170 ","End":"06:47.225","Text":"minus pi over 4,"},{"Start":"06:47.225 ","End":"06:50.445","Text":"which equals pi over 2."},{"Start":"06:50.445 ","End":"06:54.155","Text":"This is what we have to show that this yellow equals this yellow,"},{"Start":"06:54.155 ","End":"06:57.260","Text":"and indeed this is equal to pi over 2."},{"Start":"06:57.260 ","End":"07:03.510","Text":"That\u0027s the debt we owed and we are done that this is the answer."}],"ID":4525},{"Watched":false,"Name":"Exercise 9","Duration":"5m 2s","ChapterTopicVideoID":4517,"CourseChapterTopicPlaylistID":84700,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/4517.jpeg","UploadDate":"2017-01-26T13:21:39.4330000","DurationForVideoObject":"PT5M2S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.940","Text":"In this exercise, we have to compute a definite"},{"Start":"00:02.940 ","End":"00:07.620","Text":"integral from 0 to pi over 2 of this stuff dx."},{"Start":"00:07.620 ","End":"00:09.510","Text":"I\u0027ve copied it over here."},{"Start":"00:09.510 ","End":"00:13.035","Text":"We\u0027re actually going to use a trick here."},{"Start":"00:13.035 ","End":"00:17.440","Text":"We\u0027re going to rely on the fact that specifically it\u0027s from 0 to pi over 2."},{"Start":"00:17.440 ","End":"00:20.345","Text":"Let me just call this integral I,"},{"Start":"00:20.345 ","End":"00:21.950","Text":"because I\u0027ll need that later."},{"Start":"00:21.950 ","End":"00:23.510","Text":"Now, here\u0027s the trick,"},{"Start":"00:23.510 ","End":"00:26.000","Text":"not exactly a trick, but something that\u0027s not intuitive,"},{"Start":"00:26.000 ","End":"00:34.664","Text":"is to make a substitution where x is equal to pi over 2 minus t. Therefore,"},{"Start":"00:34.664 ","End":"00:40.420","Text":"dx will equal minus 1 dt or just minus dt."},{"Start":"00:40.420 ","End":"00:42.050","Text":"But when we substitute,"},{"Start":"00:42.050 ","End":"00:44.620","Text":"we also have to substitute the limits."},{"Start":"00:44.620 ","End":"00:47.475","Text":"When x is 0,"},{"Start":"00:47.475 ","End":"00:53.535","Text":"then t is pi over 2 minus 0 is pi over 2."},{"Start":"00:53.535 ","End":"00:56.429","Text":"If x is pi over 2,"},{"Start":"00:56.429 ","End":"00:59.580","Text":"then t is equal to 0."},{"Start":"00:59.580 ","End":"01:02.145","Text":"We get the integral,"},{"Start":"01:02.145 ","End":"01:12.690","Text":"but from pi over 2 to 0 of the 4th root of sine of pi over 2 minus t,"},{"Start":"01:12.690 ","End":"01:19.439","Text":"over the 4th root of sine pi over 2 minus t,"},{"Start":"01:19.439 ","End":"01:27.285","Text":"plus the 4th root of cosine of pi over 2 minus t,"},{"Start":"01:27.285 ","End":"01:31.440","Text":"and all this dx is minus dt."},{"Start":"01:31.440 ","End":"01:35.960","Text":"I\u0027m going to do several things now to simplify this."},{"Start":"01:35.960 ","End":"01:38.540","Text":"The first thing I\u0027m going to do is that if we have the"},{"Start":"01:38.540 ","End":"01:41.990","Text":"integral of some function from a to b,"},{"Start":"01:41.990 ","End":"01:47.420","Text":"then this is the same thing as minus the integral from b to a of"},{"Start":"01:47.420 ","End":"01:53.030","Text":"that same f. I\u0027m going to basically switch these 2 and get rid of this minus."},{"Start":"01:53.030 ","End":"01:58.010","Text":"The second thing is that pi over 2 minus an angle is a complimentary angle."},{"Start":"01:58.010 ","End":"02:02.930","Text":"We know that the sine of an angle is equal to the cosine of the complimentary angle."},{"Start":"02:02.930 ","End":"02:09.575","Text":"Basically, what I need to know is that sine of pi over 2 minus t is cosine t."},{"Start":"02:09.575 ","End":"02:16.915","Text":"Cosine of pi over 2 minus t is sine t. Eventually I get,"},{"Start":"02:16.915 ","End":"02:21.015","Text":"I\u0027ve switched the order 0 to pi over 2."},{"Start":"02:21.015 ","End":"02:26.990","Text":"Here I have the 4th root of cosine t from what I wrote here,"},{"Start":"02:26.990 ","End":"02:32.220","Text":"and here the 4th root of sine t and plus"},{"Start":"02:32.220 ","End":"02:40.594","Text":"the 4th root of cosine t. What I\u0027d like to do is replace t by x."},{"Start":"02:40.594 ","End":"02:42.990","Text":"The actual letter is not important,"},{"Start":"02:42.990 ","End":"02:45.065","Text":"it\u0027s still going to be the same thing."},{"Start":"02:45.065 ","End":"02:50.015","Text":"This is equal to the integral from 0 to pi over 2,"},{"Start":"02:50.015 ","End":"02:52.430","Text":"just a letter replacement or if you want to think of"},{"Start":"02:52.430 ","End":"02:54.920","Text":"it as a substitution, that\u0027s okay too,"},{"Start":"02:54.920 ","End":"03:00.930","Text":"of the 4th root of cosine x over the 4th root of"},{"Start":"03:00.930 ","End":"03:07.810","Text":"sine x plus the 4th root of cosine x dx."},{"Start":"03:07.810 ","End":"03:12.455","Text":"Now this has got to also equal I because I started from I,"},{"Start":"03:12.455 ","End":"03:14.890","Text":"so this is also equal to I."},{"Start":"03:14.890 ","End":"03:17.555","Text":"But now you\u0027ll see where the trick comes in."},{"Start":"03:17.555 ","End":"03:20.240","Text":"If I say what is I plus I,"},{"Start":"03:20.240 ","End":"03:23.915","Text":"I can get that 2I is equal to,"},{"Start":"03:23.915 ","End":"03:27.730","Text":"so I plus I I can take this plus this."},{"Start":"03:27.730 ","End":"03:29.600","Text":"I don\u0027t have to take twice this or twice this,"},{"Start":"03:29.600 ","End":"03:32.705","Text":"I can take 1 of these and 1 of these each of them is equal to I."},{"Start":"03:32.705 ","End":"03:36.315","Text":"It\u0027s the integral from 0 to pi over 2."},{"Start":"03:36.315 ","End":"03:37.880","Text":"You know what? I\u0027ll use copy paste."},{"Start":"03:37.880 ","End":"03:40.040","Text":"The first one I copied here."},{"Start":"03:40.040 ","End":"03:42.230","Text":"This form I copied here,"},{"Start":"03:42.230 ","End":"03:44.065","Text":"and that\u0027s equal to 2I."},{"Start":"03:44.065 ","End":"03:48.090","Text":"Now look, these 2 integrals have the same upper and lower limit,"},{"Start":"03:48.090 ","End":"03:50.285","Text":"so I can add the 2 functions."},{"Start":"03:50.285 ","End":"03:51.695","Text":"I can add this to this."},{"Start":"03:51.695 ","End":"03:53.740","Text":"Now they have the same denominator."},{"Start":"03:53.740 ","End":"04:00.980","Text":"What I\u0027m going to get is the integral from 0 to pi over 2 of the sum of these 2 things."},{"Start":"04:00.980 ","End":"04:09.860","Text":"Now the denominator is just the 4th root of sine x plus the fourth root of cosine x."},{"Start":"04:09.860 ","End":"04:12.320","Text":"Then I add the numerators from here,"},{"Start":"04:12.320 ","End":"04:15.455","Text":"the 4th root of sine x,"},{"Start":"04:15.455 ","End":"04:20.825","Text":"from here, the 4th root of cosine x and dx."},{"Start":"04:20.825 ","End":"04:24.935","Text":"But look, this fraction has the same numerator as the denominator."},{"Start":"04:24.935 ","End":"04:26.975","Text":"That\u0027s got to be equal to 1."},{"Start":"04:26.975 ","End":"04:32.565","Text":"It\u0027s equal to the integral from 0 to pi over 2 of 1 dx."},{"Start":"04:32.565 ","End":"04:38.850","Text":"The integral of 1 is just x taken from 0 to pi over 2."},{"Start":"04:38.850 ","End":"04:40.500","Text":"If I put pi over 2,"},{"Start":"04:40.500 ","End":"04:41.910","Text":"it\u0027s pi over 2."},{"Start":"04:41.910 ","End":"04:44.160","Text":"If x is 0, then x is 0."},{"Start":"04:44.160 ","End":"04:47.690","Text":"All together, this is equal to pi over 2, but this is 2I."},{"Start":"04:47.690 ","End":"04:49.010","Text":"I\u0027ll just copy it over here."},{"Start":"04:49.010 ","End":"04:50.950","Text":"This is 2I."},{"Start":"04:50.950 ","End":"04:54.889","Text":"Finally, I which is the integral itself,"},{"Start":"04:54.889 ","End":"05:03.480","Text":"is equal to pi over 4 because I divide this by 2 and this will be my answer. We are done."}],"ID":4526},{"Watched":false,"Name":"Exercise 10 part a","Duration":"3m 40s","ChapterTopicVideoID":4518,"CourseChapterTopicPlaylistID":84700,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/4518.jpeg","UploadDate":"2017-01-26T13:22:49.6170000","DurationForVideoObject":"PT3M40S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.795","Text":"In this exercise, f is a continuous function and we have to prove 2 things."},{"Start":"00:06.795 ","End":"00:09.119","Text":"A, if f is even,"},{"Start":"00:09.119 ","End":"00:15.750","Text":"then the integral from minus a to a of f is twice the integral from 0 to a."},{"Start":"00:15.750 ","End":"00:17.430","Text":"Let\u0027s leave b for later,"},{"Start":"00:17.430 ","End":"00:22.360","Text":"and let\u0027s start off with a. I\u0027d like to remind you what even means."},{"Start":"00:22.360 ","End":"00:31.144","Text":"Means in general that f of minus x is the same as f of x for all x."},{"Start":"00:31.144 ","End":"00:33.079","Text":"Okay. If that\u0027s the case,"},{"Start":"00:33.079 ","End":"00:38.680","Text":"then what we have to do is split this up into 2 integrals."},{"Start":"00:38.680 ","End":"00:46.940","Text":"The integral from minus a to a of f of x dx x equals 0 is a mirror image, it\u0027s an axis."},{"Start":"00:46.940 ","End":"00:49.820","Text":"What we\u0027re going to do is take the interval from minus a to"},{"Start":"00:49.820 ","End":"00:53.755","Text":"a and break it up into 2 intervals from minus a to 0,"},{"Start":"00:53.755 ","End":"00:57.045","Text":"and then from 0 to a."},{"Start":"00:57.045 ","End":"00:59.950","Text":"Now here we have f of x dx,"},{"Start":"00:59.950 ","End":"01:04.720","Text":"and here we have also f of x dx."},{"Start":"01:04.720 ","End":"01:09.890","Text":"Now, what I\u0027m going to do is use the fact that in the negative interval,"},{"Start":"01:09.890 ","End":"01:11.930","Text":"the even property holds."},{"Start":"01:11.930 ","End":"01:16.280","Text":"This is equal to the integral from minus a to"},{"Start":"01:16.280 ","End":"01:22.980","Text":"0 of f of minus x dx plus same thing,"},{"Start":"01:22.980 ","End":"01:26.995","Text":"integral 0 to a f of x dx."},{"Start":"01:26.995 ","End":"01:29.870","Text":"Now here I\u0027d like to make a substitution."},{"Start":"01:29.870 ","End":"01:38.780","Text":"I\u0027d like to say that t is going to be minus x. T is minus x,"},{"Start":"01:38.780 ","End":"01:43.640","Text":"and so dt is minus dx,"},{"Start":"01:43.640 ","End":"01:53.795","Text":"and the limits 0 for x goes to 0 for t. If I put x equals minus a,"},{"Start":"01:53.795 ","End":"01:56.090","Text":"then t becomes a."},{"Start":"01:56.090 ","End":"01:59.660","Text":"In other words, these are the x values and these are the t values."},{"Start":"01:59.660 ","End":"02:07.815","Text":"We have to substitute everything and what I get is this equals the integral."},{"Start":"02:07.815 ","End":"02:14.855","Text":"Now, minus a to 0 becomes a to 0 minus x is t,"},{"Start":"02:14.855 ","End":"02:16.940","Text":"so that\u0027s f of t,"},{"Start":"02:16.940 ","End":"02:20.165","Text":"and if dt is minus dx,"},{"Start":"02:20.165 ","End":"02:27.335","Text":"I could put the dx on the other side and say this is minus dt plus other side the same."},{"Start":"02:27.335 ","End":"02:30.920","Text":"Now we have our usual trick that if we"},{"Start":"02:30.920 ","End":"02:35.360","Text":"reverse the top and bottom limits and make it from 0 to a,"},{"Start":"02:35.360 ","End":"02:37.390","Text":"we have to throw it a minus somewhere,"},{"Start":"02:37.390 ","End":"02:41.255","Text":"so I can just get rid of this minus and say this is f of t,"},{"Start":"02:41.255 ","End":"02:44.590","Text":"dt plus the same."},{"Start":"02:44.590 ","End":"02:46.785","Text":"Now here\u0027s the thing,"},{"Start":"02:46.785 ","End":"02:52.055","Text":"there\u0027s no significance to the particular letter t. If I put here f of u du,"},{"Start":"02:52.055 ","End":"02:54.110","Text":"or f of z dz,"},{"Start":"02:54.110 ","End":"02:55.430","Text":"or f of x dx,"},{"Start":"02:55.430 ","End":"02:56.810","Text":"even it would be the same thing,"},{"Start":"02:56.810 ","End":"02:58.120","Text":"the letter doesn\u0027t matter."},{"Start":"02:58.120 ","End":"02:59.990","Text":"I\u0027m going to go back to x."},{"Start":"02:59.990 ","End":"03:05.750","Text":"This is equal to the integral from 0 to a of f of x dx plus,"},{"Start":"03:05.750 ","End":"03:11.500","Text":"and now I will copy it out in full 0 to a f of x dx,"},{"Start":"03:11.500 ","End":"03:14.150","Text":"but look, this expression is the same as this expression,"},{"Start":"03:14.150 ","End":"03:15.665","Text":"so we have 2 of these."},{"Start":"03:15.665 ","End":"03:22.525","Text":"This is twice the integral from 0 to a of f of x dx."},{"Start":"03:22.525 ","End":"03:25.205","Text":"Let\u0027s see what we were asked to prove."},{"Start":"03:25.205 ","End":"03:30.365","Text":"You were asked to prove that this integral is twice the integral from 0 to a,"},{"Start":"03:30.365 ","End":"03:32.420","Text":"and I think this is what we have here,"},{"Start":"03:32.420 ","End":"03:37.760","Text":"so this is it what we had to prove for part a."},{"Start":"03:37.760 ","End":"03:41.010","Text":"Now let\u0027s go to part b."}],"ID":4527},{"Watched":false,"Name":"Exercise 10 part b","Duration":"2m 50s","ChapterTopicVideoID":4519,"CourseChapterTopicPlaylistID":84700,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/4519.jpeg","UploadDate":"2017-03-13T05:32:01.1670000","DurationForVideoObject":"PT2M50S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"We just finished Part A,"},{"Start":"00:01.890 ","End":"00:03.480","Text":"now onto Part B."},{"Start":"00:03.480 ","End":"00:06.600","Text":"Previously, it was where f was even and now,"},{"Start":"00:06.600 ","End":"00:08.250","Text":"we have the case where f is odd,"},{"Start":"00:08.250 ","End":"00:11.685","Text":"and I\u0027ll remind you that odd means that f of minus x,"},{"Start":"00:11.685 ","End":"00:14.430","Text":"in general, is equal to minus f of x."},{"Start":"00:14.430 ","End":"00:15.830","Text":"What I\u0027m going to do is just,"},{"Start":"00:15.830 ","End":"00:17.535","Text":"first of all, copy this,"},{"Start":"00:17.535 ","End":"00:21.020","Text":"minus a to a of f of x dx,"},{"Start":"00:21.020 ","End":"00:24.555","Text":"and just like before, we split it up into 2 integrals."},{"Start":"00:24.555 ","End":"00:31.515","Text":"We first go from minus a to 0 and then from 0 to a."},{"Start":"00:31.515 ","End":"00:35.760","Text":"This here, I\u0027m going to do by a substitution."},{"Start":"00:35.760 ","End":"00:37.909","Text":"The first thing before the substitution,"},{"Start":"00:37.909 ","End":"00:42.445","Text":"I\u0027m going to say that this is the integral from minus a to 0,"},{"Start":"00:42.445 ","End":"00:49.580","Text":"and f of x is equal to minus f of minus x. I could put this minus on"},{"Start":"00:49.580 ","End":"00:57.855","Text":"the other side and say that this is minus f of minus x dx plus the same bit."},{"Start":"00:57.855 ","End":"01:00.450","Text":"You know what? I\u0027ll just put it like ditto sign."},{"Start":"01:00.450 ","End":"01:05.855","Text":"At this point, I\u0027m going to make a substitution for t to be equal to minus x,"},{"Start":"01:05.855 ","End":"01:09.530","Text":"so if t is equal to minus x,"},{"Start":"01:09.530 ","End":"01:13.140","Text":"then dt is minus 1dx."},{"Start":"01:13.140 ","End":"01:18.720","Text":"The limits. When x is equal to minus a,"},{"Start":"01:18.720 ","End":"01:22.560","Text":"then the t, which is minus x, is plus a,"},{"Start":"01:22.560 ","End":"01:26.960","Text":"and when x is 0 and t is minus 0,"},{"Start":"01:26.960 ","End":"01:34.915","Text":"which is also 0, so what I get is the integral from minus a to 0."},{"Start":"01:34.915 ","End":"01:41.970","Text":"The minus with the dx could be dt and minus x is t,"},{"Start":"01:41.970 ","End":"01:45.670","Text":"so I get f of t dt."},{"Start":"01:45.830 ","End":"01:50.580","Text":"I\u0027m going to switch upper and lower limits,"},{"Start":"01:50.580 ","End":"01:55.205","Text":"and we can do that provided that we introduce a minus sign."},{"Start":"01:55.205 ","End":"01:57.095","Text":"Why did I write minus a?"},{"Start":"01:57.095 ","End":"01:59.210","Text":"This is a. I fixed that,"},{"Start":"01:59.210 ","End":"02:00.815","Text":"and I completed the line."},{"Start":"02:00.815 ","End":"02:05.995","Text":"What I did is switch the 0 and the a and introduce an extra minus sign,"},{"Start":"02:05.995 ","End":"02:07.755","Text":"and this thing, all along,"},{"Start":"02:07.755 ","End":"02:12.475","Text":"have been dragging this thing and this thing from here."},{"Start":"02:12.475 ","End":"02:15.320","Text":"I\u0027m going to write it out in full again,"},{"Start":"02:15.320 ","End":"02:20.705","Text":"integral from 0 to a of f of x dx,"},{"Start":"02:20.705 ","End":"02:22.890","Text":"and this part, just like in Part A,"},{"Start":"02:22.890 ","End":"02:28.295","Text":"there\u0027s no special meaning to a letter t. I could make it back to x again."},{"Start":"02:28.295 ","End":"02:30.920","Text":"The letter itself has no particular significance,"},{"Start":"02:30.920 ","End":"02:35.030","Text":"so I get minus the integral of 0 to a of f"},{"Start":"02:35.030 ","End":"02:39.385","Text":"of x dx by just simply switching the name of the letter."},{"Start":"02:39.385 ","End":"02:43.775","Text":"Examine this. Here, I have minus something plus the same something."},{"Start":"02:43.775 ","End":"02:46.130","Text":"That is equal to 0,"},{"Start":"02:46.130 ","End":"02:48.725","Text":"and that\u0027s what we have to prove in Part B,"},{"Start":"02:48.725 ","End":"02:50.970","Text":"and now we\u0027re done."}],"ID":4528},{"Watched":false,"Name":"Exercise 11","Duration":"3m 27s","ChapterTopicVideoID":4520,"CourseChapterTopicPlaylistID":84700,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/4520.jpeg","UploadDate":"2017-03-13T05:48:56.4170000","DurationForVideoObject":"PT3M27S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.060","Text":"In this exercise, which is really 2 exercises,"},{"Start":"00:03.060 ","End":"00:04.904","Text":"we have to compute the integrals."},{"Start":"00:04.904 ","End":"00:08.910","Text":"Now the previous exercise was all about odd and even functions."},{"Start":"00:08.910 ","End":"00:11.430","Text":"We\u0027re going to use the results of that here."},{"Start":"00:11.430 ","End":"00:17.640","Text":"In part 1, we have the integral from minus 4^4 of"},{"Start":"00:17.640 ","End":"00:24.135","Text":"cosine x over x cubed plus x^5 dx."},{"Start":"00:24.135 ","End":"00:27.145","Text":"Now let\u0027s look at this function, the integrant."},{"Start":"00:27.145 ","End":"00:31.355","Text":"I\u0027m claiming that it\u0027s an odd function because the numerator,"},{"Start":"00:31.355 ","End":"00:34.384","Text":"this cosine x is even."},{"Start":"00:34.384 ","End":"00:38.780","Text":"X cubed is odd and x^5 is odd."},{"Start":"00:38.780 ","End":"00:40.250","Text":"That\u0027s easy to see."},{"Start":"00:40.250 ","End":"00:41.990","Text":"If I replace x by minus x,"},{"Start":"00:41.990 ","End":"00:44.865","Text":"I get minus^5 or minus^3,"},{"Start":"00:44.865 ","End":"00:46.610","Text":"because these are odd numbers."},{"Start":"00:46.610 ","End":"00:51.710","Text":"This is odd, this is odd and the sum of odd is odd and even over odd is odd."},{"Start":"00:51.710 ","End":"00:56.135","Text":"Altogether, I get that this function is odd."},{"Start":"00:56.135 ","End":"01:01.220","Text":"Now we have the result that if f is odd,"},{"Start":"01:01.220 ","End":"01:09.515","Text":"then the integral from minus a^a of f of x dx is equal to 0."},{"Start":"01:09.515 ","End":"01:15.430","Text":"This is what we exactly have here because this whole thing is an odd function."},{"Start":"01:15.430 ","End":"01:18.655","Text":"Using this theorem, where a is 4,"},{"Start":"01:18.655 ","End":"01:25.595","Text":"we just get that the answer is that this thing is equal to 0, and we\u0027re done."},{"Start":"01:25.595 ","End":"01:28.095","Text":"Now, let\u0027s see part 2."},{"Start":"01:28.095 ","End":"01:38.615","Text":"We have the integral from minus 1^1 of sine x plus 1 over x squared plus 1 dx."},{"Start":"01:38.615 ","End":"01:41.015","Text":"Now, here it\u0027s a little bit trickier."},{"Start":"01:41.015 ","End":"01:45.395","Text":"The denominator is certainly an even function."},{"Start":"01:45.395 ","End":"01:51.760","Text":"The thing is about the numerator is that this is odd and this is even."},{"Start":"01:51.760 ","End":"01:56.315","Text":"I don\u0027t know what an odd plus an even is but if I break it up into 2,"},{"Start":"01:56.315 ","End":"01:57.710","Text":"that will help me more."},{"Start":"01:57.710 ","End":"02:01.685","Text":"Let\u0027s break it up into 2 separate integrals."},{"Start":"02:01.685 ","End":"02:08.480","Text":"Then we have minus 1^1 of sine x over x squared plus"},{"Start":"02:08.480 ","End":"02:12.290","Text":"1 dx plus the integral from"},{"Start":"02:12.290 ","End":"02:19.550","Text":"minus 1^1 of 1 over x squared plus 1 dx."},{"Start":"02:19.550 ","End":"02:26.835","Text":"In this case, I have that this is still even and this is still odd."},{"Start":"02:26.835 ","End":"02:32.985","Text":"That makes this whole thing odd a and minus a is 1, and minus 1."},{"Start":"02:32.985 ","End":"02:36.860","Text":"Then we get this using the same theorem that this is equal to 0."},{"Start":"02:36.860 ","End":"02:39.154","Text":"I still have this part."},{"Start":"02:39.154 ","End":"02:40.680","Text":"Now this is an even function,"},{"Start":"02:40.680 ","End":"02:43.910","Text":"and there is a similar rule for even functions,"},{"Start":"02:43.910 ","End":"02:45.590","Text":"but it\u0027s not really going to help me."},{"Start":"02:45.590 ","End":"02:47.945","Text":"I\u0027m just going to carry on with this."},{"Start":"02:47.945 ","End":"02:51.440","Text":"Now this equals and I\u0027m just continuing with this part."},{"Start":"02:51.440 ","End":"02:59.000","Text":"Immediate integral is arctangent of x and I have to take this between minus 1 and 1."},{"Start":"02:59.000 ","End":"03:06.620","Text":"Now, the arctangent of 1 minus the arctangent of minus 1."},{"Start":"03:06.620 ","End":"03:08.810","Text":"Now if you remember your trigonometry,"},{"Start":"03:08.810 ","End":"03:14.570","Text":"then you\u0027d remember that the arctangent of 1 is 45 degrees or Pi over"},{"Start":"03:14.570 ","End":"03:21.945","Text":"4 and the arctangent of minus 1 is minus Pi over 4 minus 45 degrees."},{"Start":"03:21.945 ","End":"03:25.710","Text":"All together we get Pi over 2."},{"Start":"03:25.710 ","End":"03:28.120","Text":"That\u0027s the answer"}],"ID":4529}],"Thumbnail":null,"ID":84700},{"Name":"Inequalities","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Inequalities - Part 1","Duration":"4m 40s","ChapterTopicVideoID":8330,"CourseChapterTopicPlaylistID":84701,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8330.jpeg","UploadDate":"2019-12-11T21:04:13.5300000","DurationForVideoObject":"PT4M40S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.910","Text":"In this clip, I\u0027ll be talking about some"},{"Start":"00:02.910 ","End":"00:06.300","Text":"inequalities involving the indefinite integral."},{"Start":"00:06.300 ","End":"00:08.070","Text":"These are particularly useful when"},{"Start":"00:08.070 ","End":"00:11.115","Text":"we can evaluate the indefinite integral."},{"Start":"00:11.115 ","End":"00:12.360","Text":"At least not exactly,"},{"Start":"00:12.360 ","End":"00:14.190","Text":"but we can give an estimate from above"},{"Start":"00:14.190 ","End":"00:16.170","Text":"or below as to what it might be."},{"Start":"00:16.170 ","End":"00:19.800","Text":"I\u0027m going to be giving 2 inequalities."},{"Start":"00:19.800 ","End":"00:21.800","Text":"I\u0027ll write them each 1,"},{"Start":"00:21.800 ","End":"00:23.180","Text":"and then I\u0027ll explain."},{"Start":"00:23.180 ","End":"00:25.820","Text":"Here\u0027s the first formula which"},{"Start":"00:25.820 ","End":"00:28.280","Text":"gives us an estimate of an integral."},{"Start":"00:28.280 ","End":"00:30.380","Text":"It gives an upper bound and a lower bound."},{"Start":"00:30.380 ","End":"00:32.030","Text":"It says that if we have the integral"},{"Start":"00:32.030 ","End":"00:33.995","Text":"from a to b of a function,"},{"Start":"00:33.995 ","End":"00:35.960","Text":"then it\u0027s less than or equal to"},{"Start":"00:35.960 ","End":"00:38.435","Text":"big M times b minus a."},{"Start":"00:38.435 ","End":"00:39.620","Text":"This is b and this is a."},{"Start":"00:39.620 ","End":"00:41.510","Text":"In a moment I\u0027ll say what M is,"},{"Start":"00:41.510 ","End":"00:44.480","Text":"and it\u0027s bigger or equal to little m times b minus a."},{"Start":"00:44.480 ","End":"00:47.540","Text":"Big M is the absolute maximum"},{"Start":"00:47.540 ","End":"00:50.087","Text":"of the function on the interval a,b,"},{"Start":"00:50.087 ","End":"00:52.820","Text":"means when x is between a and b inclusive,"},{"Start":"00:52.820 ","End":"00:55.390","Text":"and m is the absolute minimum."},{"Start":"00:55.390 ","End":"00:57.570","Text":"Here is a couple of axes."},{"Start":"00:57.570 ","End":"01:00.420","Text":"Let\u0027s take a and b along the x-axis."},{"Start":"01:00.420 ","End":"01:02.690","Text":"Let\u0027s say that this here is a,"},{"Start":"01:02.690 ","End":"01:05.805","Text":"and this here is b."},{"Start":"01:05.805 ","End":"01:08.510","Text":"Let\u0027s take some function f of x,"},{"Start":"01:08.510 ","End":"01:11.195","Text":"y equals f of x."},{"Start":"01:11.195 ","End":"01:12.220","Text":"Here\u0027s a."},{"Start":"01:12.220 ","End":"01:15.395","Text":"I want to take a vertical line from a,"},{"Start":"01:15.395 ","End":"01:18.530","Text":"and I want to take a vertical line from b."},{"Start":"01:18.530 ","End":"01:22.280","Text":"I want to shade the area under the curve."},{"Start":"01:22.280 ","End":"01:24.770","Text":"This green area actually represents"},{"Start":"01:24.770 ","End":"01:27.920","Text":"the integral from a to b of f of x/dx."},{"Start":"01:27.920 ","End":"01:32.390","Text":"Let me draw the absolute maximum in the interval a,b."},{"Start":"01:32.390 ","End":"01:34.850","Text":"I can see that the maximum is here,"},{"Start":"01:34.850 ","End":"01:38.914","Text":"and the absolute minimum would be here."},{"Start":"01:38.914 ","End":"01:44.780","Text":"This will be the absolute maximum of f of x"},{"Start":"01:44.780 ","End":"01:48.455","Text":"from a to b on the interval a,b"},{"Start":"01:48.455 ","End":"01:52.730","Text":"and this here will be the absolute minimum"},{"Start":"01:52.730 ","End":"01:56.330","Text":"on the interval a,b of the function f."},{"Start":"01:56.330 ","End":"02:00.125","Text":"I want to estimate the area under the curve."},{"Start":"02:00.125 ","End":"02:02.510","Text":"I want to estimate it from above and from below."},{"Start":"02:02.510 ","End":"02:05.524","Text":"If I draw a line here,"},{"Start":"02:05.524 ","End":"02:08.395","Text":"the area of this rectangle,"},{"Start":"02:08.395 ","End":"02:10.035","Text":"that I\u0027ve outlined in blue,"},{"Start":"02:10.035 ","End":"02:13.340","Text":"certainly its area is less than the area"},{"Start":"02:13.340 ","End":"02:16.820","Text":"of the shaded green because it\u0027s part of it."},{"Start":"02:16.820 ","End":"02:21.410","Text":"If I abbreviate this as little m"},{"Start":"02:21.410 ","End":"02:24.230","Text":"and this 1 big M,"},{"Start":"02:24.230 ","End":"02:27.950","Text":"then the area of this rectangle here"},{"Start":"02:27.950 ","End":"02:30.320","Text":"is going to be base times height"},{"Start":"02:30.320 ","End":"02:32.015","Text":"or height times base."},{"Start":"02:32.015 ","End":"02:33.919","Text":"The height is m,"},{"Start":"02:33.919 ","End":"02:37.685","Text":"and the base is b minus a."},{"Start":"02:37.685 ","End":"02:39.500","Text":"The area under the curve"},{"Start":"02:39.500 ","End":"02:41.000","Text":"which is shaded in green as we said,"},{"Start":"02:41.000 ","End":"02:48.920","Text":"is the integral from a to b of f of x/dx."},{"Start":"02:48.920 ","End":"02:51.485","Text":"This is less than or equal to this,"},{"Start":"02:51.485 ","End":"02:52.850","Text":"and before I write it,"},{"Start":"02:52.850 ","End":"02:54.470","Text":"I want to go for the third thing."},{"Start":"02:54.470 ","End":"02:56.390","Text":"This third thing I\u0027ll sketch in a moment"},{"Start":"02:56.390 ","End":"02:58.460","Text":"will be the rectangle that goes up"},{"Start":"02:58.460 ","End":"03:01.090","Text":"to the maximum which is in red."},{"Start":"03:01.090 ","End":"03:02.420","Text":"Here\u0027s the rectangle I mean,"},{"Start":"03:02.420 ","End":"03:05.660","Text":"and this rectangle certainly has an area"},{"Start":"03:05.660 ","End":"03:08.210","Text":"bigger or equal to the green area,"},{"Start":"03:08.210 ","End":"03:10.490","Text":"because the green area is included in it."},{"Start":"03:10.490 ","End":"03:12.050","Text":"What I can say is that"},{"Start":"03:12.050 ","End":"03:16.445","Text":"the red rectangle has an area of,"},{"Start":"03:16.445 ","End":"03:17.870","Text":"again, base times height"},{"Start":"03:17.870 ","End":"03:18.980","Text":"or height times base."},{"Start":"03:18.980 ","End":"03:21.020","Text":"In this case, the height is big M,"},{"Start":"03:21.020 ","End":"03:26.720","Text":"and the base is also b minus a."},{"Start":"03:26.720 ","End":"03:28.010","Text":"Now, let\u0027s put together"},{"Start":"03:28.010 ","End":"03:29.585","Text":"what I\u0027ve said so far."},{"Start":"03:29.585 ","End":"03:30.800","Text":"Then we have that"},{"Start":"03:30.800 ","End":"03:33.665","Text":"the smallest is the blue rectangle,"},{"Start":"03:33.665 ","End":"03:38.004","Text":"little m times b minus a,"},{"Start":"03:38.004 ","End":"03:40.035","Text":"less than or equal to."},{"Start":"03:40.035 ","End":"03:42.260","Text":"In this diagram, it\u0027s clearly less than."},{"Start":"03:42.260 ","End":"03:43.790","Text":"But it could be that it\u0027s less than"},{"Start":"03:43.790 ","End":"03:45.934","Text":"or equal to the green,"},{"Start":"03:45.934 ","End":"03:48.260","Text":"which is the area under the curve,"},{"Start":"03:48.260 ","End":"03:49.550","Text":"and that\u0027s the integral"},{"Start":"03:49.550 ","End":"03:54.535","Text":"from a to b of f of x/dx."},{"Start":"03:54.535 ","End":"03:57.410","Text":"The biggest of all is the red 1,"},{"Start":"03:57.410 ","End":"04:02.660","Text":"which is big M times b minus a."},{"Start":"04:02.660 ","End":"04:08.120","Text":"Little m, is the minimum of f of x."},{"Start":"04:08.120 ","End":"04:11.120","Text":"When x goes on the interval a,b,"},{"Start":"04:11.120 ","End":"04:14.990","Text":"and big M is the maximum"},{"Start":"04:14.990 ","End":"04:19.800","Text":"on the interval a,b of f of x."},{"Start":"04:19.800 ","End":"04:22.500","Text":"Of course, I mean the absolute minimum"},{"Start":"04:22.500 ","End":"04:24.075","Text":"and absolute maximum."},{"Start":"04:24.075 ","End":"04:26.040","Text":"Basically, we\u0027ve proven"},{"Start":"04:26.040 ","End":"04:27.990","Text":"what we set out to do,"},{"Start":"04:27.990 ","End":"04:30.320","Text":"and all that I would add here"},{"Start":"04:30.320 ","End":"04:31.820","Text":"would be an example."},{"Start":"04:31.820 ","End":"04:33.440","Text":"But there are examples"},{"Start":"04:33.440 ","End":"04:36.005","Text":"right after this theoretical clip."},{"Start":"04:36.005 ","End":"04:40.350","Text":"Let\u0027s get to the second inequality."}],"ID":8607},{"Watched":false,"Name":"Inequalities - Part 2","Duration":"3m 26s","ChapterTopicVideoID":8331,"CourseChapterTopicPlaylistID":84701,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8331.jpeg","UploadDate":"2019-12-11T21:05:21.6570000","DurationForVideoObject":"PT3M26S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.195","Text":"Here\u0027s our second inequality."},{"Start":"00:03.195 ","End":"00:06.420","Text":"I\u0027ve just written it and now explain it."},{"Start":"00:06.420 ","End":"00:10.290","Text":"It says that if we have 2 functions, f and g,"},{"Start":"00:10.290 ","End":"00:16.335","Text":"and f of x is less than or equal to g of x for all x between a and b inclusive,"},{"Start":"00:16.335 ","End":"00:19.830","Text":"then the integral and that same interval from a to b of"},{"Start":"00:19.830 ","End":"00:23.790","Text":"f is less than or equal to the integral of g. Other words,"},{"Start":"00:23.790 ","End":"00:25.515","Text":"if f is less than or equal to g,"},{"Start":"00:25.515 ","End":"00:27.960","Text":"then the integral of f is also less than or equal to the"},{"Start":"00:27.960 ","End":"00:30.615","Text":"integral of g. But of course, they have to match up."},{"Start":"00:30.615 ","End":"00:32.550","Text":"This has to be true for the interval a b,"},{"Start":"00:32.550 ","End":"00:34.650","Text":"and this is the integral from a to b."},{"Start":"00:34.650 ","End":"00:38.025","Text":"Here it is, and now I\u0027ll explain why this is so."},{"Start":"00:38.025 ","End":"00:41.225","Text":"Let\u0027s start with 1 of the functions f of x,"},{"Start":"00:41.225 ","End":"00:42.825","Text":"and I\u0027ve sketched it here actually,"},{"Start":"00:42.825 ","End":"00:45.275","Text":"I borrowed the sketch from the previous 1,"},{"Start":"00:45.275 ","End":"00:48.775","Text":"and now let\u0027s add g of x into the picture."},{"Start":"00:48.775 ","End":"00:50.300","Text":"Here is our second function,"},{"Start":"00:50.300 ","End":"00:51.550","Text":"y equals g of x,"},{"Start":"00:51.550 ","End":"00:54.080","Text":"and now I want to do a bit more shading because you"},{"Start":"00:54.080 ","End":"00:56.930","Text":"see the green represents the integral of"},{"Start":"00:56.930 ","End":"01:02.340","Text":"f of x. I\u0027ve shaded the area under the curve g of x between a and b."},{"Start":"01:02.340 ","End":"01:07.340","Text":"It\u0027s clear that the area that\u0027s highlighted in green is"},{"Start":"01:07.340 ","End":"01:13.135","Text":"less than or equal to the area that is marked with diagonal stripes."},{"Start":"01:13.135 ","End":"01:14.735","Text":"Here\u0027s what I mean."},{"Start":"01:14.735 ","End":"01:21.185","Text":"If I look at the diagonal purple lines that represents the integral of g of x dx."},{"Start":"01:21.185 ","End":"01:24.470","Text":"The green represents the integral of f of x dx,"},{"Start":"01:24.470 ","End":"01:28.775","Text":"and clearly because g of x is always above f of x,"},{"Start":"01:28.775 ","End":"01:30.770","Text":"at least on the interval a b,"},{"Start":"01:30.770 ","End":"01:36.680","Text":"then we can see that the green is less than or equal to the purple diagonal shaded."},{"Start":"01:36.680 ","End":"01:39.780","Text":"This means that this is less than or equal to this."},{"Start":"01:39.780 ","End":"01:44.090","Text":"Which says that the integral from a to b of f of"},{"Start":"01:44.090 ","End":"01:51.590","Text":"x dx is less than or equal to the integral from a to b of g of x dx."},{"Start":"01:51.590 ","End":"01:57.920","Text":"By the way, if we look over here that actually g at some point is less than or"},{"Start":"01:57.920 ","End":"02:00.890","Text":"equal to f. But that doesn\u0027t matter because we only"},{"Start":"02:00.890 ","End":"02:04.670","Text":"care about what happens in the interval from a to b."},{"Start":"02:04.670 ","End":"02:07.910","Text":"This is an illustration of what I wrote here,"},{"Start":"02:07.910 ","End":"02:09.995","Text":"and I think it\u0027s fairly straightforward."},{"Start":"02:09.995 ","End":"02:13.990","Text":"Of course, it\u0027ll be clear in the examples that follow,"},{"Start":"02:13.990 ","End":"02:16.570","Text":"there at least a couple of examples that use this."},{"Start":"02:16.570 ","End":"02:19.460","Text":"This can be used for an inequality on an integral."},{"Start":"02:19.460 ","End":"02:23.270","Text":"For example, if we have difficulty in computing f of x,"},{"Start":"02:23.270 ","End":"02:24.920","Text":"we can compute g of x,"},{"Start":"02:24.920 ","End":"02:28.175","Text":"then we can get an upper bound for f, and the other hand,"},{"Start":"02:28.175 ","End":"02:32.870","Text":"if we can\u0027t do g of x and we know how to compute f of x so we can get a lower bound for"},{"Start":"02:32.870 ","End":"02:35.210","Text":"g. It\u0027s good for either getting an upper bound"},{"Start":"02:35.210 ","End":"02:37.805","Text":"or a lower bound depending on which 1 we know."},{"Start":"02:37.805 ","End":"02:41.390","Text":"I just want to say another thing that it works also with"},{"Start":"02:41.390 ","End":"02:45.275","Text":"3 integrals and I\u0027ll just write it down."},{"Start":"02:45.275 ","End":"02:48.589","Text":"Here is the generalization to 3 functions."},{"Start":"02:48.589 ","End":"02:51.160","Text":"If instead of f and g have f and g and h,"},{"Start":"02:51.160 ","End":"02:52.940","Text":"and this is less than or equal to this,"},{"Start":"02:52.940 ","End":"02:57.230","Text":"less than or equal to this on the interval from a to b inclusive."},{"Start":"02:57.230 ","End":"03:02.030","Text":"Then we can conclude that the integral of f is less than or equal to integral of g,"},{"Start":"03:02.030 ","End":"03:05.630","Text":"which is less than or equal to the integral of h. This is often used"},{"Start":"03:05.630 ","End":"03:09.530","Text":"when the middle 1 g is unknown or difficult to compute."},{"Start":"03:09.530 ","End":"03:15.065","Text":"But we can compute f and h. Then we get an estimate on g of x."},{"Start":"03:15.065 ","End":"03:17.585","Text":"It\u0027s integral between something and something,"},{"Start":"03:17.585 ","End":"03:20.945","Text":"and you\u0027ll see there\u0027s these 1 exercise they\u0027re using this."},{"Start":"03:20.945 ","End":"03:23.990","Text":"Other than telling you to go ahead and do the exercises,"},{"Start":"03:23.990 ","End":"03:27.000","Text":"the solved ones. I\u0027m done here."}],"ID":8608},{"Watched":false,"Name":"Exercise 1","Duration":"3m 26s","ChapterTopicVideoID":8332,"CourseChapterTopicPlaylistID":84701,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8332.jpeg","UploadDate":"2017-01-28T18:21:21.8930000","DurationForVideoObject":"PT3M26S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.520","Text":"In this exercise, we have to prove certain inequalities that some definite"},{"Start":"00:05.520 ","End":"00:11.355","Text":"integral is less than or equal to a number and greater or equal to something else."},{"Start":"00:11.355 ","End":"00:16.015","Text":"This exercise is usually solved by means of the following theorem"},{"Start":"00:16.015 ","End":"00:20.255","Text":"that whenever we have a definite integral on an interval,"},{"Start":"00:20.255 ","End":"00:25.400","Text":"then we can bound it above by b minus a times capital M"},{"Start":"00:25.400 ","End":"00:28.080","Text":"and below by b minus a times little m,"},{"Start":"00:28.080 ","End":"00:30.165","Text":"where the little m and big M"},{"Start":"00:30.165 ","End":"00:33.740","Text":"are just the minimum and the maximum of the function on the interval."},{"Start":"00:33.740 ","End":"00:37.025","Text":"So we already have some of the things here."},{"Start":"00:37.025 ","End":"00:38.965","Text":"We know what f of x is."},{"Start":"00:38.965 ","End":"00:44.450","Text":"Our function is 1 over 1 plus x^4. We also have"},{"Start":"00:44.450 ","End":"00:51.360","Text":"a, which is minus 1, and we have that b is equal to 3."},{"Start":"00:51.360 ","End":"00:55.660","Text":"What we need are little m and big M."}],"ID":8609},{"Watched":false,"Name":"Exercise 2","Duration":"3m 19s","ChapterTopicVideoID":8333,"CourseChapterTopicPlaylistID":84701,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8333.jpeg","UploadDate":"2017-01-28T18:21:29.5400000","DurationForVideoObject":"PT3M19S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.930","Text":"In this exercise, we\u0027re given a definite integral and we have"},{"Start":"00:03.930 ","End":"00:07.424","Text":"to show that it\u0027s between something and something."},{"Start":"00:07.424 ","End":"00:10.005","Text":"In other words, we possibly can\u0027t compute it,"},{"Start":"00:10.005 ","End":"00:12.990","Text":"but we can still give an estimate of what it can be at most,"},{"Start":"00:12.990 ","End":"00:14.610","Text":"and what good can be at least."},{"Start":"00:14.610 ","End":"00:17.940","Text":"This is based on the theorem that,"},{"Start":"00:17.940 ","End":"00:20.835","Text":"if we have a function on a, b,"},{"Start":"00:20.835 ","End":"00:24.750","Text":"then the integral is going to be less than or equal to,"},{"Start":"00:24.750 ","End":"00:30.390","Text":"b minus a times big m and bigger or equal to b minus a times little m,"},{"Start":"00:30.390 ","End":"00:33.060","Text":"little m and big m are just shortcuts for"},{"Start":"00:33.060 ","End":"00:37.630","Text":"the minimum and the maximum of f of x on the interval a."}],"ID":8610},{"Watched":false,"Name":"Exercise 3","Duration":"2m 31s","ChapterTopicVideoID":8334,"CourseChapterTopicPlaylistID":84701,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8334.jpeg","UploadDate":"2017-01-28T18:21:52.2470000","DurationForVideoObject":"PT2M31S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.340","Text":"Here we have to prove that this definite integral satisfies a couple of inequalities,"},{"Start":"00:05.340 ","End":"00:08.250","Text":"that it\u0027s less than or equal to this and bigger or equal to this."},{"Start":"00:08.250 ","End":"00:13.395","Text":"The theory we need here is this 1 that I\u0027ve already put on the page,"},{"Start":"00:13.395 ","End":"00:17.025","Text":"is that the integral is less than or equal to,"},{"Start":"00:17.025 ","End":"00:22.875","Text":"basically it\u0027s the difference between the upper and lower limits times both M and m,"},{"Start":"00:22.875 ","End":"00:29.100","Text":"where M is the maximum and m is the minimum of the function f on the interval a,"},{"Start":"00:29.100 ","End":"00:31.050","Text":"b in our case 0 and 2."},{"Start":"00:31.050 ","End":"00:35.610","Text":"In our case, the f of x is e^ x squared."},{"Start":"00:35.610 ","End":"00:36.894","Text":"That\u0027s the function."},{"Start":"00:36.894 ","End":"00:41.030","Text":"The end points a is 0, b is 2,"},{"Start":"00:41.030 ","End":"00:47.030","Text":"and I guess it\u0027s useful to write down what is b minus a, which is 2."},{"Start":"00:47.030 ","End":"00:49.445","Text":"Now, the way I look for extrema,"},{"Start":"00:49.445 ","End":"00:50.989","Text":"which is minimum and maximum,"},{"Start":"00:50.989 ","End":"00:52.130","Text":"is in 2 places."},{"Start":"00:52.130 ","End":"00:56.930","Text":"I first differentiate f prime of x and set it to"},{"Start":"00:56.930 ","End":"00:59.870","Text":"0 and see what x that gives us and we\u0027re looking"},{"Start":"00:59.870 ","End":"01:03.050","Text":"for x is in the interior between 0 and 2 and secondly,"},{"Start":"01:03.050 ","End":"01:04.850","Text":"we\u0027ll take the end points 0 and 2."},{"Start":"01:04.850 ","End":"01:08.160","Text":"First, this suspect for critical point."},{"Start":"01:08.160 ","End":"01:11.940","Text":"We get the derivative of e^ x squared,"},{"Start":"01:11.940 ","End":"01:19.685","Text":"is e^ x squared times inner derivative 2x is equal to 0."},{"Start":"01:19.685 ","End":"01:22.459","Text":"Now e to the something is never 0, always positive."},{"Start":"01:22.459 ","End":"01:25.475","Text":"This can only happen when x equals 0,"},{"Start":"01:25.475 ","End":"01:27.889","Text":"but it\u0027s not in the interior,"},{"Start":"01:27.889 ","End":"01:30.020","Text":"it\u0027s on the endpoints, so I can ignore it,"},{"Start":"01:30.020 ","End":"01:33.590","Text":"but I still get it because I get it by taking the endpoints."},{"Start":"01:33.590 ","End":"01:36.519","Text":"What I need to check is f of 0"},{"Start":"01:36.519 ","End":"01:42.050","Text":"and f of 2 and let\u0027s see which of these is the largest and which is the smallest?"},{"Start":"01:42.050 ","End":"01:46.790","Text":"If x is 0, f of x is e^0, which is 1."},{"Start":"01:46.790 ","End":"01:51.620","Text":"If f is 2, we get 2 squared is 4, e^4."},{"Start":"01:51.620 ","End":"01:56.640","Text":"Now, it\u0027s clear that the e^4 is much bigger than 1,"},{"Start":"01:56.640 ","End":"02:01.350","Text":"so this 1 is going to be my M and 1 will be"},{"Start":"02:01.350 ","End":"02:07.195","Text":"m. All I have left to do is to substitute in this formula here."},{"Start":"02:07.195 ","End":"02:16.880","Text":"What I get is b minus a is 2 times m. I get 2 times 1 is less than or equal to"},{"Start":"02:16.880 ","End":"02:23.240","Text":"the integral e^ x squared dx and it\u0027s less than or equal to"},{"Start":"02:23.240 ","End":"02:29.990","Text":"the same 2 times M and M is e^4 and now I\u0027ve got exactly what\u0027s written here,"},{"Start":"02:29.990 ","End":"02:32.340","Text":"so we are done."}],"ID":8611},{"Watched":false,"Name":"Exercise 4","Duration":"4m 16s","ChapterTopicVideoID":8335,"CourseChapterTopicPlaylistID":84701,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8335.jpeg","UploadDate":"2017-01-28T18:22:41.6400000","DurationForVideoObject":"PT4M16S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.010","Text":"In this exercise, we have to prove that"},{"Start":"00:02.010 ","End":"00:07.035","Text":"this definite integral satisfies certain inequalities."},{"Start":"00:07.035 ","End":"00:09.330","Text":"We have to show that it\u0027s less than 1"},{"Start":"00:09.330 ","End":"00:14.010","Text":"and bigger than 1/2 e to the minus 10."},{"Start":"00:14.010 ","End":"00:17.025","Text":"Then it fits the pattern of the theorem,"},{"Start":"00:17.025 ","End":"00:19.335","Text":"which says that when we have a definite integral,"},{"Start":"00:19.335 ","End":"00:21.090","Text":"we can estimate it to be less than"},{"Start":"00:21.090 ","End":"00:22.470","Text":"or equal to something and bigger"},{"Start":"00:22.470 ","End":"00:23.745","Text":"or equal to something else."},{"Start":"00:23.745 ","End":"00:27.090","Text":"Here, the a and the b are 0 and 10."},{"Start":"00:27.090 ","End":"00:30.240","Text":"The m and the big M are going to be"},{"Start":"00:30.240 ","End":"00:32.730","Text":"the minimum and maximum of the function"},{"Start":"00:32.730 ","End":"00:34.440","Text":"on the interval from 0 to 10."},{"Start":"00:34.440 ","End":"00:36.000","Text":"We\u0027re going to find those out."},{"Start":"00:36.000 ","End":"00:37.860","Text":"Well, let\u0027s write down what we have."},{"Start":"00:37.860 ","End":"00:40.790","Text":"The function of x that we have is"},{"Start":"00:40.790 ","End":"00:45.455","Text":"e to the minus x over x plus 10."},{"Start":"00:45.455 ","End":"00:48.740","Text":"We have a lower limit a which is 0,"},{"Start":"00:48.740 ","End":"00:52.309","Text":"and upper limit b which is 10."},{"Start":"00:52.309 ","End":"00:54.440","Text":"We also might as well"},{"Start":"00:54.440 ","End":"00:56.510","Text":"compute b minus a already,"},{"Start":"00:56.510 ","End":"01:00.410","Text":"we\u0027ll need that 10 minus 0 is 10."},{"Start":"01:00.410 ","End":"01:01.910","Text":"What remains is to find"},{"Start":"01:01.910 ","End":"01:03.020","Text":"the minimum and the maximum."},{"Start":"01:03.020 ","End":"01:04.235","Text":"In other words, the extrema."},{"Start":"01:04.235 ","End":"01:05.510","Text":"The way we find the extrema"},{"Start":"01:05.510 ","End":"01:08.690","Text":"is first we look for derivative equal to 0."},{"Start":"01:08.690 ","End":"01:11.750","Text":"It has to be in the interior between 0 and 10."},{"Start":"01:11.750 ","End":"01:15.184","Text":"Let\u0027s see, we have to differentiate a quotient."},{"Start":"01:15.184 ","End":"01:17.120","Text":"Better write the quotient rule quickly,"},{"Start":"01:17.120 ","End":"01:19.310","Text":"u is e to the minus x,"},{"Start":"01:19.310 ","End":"01:20.630","Text":"v is x plus 10."},{"Start":"01:20.630 ","End":"01:28.670","Text":"I get u prime is minus e to the minus x times v,"},{"Start":"01:28.670 ","End":"01:31.940","Text":"which is x plus 10 minus u,"},{"Start":"01:31.940 ","End":"01:33.770","Text":"which is e to the minus x"},{"Start":"01:33.770 ","End":"01:37.009","Text":"times v prime which is 1,"},{"Start":"01:37.009 ","End":"01:41.265","Text":"all this over x plus 10 squared."},{"Start":"01:41.265 ","End":"01:43.385","Text":"This has to equal 0."},{"Start":"01:43.385 ","End":"01:45.410","Text":"Well, the denominator"},{"Start":"01:45.410 ","End":"01:47.330","Text":"is positive and defined,"},{"Start":"01:47.330 ","End":"01:49.250","Text":"so it must be that the numerator\u0027s 0."},{"Start":"01:49.250 ","End":"01:53.450","Text":"The numerator equals 0 minus e to the minus x."},{"Start":"01:53.450 ","End":"01:53.930","Text":"You know what?"},{"Start":"01:53.930 ","End":"01:56.075","Text":"I can take this outside the brackets."},{"Start":"01:56.075 ","End":"02:01.295","Text":"I get x plus 10 plus 1 equals 0."},{"Start":"02:01.295 ","End":"02:03.200","Text":"Now, this is never 0,"},{"Start":"02:03.200 ","End":"02:05.180","Text":"e to the something is always positive."},{"Start":"02:05.180 ","End":"02:06.350","Text":"So x plus 10 plus 1,"},{"Start":"02:06.350 ","End":"02:08.659","Text":"which is x plus 11, must be 0;"},{"Start":"02:08.659 ","End":"02:12.050","Text":"x equals minus 11."},{"Start":"02:12.050 ","End":"02:15.710","Text":"Minus 11 is completely out of range."},{"Start":"02:15.710 ","End":"02:17.720","Text":"It\u0027s not between 0 and 10."},{"Start":"02:17.720 ","End":"02:20.150","Text":"So this is no good for me."},{"Start":"02:20.150 ","End":"02:23.780","Text":"My only suspects or possible extrema"},{"Start":"02:23.780 ","End":"02:26.195","Text":"are the endpoints, 0 and 10."},{"Start":"02:26.195 ","End":"02:29.195","Text":"What I need to do is check what is f of 0."},{"Start":"02:29.195 ","End":"02:31.580","Text":"I need to check what is f of 10"},{"Start":"02:31.580 ","End":"02:33.650","Text":"and see which is the big 1,"},{"Start":"02:33.650 ","End":"02:34.730","Text":"which is a small 1,"},{"Start":"02:34.730 ","End":"02:37.250","Text":"which is little m and which is big M."},{"Start":"02:37.250 ","End":"02:40.490","Text":"Now, f of 0, I plug in 0 here."},{"Start":"02:40.490 ","End":"02:45.565","Text":"So f of 0 is 1/0 plus 10, which is 1/10;"},{"Start":"02:45.565 ","End":"02:49.460","Text":"f of 10 is e to the minus 10"},{"Start":"02:49.460 ","End":"02:53.015","Text":"over 10 plus 10, which is 20."},{"Start":"02:53.015 ","End":"02:54.620","Text":"Now, which is smaller?"},{"Start":"02:54.620 ","End":"02:57.980","Text":"Well, e to the minus 10 is less than 1."},{"Start":"02:57.980 ","End":"03:00.665","Text":"This is less than a 20th."},{"Start":"03:00.665 ","End":"03:03.740","Text":"In fact, this is an incredibly small number."},{"Start":"03:03.740 ","End":"03:08.000","Text":"This is my minimum called little m,"},{"Start":"03:08.000 ","End":"03:10.850","Text":"and this is the maximum, big M."},{"Start":"03:10.850 ","End":"03:14.465","Text":"Now, I can just plug in the equation here."},{"Start":"03:14.465 ","End":"03:19.350","Text":"What I get is that b minus a is 10."},{"Start":"03:19.350 ","End":"03:23.030","Text":"I get 10 times little m,"},{"Start":"03:23.030 ","End":"03:26.540","Text":"which is e to the minus 10/20"},{"Start":"03:26.540 ","End":"03:31.695","Text":"is less than or equal to the integral."},{"Start":"03:31.695 ","End":"03:33.750","Text":"I\u0027ll write that in a second."},{"Start":"03:33.750 ","End":"03:37.100","Text":"This will be less than or equal to big M,"},{"Start":"03:37.100 ","End":"03:39.890","Text":"also 10, which is the b minus a,"},{"Start":"03:39.890 ","End":"03:43.250","Text":"and big M which is 1/10."},{"Start":"03:43.250 ","End":"03:49.165","Text":"The integral is e to the minus x over x plus 10."},{"Start":"03:49.165 ","End":"03:52.190","Text":"I don\u0027t want to write this whole integral again."},{"Start":"03:52.190 ","End":"03:53.795","Text":"Let\u0027s just simplify this."},{"Start":"03:53.795 ","End":"03:58.970","Text":"The 10 with the 20 goes twice,"},{"Start":"03:58.970 ","End":"04:02.270","Text":"and 10 with the 10 disappears."},{"Start":"04:02.270 ","End":"04:05.360","Text":"Well, you can see that this is 1/2 e"},{"Start":"04:05.360 ","End":"04:09.465","Text":"to the minus 10 and this is 1."},{"Start":"04:09.465 ","End":"04:13.310","Text":"I think we can give ourselves a checkmark"},{"Start":"04:13.310 ","End":"04:15.230","Text":"that we\u0027ve succeeded in proving it,"},{"Start":"04:15.230 ","End":"04:17.670","Text":"and we are done."}],"ID":8612},{"Watched":false,"Name":"Exercise 5","Duration":"4m 52s","ChapterTopicVideoID":8345,"CourseChapterTopicPlaylistID":84701,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8345.jpeg","UploadDate":"2017-01-28T18:24:48.8530000","DurationForVideoObject":"PT4M52S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.060","Text":"Here we have another 1 of those exercises where we\u0027re given an integral,"},{"Start":"00:03.060 ","End":"00:07.920","Text":"and we have to show that it\u0027s bounded above and below by certain constants."},{"Start":"00:07.920 ","End":"00:12.210","Text":"We use the theorem that the definite integral is"},{"Start":"00:12.210 ","End":"00:17.700","Text":"bounded below by b minus a times the minimum of the function,"},{"Start":"00:17.700 ","End":"00:21.090","Text":"and above by b minus a times the maximum of the function,"},{"Start":"00:21.090 ","End":"00:22.730","Text":"where b and a are the limits."},{"Start":"00:22.730 ","End":"00:24.625","Text":"I\u0027m going to use this."},{"Start":"00:24.625 ","End":"00:28.590","Text":"In our case, f of x is 1,"},{"Start":"00:28.590 ","End":"00:33.405","Text":"over 3 plus 4 sine squared of x."},{"Start":"00:33.405 ","End":"00:38.610","Text":"Our a is equal to 0, the lower limit."},{"Start":"00:38.610 ","End":"00:41.534","Text":"B is Pi over 2."},{"Start":"00:41.534 ","End":"00:44.900","Text":"I guess it\u0027s useful to write down b minus a,"},{"Start":"00:44.900 ","End":"00:49.925","Text":"also, so b minus a just squeeze it in."},{"Start":"00:49.925 ","End":"00:53.750","Text":"Pi over 2, minus 0 is Pi over 2."},{"Start":"00:53.750 ","End":"00:56.360","Text":"Now how do we look for minimum and maximum,"},{"Start":"00:56.360 ","End":"00:57.755","Text":"also known as extrema?"},{"Start":"00:57.755 ","End":"01:00.980","Text":"There are 2 places we look on a closed interval."},{"Start":"01:00.980 ","End":"01:05.900","Text":"First of all, we try differentiating and setting to 0 and see if these points,"},{"Start":"01:05.900 ","End":"01:06.995","Text":"which are called critical points,"},{"Start":"01:06.995 ","End":"01:09.920","Text":"are in the interval in the interior,"},{"Start":"01:09.920 ","End":"01:11.665","Text":"and we also take the endpoints."},{"Start":"01:11.665 ","End":"01:17.810","Text":"Let\u0027s first start with finding critical points f prime of x equals 0."},{"Start":"01:17.810 ","End":"01:19.815","Text":"Let\u0027s see. I need to differentiate."},{"Start":"01:19.815 ","End":"01:26.580","Text":"The derivative of 1 over something is minus 1 over that something squared."},{"Start":"01:26.580 ","End":"01:31.785","Text":"I\u0027ll start with 3 plus 4 sine squared x,"},{"Start":"01:31.785 ","End":"01:38.985","Text":"all squared, minus, above I need the derivative of the denominator."},{"Start":"01:38.985 ","End":"01:41.060","Text":"The derivative would be,"},{"Start":"01:41.060 ","End":"01:43.475","Text":"let\u0027s see, 3 gives me nothing."},{"Start":"01:43.475 ","End":"01:45.350","Text":"This gives me 4."},{"Start":"01:45.350 ","End":"01:50.180","Text":"Then sine squared gives me 2 sine x,"},{"Start":"01:50.180 ","End":"01:52.940","Text":"but times the derivative of sine x,"},{"Start":"01:52.940 ","End":"01:56.315","Text":"which is cosine x."},{"Start":"01:56.315 ","End":"02:00.370","Text":"This has to be equal to 0 now."},{"Start":"02:00.370 ","End":"02:02.765","Text":"The numerator has to be 0."},{"Start":"02:02.765 ","End":"02:04.970","Text":"I can ignore the constant."},{"Start":"02:04.970 ","End":"02:10.444","Text":"Basically, what it tells me is that sine squared x cosine x is 0."},{"Start":"02:10.444 ","End":"02:13.100","Text":"Now, 1 of these has to be 0,"},{"Start":"02:13.100 ","End":"02:15.470","Text":"so either sine squared x is 0,"},{"Start":"02:15.470 ","End":"02:17.420","Text":"in which case sine x is 0."},{"Start":"02:17.420 ","End":"02:24.560","Text":"I say that sine x equals 0 or cosine x equals 0."},{"Start":"02:24.560 ","End":"02:29.210","Text":"But I\u0027m looking for x in the interval from 0 to Pi over 2."},{"Start":"02:29.210 ","End":"02:34.430","Text":"In other words, I\u0027ll just remind you that x has to be between 0 and Pi over 2."},{"Start":"02:34.430 ","End":"02:37.145","Text":"Or if you like this is 90 degrees."},{"Start":"02:37.145 ","End":"02:39.800","Text":"When is the sine equal to 0,"},{"Start":"02:39.800 ","End":"02:44.385","Text":"and when is the cosine equal to 0 from 0-90?"},{"Start":"02:44.385 ","End":"02:46.175","Text":"Well, this we know already."},{"Start":"02:46.175 ","End":"02:47.910","Text":"Sine of 0 is 0,"},{"Start":"02:47.910 ","End":"02:50.130","Text":"and cosine of 90 is 0."},{"Start":"02:50.130 ","End":"02:53.730","Text":"Basically, x has to be either 0 or Pi over 2."},{"Start":"02:53.730 ","End":"03:00.515","Text":"This gives us that x equals 0 or x equals Pi over 2."},{"Start":"03:00.515 ","End":"03:02.210","Text":"Now, these are not in the interior."},{"Start":"03:02.210 ","End":"03:04.100","Text":"They\u0027re on the edge, so we ignore them."},{"Start":"03:04.100 ","End":"03:08.410","Text":"But ironically, these are exactly the points that we take when we take the endpoints."},{"Start":"03:08.410 ","End":"03:12.590","Text":"Whether you choose to take them as critical points or as endpoints, these are the points,"},{"Start":"03:12.590 ","End":"03:14.660","Text":"0 and Pi over 2,"},{"Start":"03:14.660 ","End":"03:19.135","Text":"and we just substitute them into the function to see which is larger, which is smaller."},{"Start":"03:19.135 ","End":"03:21.360","Text":"Let\u0027s see. F of 0 equals,"},{"Start":"03:21.360 ","End":"03:25.680","Text":"and I want f of Pi over 2 equals."},{"Start":"03:25.680 ","End":"03:27.075","Text":"Start with the 0."},{"Start":"03:27.075 ","End":"03:31.770","Text":"If x is 0, sine of 0 is 0 times 4,"},{"Start":"03:31.770 ","End":"03:35.380","Text":"is still 0, so we get 1/3."},{"Start":"03:35.630 ","End":"03:39.555","Text":"Let\u0027s see. F of Pi over 2,"},{"Start":"03:39.555 ","End":"03:42.780","Text":"sine of Pi over 2 is 1."},{"Start":"03:42.780 ","End":"03:47.775","Text":"Here we get 3 plus 4 is 7."},{"Start":"03:47.775 ","End":"03:51.850","Text":"Here we get 1 over 7."},{"Start":"03:52.010 ","End":"03:55.060","Text":"This 1 is obviously smaller than this,"},{"Start":"03:55.060 ","End":"03:58.735","Text":"so this 1 is going to be my little m,"},{"Start":"03:58.735 ","End":"04:03.755","Text":"and this 1 is going to be big M, minimum, maximum."},{"Start":"04:03.755 ","End":"04:06.245","Text":"Now we\u0027ll have to do is plug it in here,"},{"Start":"04:06.245 ","End":"04:08.825","Text":"where b minus a is Pi over 2."},{"Start":"04:08.825 ","End":"04:11.015","Text":"What I get is Pi over 2,"},{"Start":"04:11.015 ","End":"04:19.295","Text":"times little m is 1/17 is less than or equal to the integral of f of x d_x,"},{"Start":"04:19.295 ","End":"04:26.350","Text":"which is d_x over 3 plus 4 sine squared x."},{"Start":"04:26.350 ","End":"04:30.415","Text":"That\u0027s less than or equal to the same Pi over 2."},{"Start":"04:30.415 ","End":"04:33.740","Text":"But this time times big M, which is 1/3."},{"Start":"04:33.740 ","End":"04:40.070","Text":"Now, I\u0027m not going to continue with this simplification because obviously Pi over 2,"},{"Start":"04:40.070 ","End":"04:43.910","Text":"times 1/17 is the same as Pi over 14,"},{"Start":"04:43.910 ","End":"04:45.290","Text":"and this Pi over 2, times 1,"},{"Start":"04:45.290 ","End":"04:48.040","Text":"over 3 is Pi over 6."},{"Start":"04:48.040 ","End":"04:53.280","Text":"Looks like we\u0027ve exactly proved what we had to prove. We\u0027re done."}],"ID":8613},{"Watched":false,"Name":"Exercise 6","Duration":"3m 26s","ChapterTopicVideoID":8346,"CourseChapterTopicPlaylistID":84701,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8346.jpeg","UploadDate":"2017-01-28T18:25:05.9700000","DurationForVideoObject":"PT3M26S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.560","Text":"In this exercise, we have one of those inequalities of integrals to show"},{"Start":"00:04.560 ","End":"00:06.240","Text":"that the integral is less than or equal to"},{"Start":"00:06.240 ","End":"00:08.940","Text":"something, and bigger or equal to something else."},{"Start":"00:08.940 ","End":"00:11.280","Text":"Previously we used a certain theorem,"},{"Start":"00:11.280 ","End":"00:13.875","Text":"but here we\u0027re going to use a different tool to solve."},{"Start":"00:13.875 ","End":"00:18.180","Text":"We\u0027re going to use the theorem that if you have 2 functions, f and g,"},{"Start":"00:18.180 ","End":"00:21.270","Text":"and one of them is less than or equal to the other on an interval,"},{"Start":"00:21.270 ","End":"00:23.325","Text":"say x goes from a to b,"},{"Start":"00:23.325 ","End":"00:27.675","Text":"then the same inequality less than or equal to will hold for the integrals."},{"Start":"00:27.675 ","End":"00:29.280","Text":"In other words, the set of f and g,"},{"Start":"00:29.280 ","End":"00:34.120","Text":"I can put the integral from a to b of f of x and the integral from a to b of g of x."},{"Start":"00:34.120 ","End":"00:38.975","Text":"Now, we can extend this theorem to say that if we had a third one, h of x,"},{"Start":"00:38.975 ","End":"00:43.830","Text":"then that would also hold true, because g would be like my f and h would be like"},{"Start":"00:43.830 ","End":"00:49.730","Text":"my g. I could put that this is equal to a to b of h of x dx."},{"Start":"00:49.730 ","End":"00:52.075","Text":"It works for 3 or any number in the chain."},{"Start":"00:52.075 ","End":"00:55.430","Text":"Now, what I need to decide is, what is going to be my f,"},{"Start":"00:55.430 ","End":"00:57.095","Text":"my g, and my h."},{"Start":"00:57.095 ","End":"01:00.860","Text":"The first thing is that the sign, and I don\u0027t care what\u0027s"},{"Start":"01:00.860 ","End":"01:05.210","Text":"inside of it, is always between 1 and minus 1."},{"Start":"01:05.210 ","End":"01:07.805","Text":"I\u0027m going to write that also here is a useful thing."},{"Start":"01:07.805 ","End":"01:14.145","Text":"The sine of an angle Alpha is always between minus 1 and 1, whatever Alpha is."},{"Start":"01:14.145 ","End":"01:16.350","Text":"If I use that fact here,"},{"Start":"01:16.350 ","End":"01:18.965","Text":"then what I\u0027m going to take as my g,"},{"Start":"01:18.965 ","End":"01:25.160","Text":"the middle function will be sine of natural log of x over x plus 1,"},{"Start":"01:25.160 ","End":"01:31.880","Text":"is going to be always less than or equal to 1 and bigger or equal to minus 1."},{"Start":"01:31.880 ","End":"01:36.140","Text":"My apologies. I think I meant to put x plus 1 here,"},{"Start":"01:36.140 ","End":"01:37.810","Text":"so let me just change that."},{"Start":"01:37.810 ","End":"01:40.315","Text":"Now, we\u0027re okay with the definition."},{"Start":"01:40.315 ","End":"01:42.890","Text":"Let\u0027s see what we can do with this."},{"Start":"01:42.890 ","End":"01:45.935","Text":"I want to get from here to this integral."},{"Start":"01:45.935 ","End":"01:49.760","Text":"I\u0027m almost there, I just need an extra X in front."},{"Start":"01:49.760 ","End":"01:55.760","Text":"We can multiply an inequality by a positive quantity without changing the inequality."},{"Start":"01:55.760 ","End":"01:58.445","Text":"I\u0027m going to multiply everything by x,"},{"Start":"01:58.445 ","End":"02:01.250","Text":"so I get minus x is"},{"Start":"02:01.250 ","End":"02:10.355","Text":"less than or equal to x sine of natural log of x plus 1 over x plus 1,"},{"Start":"02:10.355 ","End":"02:12.530","Text":"which is less than or equal to x."},{"Start":"02:12.530 ","End":"02:15.910","Text":"It\u0027s actually true from the 0 or positive."},{"Start":"02:15.910 ","End":"02:17.900","Text":"Now, this is going to be my f, g,"},{"Start":"02:17.900 ","End":"02:21.690","Text":"and h. Let\u0027s take the integral of all of these from 0-1,"},{"Start":"02:21.690 ","End":"02:28.220","Text":"so we get the integral of minus x dx from 0-1 is"},{"Start":"02:28.220 ","End":"02:35.645","Text":"less than or equal to the integral from 0-1 of x sine of natural log of x plus 1,"},{"Start":"02:35.645 ","End":"02:38.900","Text":"over x plus 1 dx,"},{"Start":"02:38.900 ","End":"02:44.965","Text":"which is less than or equal to the integral of x dx from 0-1."},{"Start":"02:44.965 ","End":"02:47.930","Text":"Let\u0027s see what each of these things at the side is."},{"Start":"02:47.930 ","End":"02:55.845","Text":"The integral of minus x is equal to minus x squared over 2,"},{"Start":"02:55.845 ","End":"02:59.515","Text":"and this goes from 0-1."},{"Start":"02:59.515 ","End":"03:02.010","Text":"When we put 0, that\u0027s 0 and put 1,"},{"Start":"03:02.010 ","End":"03:05.610","Text":"it\u0027s minus 1/2, so this is minus 1/2."},{"Start":"03:05.610 ","End":"03:10.620","Text":"Here, I get not minus x squared over 2 plus x squared over 2."},{"Start":"03:10.620 ","End":"03:12.150","Text":"From 0 to 1,"},{"Start":"03:12.150 ","End":"03:14.850","Text":"this gives me 1/2 minus 0,"},{"Start":"03:14.850 ","End":"03:17.265","Text":"so this is 1/2."},{"Start":"03:17.265 ","End":"03:19.415","Text":"Now, this integral here,"},{"Start":"03:19.415 ","End":"03:21.170","Text":"if I just put it over here,"},{"Start":"03:21.170 ","End":"03:24.395","Text":"then I get exactly what is written here,"},{"Start":"03:24.395 ","End":"03:27.720","Text":"and so we are done."}],"ID":8614},{"Watched":false,"Name":"Exercise 7","Duration":"3m 8s","ChapterTopicVideoID":8347,"CourseChapterTopicPlaylistID":84701,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8347.jpeg","UploadDate":"2017-01-28T18:25:21.1000000","DurationForVideoObject":"PT3M8S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.900","Text":"Here, we have to show that this definite integral is less than or equal to this constant."},{"Start":"00:06.900 ","End":"00:08.970","Text":"We\u0027re going to use a theorem,"},{"Start":"00:08.970 ","End":"00:10.920","Text":"we\u0027ve used it before once,"},{"Start":"00:10.920 ","End":"00:13.635","Text":"that if we have 2 functions, f and g,"},{"Start":"00:13.635 ","End":"00:18.855","Text":"and f is less than or equal to g on a closed interval from a to b,"},{"Start":"00:18.855 ","End":"00:21.600","Text":"then the same inequality holds for the integrals."},{"Start":"00:21.600 ","End":"00:25.230","Text":"In other words, the integral from a to b of f of x dx is also going"},{"Start":"00:25.230 ","End":"00:29.055","Text":"to be less than or equal to the integral from a to b of g of x dx."},{"Start":"00:29.055 ","End":"00:30.750","Text":"We\u0027re going to use that here,"},{"Start":"00:30.750 ","End":"00:35.280","Text":"and we\u0027re going to use the fact that for any x,"},{"Start":"00:35.280 ","End":"00:42.720","Text":"that the arctangent of x is always less than or equal to pi over 2."},{"Start":"00:42.720 ","End":"00:46.745","Text":"It\u0027s actually also bigger or equal to minus pi over 2,"},{"Start":"00:46.745 ","End":"00:48.110","Text":"but I don\u0027t care about that,"},{"Start":"00:48.110 ","End":"00:50.375","Text":"I just care about this inequality."},{"Start":"00:50.375 ","End":"00:54.410","Text":"Now, x could be anything from minus infinity to infinity."},{"Start":"00:54.410 ","End":"00:57.905","Text":"In particular, it could be this mass here."},{"Start":"00:57.905 ","End":"01:02.210","Text":"What I\u0027m saying is that the arctangent of"},{"Start":"01:02.210 ","End":"01:06.560","Text":"anything is going to be less than or equal to pi over 2."},{"Start":"01:06.560 ","End":"01:10.290","Text":"In particular, the arctangent of sine x over"},{"Start":"01:10.290 ","End":"01:14.525","Text":"x plus 4 is going to be less than or equal to pi over 2."},{"Start":"01:14.525 ","End":"01:19.210","Text":"Now, I can multiply both sides by a positive number."},{"Start":"01:19.210 ","End":"01:21.095","Text":"I can get from here,"},{"Start":"01:21.095 ","End":"01:27.155","Text":"that x squared times arctangent of sine x over x plus 4"},{"Start":"01:27.155 ","End":"01:33.680","Text":"is going to be less than or equal to pi over 2 times x squared."},{"Start":"01:33.680 ","End":"01:35.340","Text":"There\u0027s a slight cheating here,"},{"Start":"01:35.340 ","End":"01:37.759","Text":"x squared is not always positive."},{"Start":"01:37.759 ","End":"01:42.440","Text":"It could be 0, but the inequality also holds if x is 0 because then we just"},{"Start":"01:42.440 ","End":"01:47.690","Text":"get that 0 is less than or equal to 0, which is obvious."},{"Start":"01:47.690 ","End":"01:52.280","Text":"Yeah, it works. x anywhere from 0 to pi, this inequality works."},{"Start":"01:52.280 ","End":"01:56.800","Text":"Then we\u0027ll take 0 and pi as our a and b here."},{"Start":"01:56.800 ","End":"02:07.265","Text":"We get that the integral from 0 to pi of x squared arctangent of sine x"},{"Start":"02:07.265 ","End":"02:12.890","Text":"over x plus 4 dx is less than or equal to the integral"},{"Start":"02:12.890 ","End":"02:19.825","Text":"from 0 to pi of pi over 2x squared dx."},{"Start":"02:19.825 ","End":"02:22.830","Text":"Let\u0027s compute this integral."},{"Start":"02:22.830 ","End":"02:24.170","Text":"This is equal to,"},{"Start":"02:24.170 ","End":"02:28.145","Text":"and I can take pi over 2 outside the brackets,"},{"Start":"02:28.145 ","End":"02:33.710","Text":"and then the integral of x squared is x cubed over 3."},{"Start":"02:33.710 ","End":"02:39.605","Text":"But I have to take this between the limits of 0 and pi and see what we get."},{"Start":"02:39.605 ","End":"02:43.130","Text":"Well, when I substitute 0, I get 0."},{"Start":"02:43.130 ","End":"02:46.104","Text":"Really all I care is to substitute pi,"},{"Start":"02:46.104 ","End":"02:51.525","Text":"so I get pi over 2 times pi cubed over 3."},{"Start":"02:51.525 ","End":"02:55.200","Text":"I\u0027ll put the minus 0 just to show you I did do a subtraction."},{"Start":"02:55.200 ","End":"02:57.915","Text":"pi times pi cubed is pi^4,"},{"Start":"02:57.915 ","End":"02:59.445","Text":"2 times 3 is 6."},{"Start":"02:59.445 ","End":"03:03.225","Text":"This gives us pi^4 over 6."},{"Start":"03:03.225 ","End":"03:05.360","Text":"It really looks a lot like this."},{"Start":"03:05.360 ","End":"03:06.530","Text":"We\u0027ve got it."},{"Start":"03:06.530 ","End":"03:09.270","Text":"I\u0027ll put a checkmark, and we\u0027re done."}],"ID":8615}],"Thumbnail":null,"ID":84701},{"Name":"Riemann Sum and Integrability","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Motivation to Riemann Sum","Duration":"12m 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39s","ChapterTopicVideoID":8359,"CourseChapterTopicPlaylistID":84703,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8359.jpeg","UploadDate":"2019-12-11T21:03:31.5830000","DurationForVideoObject":"PT3M39S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.630","Text":"There is a theorem called the fundamental theorem of the calculus,"},{"Start":"00:03.630 ","End":"00:05.520","Text":"and it has 2 parts."},{"Start":"00:05.520 ","End":"00:09.060","Text":"On this clip, we\u0027ll be talking about the first part."},{"Start":"00:09.060 ","End":"00:11.024","Text":"It goes like this."},{"Start":"00:11.024 ","End":"00:13.275","Text":"We have a function f,"},{"Start":"00:13.275 ","End":"00:17.505","Text":"which is continuous on an interval,"},{"Start":"00:17.505 ","End":"00:21.345","Text":"and we define another function,"},{"Start":"00:21.345 ","End":"00:23.790","Text":"call it I for integral,"},{"Start":"00:23.790 ","End":"00:30.540","Text":"and I of x is the integral from a to x."},{"Start":"00:30.540 ","End":"00:36.115","Text":"The interval contains some point a."},{"Start":"00:36.115 ","End":"00:44.900","Text":"We take the integral from that a to a variable x and get a function of x of f. Now,"},{"Start":"00:44.900 ","End":"00:46.910","Text":"I can\u0027t reuse the letter x,"},{"Start":"00:46.910 ","End":"00:49.535","Text":"so I\u0027ll take letter t dt."},{"Start":"00:49.535 ","End":"00:53.254","Text":"This gives us a function I of x."},{"Start":"00:53.254 ","End":"00:56.915","Text":"Then it turns out that I is differentiable,"},{"Start":"00:56.915 ","End":"01:01.280","Text":"and, in fact, the derivative I prime of"},{"Start":"01:01.280 ","End":"01:08.320","Text":"x is just equal to the original function f of x."},{"Start":"01:08.320 ","End":"01:13.520","Text":"An alternate way of phrasing this without using"},{"Start":"01:13.520 ","End":"01:17.210","Text":"the intermediate function I is just to directly say that"},{"Start":"01:17.210 ","End":"01:21.815","Text":"this integral derivative is f of x."},{"Start":"01:21.815 ","End":"01:25.160","Text":"Yet another way is to use the other notation,"},{"Start":"01:25.160 ","End":"01:28.670","Text":"the Leibniz\u0027s notation for derivative"},{"Start":"01:28.670 ","End":"01:35.120","Text":"and to say d by dx of this is also equal to f of x."},{"Start":"01:35.120 ","End":"01:38.285","Text":"Informally, what it says is, if I take an"},{"Start":"01:38.285 ","End":"01:42.050","Text":"integral of a function and then differentiate it,"},{"Start":"01:42.050 ","End":"01:44.645","Text":"I get back to the original function."},{"Start":"01:44.645 ","End":"01:54.970","Text":"As an example, let\u0027s take the integral from 1 to x of t^4 dt,"},{"Start":"01:54.970 ","End":"01:59.165","Text":"and then the derivative of this."},{"Start":"01:59.165 ","End":"02:01.670","Text":"Now, according to the theorem,"},{"Start":"02:01.670 ","End":"02:05.870","Text":"if I think of f of t as being t^4,"},{"Start":"02:05.870 ","End":"02:09.305","Text":"then what this says is that I take the same function,"},{"Start":"02:09.305 ","End":"02:15.690","Text":"but just with the letter x instead of the letter t, so this should equal x^4."},{"Start":"02:15.690 ","End":"02:18.790","Text":"Of course, in this case, we could check and say"},{"Start":"02:18.790 ","End":"02:24.300","Text":"the integral would be t^5 over 5,"},{"Start":"02:24.300 ","End":"02:30.870","Text":"evaluated from 1 to x, and then derivative,"},{"Start":"02:30.870 ","End":"02:40.830","Text":"this would be x^5 over 5 minus 1^5 over 5 derivative."},{"Start":"02:40.830 ","End":"02:47.870","Text":"This is, indeed, equal to x^4 because I get 5x^4 over 5,"},{"Start":"02:47.870 ","End":"02:49.234","Text":"and the 5 cancels,"},{"Start":"02:49.234 ","End":"02:52.010","Text":"and this is a constant, so it just drops out."},{"Start":"02:52.010 ","End":"02:57.120","Text":"Example number 2, the integral from 4 to"},{"Start":"02:57.120 ","End":"03:03.180","Text":"x of sine t over t dt."},{"Start":"03:03.180 ","End":"03:06.160","Text":"I want the derivative of that."},{"Start":"03:06.160 ","End":"03:07.790","Text":"Again, by the theorem,"},{"Start":"03:07.790 ","End":"03:10.130","Text":"I have here a function of t,"},{"Start":"03:10.130 ","End":"03:14.554","Text":"and so all I do is replace the t with x,"},{"Start":"03:14.554 ","End":"03:23.390","Text":"and so the answer will be sine x over x. I don\u0027t know if this can be checked easily."},{"Start":"03:23.390 ","End":"03:28.940","Text":"But in any event, the theorem assures us that, even if we can\u0027t check like here,"},{"Start":"03:28.940 ","End":"03:31.550","Text":"that this will be the answer."},{"Start":"03:31.550 ","End":"03:34.700","Text":"That\u0027s all I want to say for now about the fundamental theorem"},{"Start":"03:34.700 ","End":"03:38.940","Text":"of calculus, first part. So we\u0027re done."}],"ID":8580},{"Watched":false,"Name":"First Fundamental Theorem I","Duration":"5m 53s","ChapterTopicVideoID":23783,"CourseChapterTopicPlaylistID":84703,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23783.jpeg","UploadDate":"2021-01-10T18:12:56.3400000","DurationForVideoObject":"PT5M53S","Description":null,"VideoComments":[],"Subtitles":[],"ID":24710},{"Watched":false,"Name":"First Fundamental Theorem II","Duration":"10m 26s","ChapterTopicVideoID":23784,"CourseChapterTopicPlaylistID":84703,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23784.jpeg","UploadDate":"2021-01-10T18:17:11.6270000","DurationForVideoObject":"PT10M26S","Description":null,"VideoComments":[],"Subtitles":[],"ID":24711},{"Watched":false,"Name":"Second Fundamental Theorem","Duration":"8m 11s","ChapterTopicVideoID":23785,"CourseChapterTopicPlaylistID":84703,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23785.jpeg","UploadDate":"2021-01-10T18:20:46.1600000","DurationForVideoObject":"PT8M11S","Description":null,"VideoComments":[],"Subtitles":[],"ID":24712},{"Watched":false,"Name":"Exercise 1","Duration":"4m 59s","ChapterTopicVideoID":23786,"CourseChapterTopicPlaylistID":84703,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23786.jpeg","UploadDate":"2021-01-10T18:22:37.7200000","DurationForVideoObject":"PT4M59S","Description":null,"VideoComments":[],"Subtitles":[],"ID":24713},{"Watched":false,"Name":"Exercise 2","Duration":"2m 33s","ChapterTopicVideoID":23787,"CourseChapterTopicPlaylistID":84703,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23787.jpeg","UploadDate":"2021-01-10T18:24:07.1600000","DurationForVideoObject":"PT2M33S","Description":null,"VideoComments":[],"Subtitles":[],"ID":24714},{"Watched":false,"Name":"Exercise 3","Duration":"4m 15s","ChapterTopicVideoID":23788,"CourseChapterTopicPlaylistID":84703,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23788.jpeg","UploadDate":"2021-01-10T18:25:54.5800000","DurationForVideoObject":"PT4M15S","Description":null,"VideoComments":[],"Subtitles":[],"ID":24715},{"Watched":false,"Name":"Exercise 4","Duration":"2m 26s","ChapterTopicVideoID":23789,"CourseChapterTopicPlaylistID":84703,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23789.jpeg","UploadDate":"2021-01-10T18:26:14.6730000","DurationForVideoObject":"PT2M26S","Description":null,"VideoComments":[],"Subtitles":[],"ID":24716},{"Watched":false,"Name":"Exercise 5","Duration":"4m 49s","ChapterTopicVideoID":23790,"CourseChapterTopicPlaylistID":84703,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23790.jpeg","UploadDate":"2021-01-10T18:27:01.6030000","DurationForVideoObject":"PT4M49S","Description":null,"VideoComments":[],"Subtitles":[],"ID":24717},{"Watched":false,"Name":"Exercise 6","Duration":"7m ","ChapterTopicVideoID":23791,"CourseChapterTopicPlaylistID":84703,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23791.jpeg","UploadDate":"2021-01-10T18:28:01.0470000","DurationForVideoObject":"PT7M","Description":null,"VideoComments":[],"Subtitles":[],"ID":24718},{"Watched":false,"Name":"Exercise 7","Duration":"7m 16s","ChapterTopicVideoID":9283,"CourseChapterTopicPlaylistID":84703,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/9283.jpeg","UploadDate":"2017-04-26T17:13:53.9900000","DurationForVideoObject":"PT7M16S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.000","Text":"In this exercise, f is continuous and therefore"},{"Start":"00:06.000 ","End":"00:14.055","Text":"integrable and F is the integral from 0 to x of, f of t dt."},{"Start":"00:14.055 ","End":"00:22.350","Text":"We have to prove that f is odd if and only if F is even and vice versa,"},{"Start":"00:22.350 ","End":"00:24.000","Text":"which is part b."},{"Start":"00:24.000 ","End":"00:27.195","Text":"So let\u0027s start with part a,"},{"Start":"00:27.195 ","End":"00:32.670","Text":"the direction that if f is odd, F is even."},{"Start":"00:32.670 ","End":"00:41.740","Text":"F of minus x is the integral to substituting here from 0 to minus x of,"},{"Start":"00:41.740 ","End":"00:44.560","Text":"f of t dt and then make a substitution."},{"Start":"00:44.560 ","End":"00:49.735","Text":"Let t be minus s so that dt is minus ds,"},{"Start":"00:49.735 ","End":"00:52.115","Text":"and then we have an integral."},{"Start":"00:52.115 ","End":"00:53.580","Text":"Instead of t,"},{"Start":"00:53.580 ","End":"00:54.780","Text":"we have s,"},{"Start":"00:54.780 ","End":"01:03.800","Text":"this time from 0 to x and dt is minus ds and t is minus s. We have this,"},{"Start":"01:03.800 ","End":"01:10.670","Text":"but we know that f is odd so f of"},{"Start":"01:10.670 ","End":"01:14.990","Text":"minus s is minus f of s and"},{"Start":"01:14.990 ","End":"01:20.030","Text":"the minus with the minus cancel each other out so we\u0027re left with this."},{"Start":"01:20.030 ","End":"01:24.980","Text":"This is exactly the definition of F of x."},{"Start":"01:24.980 ","End":"01:27.770","Text":"Looking at this and this,"},{"Start":"01:27.770 ","End":"01:30.455","Text":"we see that F is even."},{"Start":"01:30.455 ","End":"01:38.300","Text":"Now let\u0027s do it in the other direction to show that if F is even then f is odd."},{"Start":"01:38.300 ","End":"01:42.110","Text":"Now we\u0027re going to use the fundamental theorem of the calculus,"},{"Start":"01:42.110 ","End":"01:52.820","Text":"F prime is f. So f is a derivative of F. F of minus x is the derivative of F"},{"Start":"01:52.820 ","End":"02:00.560","Text":"minus x. I can put an extra minus in here and here and why I want to do"},{"Start":"02:00.560 ","End":"02:03.830","Text":"that is because I want to say that this is"},{"Start":"02:03.830 ","End":"02:08.805","Text":"a derivative of a composite function and using the chain rule."},{"Start":"02:08.805 ","End":"02:12.280","Text":"If I take F of minus x and differentiate it,"},{"Start":"02:12.280 ","End":"02:14.480","Text":"I first differentiate the F and get"},{"Start":"02:14.480 ","End":"02:18.200","Text":"F prime but then the inner derivative of minus x gives"},{"Start":"02:18.200 ","End":"02:24.210","Text":"me an extra minus out here so this part is this."},{"Start":"02:24.210 ","End":"02:26.445","Text":"By the evenness of F,"},{"Start":"02:26.445 ","End":"02:31.805","Text":"this is equal to the derivative of F of x same thing."},{"Start":"02:31.805 ","End":"02:39.675","Text":"This is just F prime of x with a minus and finally,"},{"Start":"02:39.675 ","End":"02:43.530","Text":"we know that F prime is little"},{"Start":"02:43.530 ","End":"02:48.440","Text":"f so we have this and now if we trace the beginning and the end,"},{"Start":"02:48.440 ","End":"02:52.255","Text":"we see that f is odd."},{"Start":"02:52.255 ","End":"02:55.440","Text":"That concludes part a."},{"Start":"02:55.440 ","End":"02:56.990","Text":"Now part b,"},{"Start":"02:56.990 ","End":"02:59.450","Text":"which is very similar to part a,"},{"Start":"02:59.450 ","End":"03:04.195","Text":"f is even if and only if F is odd."},{"Start":"03:04.195 ","End":"03:07.980","Text":"Let\u0027s start with 1 direction."},{"Start":"03:07.980 ","End":"03:15.375","Text":"First of all we\u0027ll go from f being even and then we\u0027ll prove from that that F is odd."},{"Start":"03:15.375 ","End":"03:21.660","Text":"F of minus x eventually we want to get to minus f of x,"},{"Start":"03:21.660 ","End":"03:30.515","Text":"it\u0027s the integral from 0 to minus x of f. Substitute just like we did here,"},{"Start":"03:30.515 ","End":"03:37.320","Text":"t equals minus s. We get this expression then because of the evenness of"},{"Start":"03:37.320 ","End":"03:40.820","Text":"f we can get rid of this minus and this minus we can pull in"},{"Start":"03:40.820 ","End":"03:44.900","Text":"front and this is exactly minus F of x."},{"Start":"03:44.900 ","End":"03:48.950","Text":"This proves that F is odd."},{"Start":"03:48.950 ","End":"03:50.930","Text":"Now in the other direction,"},{"Start":"03:50.930 ","End":"03:53.750","Text":"we start off knowing that F is odd,"},{"Start":"03:53.750 ","End":"03:55.310","Text":"we want to show that f is even."},{"Start":"03:55.310 ","End":"03:57.395","Text":"So f of minus x,"},{"Start":"03:57.395 ","End":"03:59.975","Text":"by the fundamental theorem of the calculus,"},{"Start":"03:59.975 ","End":"04:06.340","Text":"f is a derivative of F. Put a minus here and a minus here and the reason I"},{"Start":"04:06.340 ","End":"04:13.235","Text":"want to do that is because this now is the derivative of a composite function."},{"Start":"04:13.235 ","End":"04:17.380","Text":"If I think of F of minus x as a function,"},{"Start":"04:17.380 ","End":"04:19.750","Text":"x goes to F of minus x,"},{"Start":"04:19.750 ","End":"04:21.505","Text":"and I differentiate that,"},{"Start":"04:21.505 ","End":"04:27.339","Text":"then derivative of this would be F prime of minus x times the inner derivative,"},{"Start":"04:27.339 ","End":"04:29.570","Text":"which is this minus."},{"Start":"04:29.570 ","End":"04:33.700","Text":"Then because of the oddness of F,"},{"Start":"04:33.700 ","End":"04:36.415","Text":"we can pull this minus in front."},{"Start":"04:36.415 ","End":"04:41.410","Text":"This minus becomes this minus and then the 2 minuses will cancel each other"},{"Start":"04:41.410 ","End":"04:47.215","Text":"out and we\u0027ll just get F prime of x,"},{"Start":"04:47.215 ","End":"04:55.924","Text":"and F prime is f. We\u0027ve shown from here and here that f is even."},{"Start":"04:55.924 ","End":"05:00.360","Text":"That concludes part b and the exercise."}],"ID":9593},{"Watched":false,"Name":"Exercise 8","Duration":"5m 32s","ChapterTopicVideoID":9284,"CourseChapterTopicPlaylistID":84703,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/9284.jpeg","UploadDate":"2017-04-26T17:15:20.1070000","DurationForVideoObject":"PT5M32S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.070","Text":"In this exercise, little f is a continuous function and big F is the"},{"Start":"00:05.070 ","End":"00:09.795","Text":"integral from 0 to x of little f. 2 parts."},{"Start":"00:09.795 ","End":"00:13.470","Text":"In part a, we have to prove that if big F is periodic,"},{"Start":"00:13.470 ","End":"00:17.909","Text":"then so is little f, and in part b,"},{"Start":"00:17.909 ","End":"00:24.030","Text":"we have to find an example where little f is periodic, but big F isn\u0027t,"},{"Start":"00:24.030 ","End":"00:27.390","Text":"meaning that it doesn\u0027t work the other way around."},{"Start":"00:27.390 ","End":"00:31.799","Text":"We\u0027ll start with a. Since big f is periodic,"},{"Start":"00:31.799 ","End":"00:36.510","Text":"it has a period p for some p, which is not 0,"},{"Start":"00:36.510 ","End":"00:42.250","Text":"meaning that f of x plus p equals f of x for all x,"},{"Start":"00:42.250 ","End":"00:43.535","Text":"or if you like,"},{"Start":"00:43.535 ","End":"00:50.035","Text":"the big F of x is identical to big F of x plus p as functions,"},{"Start":"00:50.035 ","End":"00:53.495","Text":"and that means we can differentiate each side."},{"Start":"00:53.495 ","End":"00:56.210","Text":"This as a composite function of x,"},{"Start":"00:56.210 ","End":"00:58.880","Text":"we can differentiate it using the chain rule,"},{"Start":"00:58.880 ","End":"01:00.530","Text":"first as the outer derivative,"},{"Start":"01:00.530 ","End":"01:01.790","Text":"which is big F prime,"},{"Start":"01:01.790 ","End":"01:03.170","Text":"and the inner derivative,"},{"Start":"01:03.170 ","End":"01:05.780","Text":"which is x plus p prime,"},{"Start":"01:05.780 ","End":"01:07.970","Text":"which is just 1."},{"Start":"01:07.970 ","End":"01:12.980","Text":"We get that big F prime of x is big F prime of"},{"Start":"01:12.980 ","End":"01:18.245","Text":"x plus p, by the first fundamental theorem of the calculus,"},{"Start":"01:18.245 ","End":"01:21.180","Text":"big F prime is little f,"},{"Start":"01:21.180 ","End":"01:24.725","Text":"so we get this for any x,"},{"Start":"01:24.725 ","End":"01:30.985","Text":"which means that little f is periodic with the same period p,"},{"Start":"01:30.985 ","End":"01:32.810","Text":"so that proves more than a."},{"Start":"01:32.810 ","End":"01:34.579","Text":"Not only is this periodic,"},{"Start":"01:34.579 ","End":"01:36.800","Text":"but it has the same period."},{"Start":"01:36.800 ","End":"01:38.660","Text":"Now onto part b,"},{"Start":"01:38.660 ","End":"01:39.950","Text":"we needed an example,"},{"Start":"01:39.950 ","End":"01:41.570","Text":"and here\u0027s the example."},{"Start":"01:41.570 ","End":"01:43.610","Text":"Could be lots of others."},{"Start":"01:43.610 ","End":"01:48.305","Text":"Take big F of x to be 2x and this is not periodic."},{"Start":"01:48.305 ","End":"01:52.235","Text":"I mean, I\u0027ll show you that it doesn\u0027t have any period p,"},{"Start":"01:52.235 ","End":"01:53.750","Text":"a period has to be non-zero."},{"Start":"01:53.750 ","End":"01:55.430","Text":"Suppose that has period p,"},{"Start":"01:55.430 ","End":"02:00.920","Text":"f of 0 equals 0 and this has to equal f of 0"},{"Start":"02:00.920 ","End":"02:07.790","Text":"plus p, but f of p is 2p and that\u0027s not 0,"},{"Start":"02:07.790 ","End":"02:10.940","Text":"showing that can\u0027t be periodic."},{"Start":"02:10.940 ","End":"02:14.525","Text":"Now, little f is"},{"Start":"02:14.525 ","End":"02:19.100","Text":"big F prime, just as above in the first fundamental theorem of the calculus,"},{"Start":"02:19.100 ","End":"02:21.685","Text":"and this is the constant function 2."},{"Start":"02:21.685 ","End":"02:24.345","Text":"Constant function is periodic."},{"Start":"02:24.345 ","End":"02:27.405","Text":"In fact, you can take any period p you want."},{"Start":"02:27.405 ","End":"02:30.515","Text":"This is an example where big F isn\u0027t,"},{"Start":"02:30.515 ","End":"02:32.105","Text":"but little f is,"},{"Start":"02:32.105 ","End":"02:34.440","Text":"and we are done."}],"ID":9594},{"Watched":false,"Name":"Exercise 9 part 1","Duration":"54s","ChapterTopicVideoID":9285,"CourseChapterTopicPlaylistID":84703,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/9285.jpeg","UploadDate":"2017-04-26T17:15:32.3700000","DurationForVideoObject":"PT54S","Description":null,"VideoComments":[],"Subtitles":[],"ID":9595},{"Watched":false,"Name":"Exercise 9 part 2","Duration":"2m 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8s","ChapterTopicVideoID":8361,"CourseChapterTopicPlaylistID":84703,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/8361.jpeg","UploadDate":"2017-01-28T18:42:28.1570000","DurationForVideoObject":"PT16M8S","Description":null,"VideoComments":[],"Subtitles":[],"ID":8590}],"Thumbnail":null,"ID":84703},{"Name":"Riemann Integration and Integrability","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Introcution, Partition and Upper,Lower Riemann sums","Duration":"11m 15s","ChapterTopicVideoID":23755,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23755.jpeg","UploadDate":"2021-01-08T10:34:16.0670000","DurationForVideoObject":"PT11M15S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.020","Text":"Starting a new topic, Riemann integration,"},{"Start":"00:04.020 ","End":"00:08.475","Text":"including Riemann integrals and integrability."},{"Start":"00:08.475 ","End":"00:11.475","Text":"This will take more than 1 clip."},{"Start":"00:11.475 ","End":"00:15.765","Text":"Most of you already know what integration is and what integrals are."},{"Start":"00:15.765 ","End":"00:17.910","Text":"Riemann was just one of many,"},{"Start":"00:17.910 ","End":"00:21.390","Text":"though he was one of the main ones who developed the concept of"},{"Start":"00:21.390 ","End":"00:27.435","Text":"integration and other names here, in particular Lebesgue."},{"Start":"00:27.435 ","End":"00:29.320","Text":"To simplify the concept,"},{"Start":"00:29.320 ","End":"00:35.450","Text":"integration is a method of finding the area under a graph of a function."},{"Start":"00:35.450 ","End":"00:41.930","Text":"We say that the integral of f from a to b is basically what\u0027s illustrated in the picture."},{"Start":"00:41.930 ","End":"00:43.790","Text":"We have a graph of a function f,"},{"Start":"00:43.790 ","End":"00:47.075","Text":"and we restrict the domain between"},{"Start":"00:47.075 ","End":"00:55.020","Text":"a and b and the area under the graph here that\u0027s shaded is the integral of f from a to b,"},{"Start":"00:55.020 ","End":"01:01.760","Text":"and the notation is as follows: the integral from a to b of f of x dx."},{"Start":"01:01.760 ","End":"01:04.145","Text":"Now, there was a short way of writing this."},{"Start":"01:04.145 ","End":"01:09.860","Text":"If we don\u0027t care about the name of the independent variable x,"},{"Start":"01:09.860 ","End":"01:15.725","Text":"then we can omit part of it and just write it as the integral from a to b of"},{"Start":"01:15.725 ","End":"01:19.460","Text":"f. This integral is a definite"},{"Start":"01:19.460 ","End":"01:24.365","Text":"integral as opposed to another kind called an indefinite integral,"},{"Start":"01:24.365 ","End":"01:27.800","Text":"which as you probably know, is a kind of antiderivative."},{"Start":"01:27.800 ","End":"01:32.600","Text":"Here we\u0027ll focus on the definite integral according to Riemann,"},{"Start":"01:32.600 ","End":"01:34.010","Text":"and like I say,"},{"Start":"01:34.010 ","End":"01:37.280","Text":"it\u0027s geometrically the area under a graph."},{"Start":"01:37.280 ","End":"01:43.745","Text":"The concept area under the graph is misleading in certain problems with that."},{"Start":"01:43.745 ","End":"01:49.805","Text":"For one thing, it assumes that the graph is above the x-axis."},{"Start":"01:49.805 ","End":"01:53.345","Text":"Really, it\u0027s more like the area"},{"Start":"01:53.345 ","End":"01:57.050","Text":"between the graph and the x-axis clipped by the vertical lines,"},{"Start":"01:57.050 ","End":"01:59.030","Text":"x equals a and x equals b."},{"Start":"01:59.030 ","End":"02:01.670","Text":"We\u0027ve got 4 parts of the boundary,"},{"Start":"02:01.670 ","End":"02:06.990","Text":"the function, 2 vertical lines, and the axis."},{"Start":"02:06.990 ","End":"02:09.305","Text":"When we say under the graph,"},{"Start":"02:09.305 ","End":"02:12.680","Text":"then usually it means that the graph is above"},{"Start":"02:12.680 ","End":"02:16.250","Text":"the axis so we use this term for non-negative functions."},{"Start":"02:16.250 ","End":"02:20.180","Text":"If we have function that\u0027s non-positive or negative,"},{"Start":"02:20.180 ","End":"02:23.914","Text":"then the integral is not the area,"},{"Start":"02:23.914 ","End":"02:26.650","Text":"but minus the area."},{"Start":"02:26.650 ","End":"02:31.354","Text":"If we wanted the integral from here to here,"},{"Start":"02:31.354 ","End":"02:34.760","Text":"then it would be minus this area."},{"Start":"02:34.760 ","End":"02:39.950","Text":"Of course, a function can have both positive and negative parts."},{"Start":"02:39.950 ","End":"02:47.385","Text":"In that case, what we do is we take the integral as the positive bits,"},{"Start":"02:47.385 ","End":"02:52.610","Text":"take away the negative bits that are below the axis,"},{"Start":"02:52.610 ","End":"02:55.480","Text":"the area above minus the area below,"},{"Start":"02:55.480 ","End":"02:58.100","Text":"and even that is problematic,"},{"Start":"02:58.100 ","End":"03:01.070","Text":"at least this geometrical description is difficult to"},{"Start":"03:01.070 ","End":"03:05.060","Text":"apply because a function can behave pretty wildly."},{"Start":"03:05.060 ","End":"03:09.695","Text":"First of all, it could be non-continuous and could also oscillate."},{"Start":"03:09.695 ","End":"03:12.680","Text":"There are various examples I could give you of crazy functions,"},{"Start":"03:12.680 ","End":"03:14.915","Text":"but this is just the general idea."},{"Start":"03:14.915 ","End":"03:16.415","Text":"This is how it started,"},{"Start":"03:16.415 ","End":"03:18.520","Text":"the concept of a definite integral,"},{"Start":"03:18.520 ","End":"03:21.905","Text":"trying to find the area under a curve."},{"Start":"03:21.905 ","End":"03:25.040","Text":"Now that we\u0027ve got that brief introduction out of the way,"},{"Start":"03:25.040 ","End":"03:28.460","Text":"we have to start getting technical and want to work"},{"Start":"03:28.460 ","End":"03:32.270","Text":"towards a formal definition of a Riemann integral."},{"Start":"03:32.270 ","End":"03:35.360","Text":"But we have to start with some intermediate concepts."},{"Start":"03:35.360 ","End":"03:37.880","Text":"First, the concept of a partition of"},{"Start":"03:37.880 ","End":"03:42.455","Text":"an interval and the concept of upper and lower Riemann sums."},{"Start":"03:42.455 ","End":"03:44.990","Text":"We\u0027ll start with partitions."},{"Start":"03:44.990 ","End":"03:47.050","Text":"We have our interval a, b."},{"Start":"03:47.050 ","End":"03:51.300","Text":"A partition, informally just means taking a bunch of"},{"Start":"03:51.300 ","End":"03:58.010","Text":"points on that interval and dividing it up into sub-intervals."},{"Start":"03:58.010 ","End":"04:01.374","Text":"It\u0027s a set of points P,"},{"Start":"04:01.374 ","End":"04:05.155","Text":"let\u0027s say it\u0027s x naught up to x_n."},{"Start":"04:05.155 ","End":"04:07.520","Text":"The first one has to be a,"},{"Start":"04:07.520 ","End":"04:10.130","Text":"the last one has to be b,"},{"Start":"04:10.130 ","End":"04:13.315","Text":"and they have to be in increasing order."},{"Start":"04:13.315 ","End":"04:18.115","Text":"There\u0027s an asterisk here that says that,"},{"Start":"04:18.115 ","End":"04:20.180","Text":"I used the term set,"},{"Start":"04:20.180 ","End":"04:25.220","Text":"but really it\u0027s a sequence because it\u0027s ordered."},{"Start":"04:25.220 ","End":"04:26.960","Text":"Although often we look at it as a set,"},{"Start":"04:26.960 ","End":"04:29.570","Text":"you might want to take the union of such things."},{"Start":"04:29.570 ","End":"04:31.400","Text":"It\u0027s a set with order,"},{"Start":"04:31.400 ","End":"04:32.795","Text":"it\u0027s like a sequence."},{"Start":"04:32.795 ","End":"04:37.085","Text":"The other asterisk is something nonstandard."},{"Start":"04:37.085 ","End":"04:44.705","Text":"In this course, we\u0027ll just assume that each x is less than or equal to the following x."},{"Start":"04:44.705 ","End":"04:49.475","Text":"But some books or some courses online or whatever,"},{"Start":"04:49.475 ","End":"04:52.790","Text":"insist on strict inequality so"},{"Start":"04:52.790 ","End":"04:57.205","Text":"that each one has to be strictly bigger than the previous one."},{"Start":"04:57.205 ","End":"04:59.600","Text":"We\u0027ll manage both ways."},{"Start":"04:59.600 ","End":"05:02.240","Text":"If we need strict inequality,"},{"Start":"05:02.240 ","End":"05:05.765","Text":"all we have to do is throw out duplicates."},{"Start":"05:05.765 ","End":"05:07.175","Text":"Whenever we see an equal,"},{"Start":"05:07.175 ","End":"05:08.785","Text":"we throw one of them out,"},{"Start":"05:08.785 ","End":"05:14.210","Text":"and it\u0027s practically the same partition but without duplicates."},{"Start":"05:14.210 ","End":"05:20.045","Text":"Some notation only go from one point to the next,"},{"Start":"05:20.045 ","End":"05:23.795","Text":"each sub-interval from x_i minus 1 to x_i,"},{"Start":"05:23.795 ","End":"05:25.070","Text":"the width of that interval,"},{"Start":"05:25.070 ","End":"05:27.470","Text":"which is x_i minus x_i minus 1,"},{"Start":"05:27.470 ","End":"05:30.335","Text":"we\u0027ll call that Delta x_i."},{"Start":"05:30.335 ","End":"05:33.530","Text":"Notice that if we add all these jumps,"},{"Start":"05:33.530 ","End":"05:39.210","Text":"these increases from 0-1 plus from 1-2,"},{"Start":"05:39.210 ","End":"05:40.860","Text":"and so on up to n,"},{"Start":"05:40.860 ","End":"05:42.500","Text":"it\u0027s just b minus a,"},{"Start":"05:42.500 ","End":"05:44.359","Text":"the width of the whole interval,"},{"Start":"05:44.359 ","End":"05:46.820","Text":"just breaking it up into sub-intervals."},{"Start":"05:46.820 ","End":"05:49.220","Text":"Now, some more notation."},{"Start":"05:49.220 ","End":"05:54.335","Text":"The normal mesh of a partition which is |P|,"},{"Start":"05:54.335 ","End":"05:59.975","Text":"is defined as the maximum of all these Delta x\u0027s."},{"Start":"05:59.975 ","End":"06:03.125","Text":"The greatest one, it could be a tie but maximum,"},{"Start":"06:03.125 ","End":"06:06.185","Text":"and we\u0027ll see what this is used for lighter."},{"Start":"06:06.185 ","End":"06:08.734","Text":"Now, let\u0027s bring in a function."},{"Start":"06:08.734 ","End":"06:13.010","Text":"This is the function that eventually will define the integral of,"},{"Start":"06:13.010 ","End":"06:16.160","Text":"and we need to assume bounded."},{"Start":"06:16.160 ","End":"06:21.290","Text":"All our functions in Riemann integrals are bounded functions."},{"Start":"06:21.290 ","End":"06:23.270","Text":"If I occasionally forget to say it,"},{"Start":"06:23.270 ","End":"06:25.505","Text":"just assume everything is bounded."},{"Start":"06:25.505 ","End":"06:30.370","Text":"Bounded means that it\u0027s sandwiched between 2 constants,"},{"Start":"06:30.370 ","End":"06:36.470","Text":"m and M, for every x on interval a, b."},{"Start":"06:36.470 ","End":"06:40.160","Text":"Sometimes it\u0027s written symmetrically as the absolute value of"},{"Start":"06:40.160 ","End":"06:44.165","Text":"f is less than or equal to a single positive constant."},{"Start":"06:44.165 ","End":"06:46.145","Text":"Either way will do."},{"Start":"06:46.145 ","End":"06:54.705","Text":"Back to our partition from x naught to x_n for each i from 1 to n,"},{"Start":"06:54.705 ","End":"06:56.100","Text":"not from 0 to n,"},{"Start":"06:56.100 ","End":"07:00.290","Text":"because each interval is from a point to its predecessor,"},{"Start":"07:00.290 ","End":"07:04.205","Text":"we\u0027ll let interval i be x_i minus 1,"},{"Start":"07:04.205 ","End":"07:07.385","Text":"x_i, and for each of these intervals,"},{"Start":"07:07.385 ","End":"07:13.105","Text":"we\u0027ll take the supremum of f on it and the infimum of f on it."},{"Start":"07:13.105 ","End":"07:17.940","Text":"This one we\u0027ll call M_i and this one m_i."},{"Start":"07:17.940 ","End":"07:21.620","Text":"Because the function is bounded, it\u0027s bounded on a,"},{"Start":"07:21.620 ","End":"07:24.260","Text":"b and therefore, bounded on a subinterval,"},{"Start":"07:24.260 ","End":"07:27.010","Text":"it will have a supremum and infimum."},{"Start":"07:27.010 ","End":"07:29.150","Text":"It may not have a maximum and a minimum,"},{"Start":"07:29.150 ","End":"07:31.840","Text":"but it will have a sup and inf."},{"Start":"07:31.840 ","End":"07:38.810","Text":"Here I just repeated what I said that M_i and m_i exist because of the boundedness,"},{"Start":"07:38.810 ","End":"07:43.115","Text":"and also note that the n is obviously going to be less than or equal to the sup,"},{"Start":"07:43.115 ","End":"07:46.460","Text":"so m_i is less than or equal to M_i."},{"Start":"07:46.460 ","End":"07:48.335","Text":"If we\u0027re being pedantic,"},{"Start":"07:48.335 ","End":"07:56.195","Text":"we would write this in full as M_i depending on the partition and the function,"},{"Start":"07:56.195 ","End":"07:57.980","Text":"but usually in any given instance,"},{"Start":"07:57.980 ","End":"08:00.590","Text":"there is only 1 partition and function in question,"},{"Start":"08:00.590 ","End":"08:03.815","Text":"so we can drop that extra information."},{"Start":"08:03.815 ","End":"08:09.050","Text":"We\u0027re getting very close to defining upper and lower Riemann sums."},{"Start":"08:09.050 ","End":"08:12.950","Text":"The upper one, U for upper,"},{"Start":"08:12.950 ","End":"08:18.240","Text":"is the sum of these products:"},{"Start":"08:18.240 ","End":"08:28.200","Text":"the width or length of the interval times the M_i or the m_i,"},{"Start":"08:28.200 ","End":"08:29.755","Text":"and then summed up."},{"Start":"08:29.755 ","End":"08:32.765","Text":"I think at this point we really need a picture."},{"Start":"08:32.765 ","End":"08:36.060","Text":"Let\u0027s get to that part already."},{"Start":"08:40.630 ","End":"08:45.605","Text":"In general, this is like a and b and we want to find the area."},{"Start":"08:45.605 ","End":"08:51.590","Text":"What we do, is we approximate it below and above by a lower sum and upper sum."},{"Start":"08:51.590 ","End":"08:53.060","Text":"We take the interval a,"},{"Start":"08:53.060 ","End":"08:57.635","Text":"b and divide it up into n sub-intervals,"},{"Start":"08:57.635 ","End":"09:00.080","Text":"not necessarily of equal length,"},{"Start":"09:00.080 ","End":"09:02.815","Text":"even though the picture implies it."},{"Start":"09:02.815 ","End":"09:06.755","Text":"Not really, this one is clearly thicker than this one."},{"Start":"09:06.755 ","End":"09:11.320","Text":"Anyway, once we have this partition of the interval a,"},{"Start":"09:11.320 ","End":"09:15.265","Text":"b, we build rectangles on each."},{"Start":"09:15.265 ","End":"09:17.080","Text":"For the lower sum,"},{"Start":"09:17.080 ","End":"09:19.750","Text":"the rectangle is the inf of the function,"},{"Start":"09:19.750 ","End":"09:25.105","Text":"is the smallest value of the function in this interval, smallest meaning inf."},{"Start":"09:25.105 ","End":"09:27.430","Text":"In the upper sum,"},{"Start":"09:27.430 ","End":"09:31.450","Text":"we take the supremum here,"},{"Start":"09:31.450 ","End":"09:32.680","Text":"here, here, and each one of them,"},{"Start":"09:32.680 ","End":"09:34.670","Text":"we take the supremum."},{"Start":"09:34.670 ","End":"09:44.500","Text":"Then the lower sum is the product of this Delta x times this m_i summed up."},{"Start":"09:44.500 ","End":"09:52.350","Text":"Similarly here, each Delta x times the corresponding M. Back here,"},{"Start":"09:52.350 ","End":"09:56.690","Text":"these are the upper and lower sums."},{"Start":"09:56.690 ","End":"09:58.430","Text":"I used the names before I wrote it."},{"Start":"09:58.430 ","End":"10:03.650","Text":"We\u0027re here, these are the upper and lower Riemann sums for the partition P,"},{"Start":"10:03.650 ","End":"10:07.140","Text":"and for the function f, of course."},{"Start":"10:08.390 ","End":"10:15.580","Text":"Notice that because the function f is bounded below by m and above by M,"},{"Start":"10:15.580 ","End":"10:19.390","Text":"that, on each interval,"},{"Start":"10:28.160 ","End":"10:38.150","Text":"the infimum on the interval is bigger or equal to the global inf from the whole a,"},{"Start":"10:38.150 ","End":"10:46.765","Text":"b, and similarly, the sup on interval i is less than or equal to sup on all of a, b."},{"Start":"10:46.765 ","End":"10:50.240","Text":"We also had another inequality here."},{"Start":"10:50.240 ","End":"10:59.280","Text":"We can put these together and get 4 things in a row: the lower bound,"},{"Start":"10:59.280 ","End":"11:02.030","Text":"the inf on each interval,"},{"Start":"11:02.030 ","End":"11:05.150","Text":"the sup on each interval and the upper bound,"},{"Start":"11:05.150 ","End":"11:14.820","Text":"and this is the order for all i from 1 to n. I think it\u0027s time for a break now."}],"ID":24682},{"Watched":false,"Name":"Examples - Upper,Lower Riemann sums","Duration":"10m 1s","ChapterTopicVideoID":23756,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23756.jpeg","UploadDate":"2021-01-08T10:36:01.0530000","DurationForVideoObject":"PT10M1S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.335","Text":"Time now for an example on upper and lower Riemann sums."},{"Start":"00:04.335 ","End":"00:07.620","Text":"Let\u0027s take the function f of x equals x squared,"},{"Start":"00:07.620 ","End":"00:11.505","Text":"and we\u0027ll restrict it to the interval from 0-2,"},{"Start":"00:11.505 ","End":"00:16.709","Text":"and note that f is increasing on this interval,"},{"Start":"00:16.709 ","End":"00:20.655","Text":"and we\u0027ll take a partition in general."},{"Start":"00:20.655 ","End":"00:23.520","Text":"We\u0027ll divide it up into n equal pieces."},{"Start":"00:23.520 ","End":"00:26.040","Text":"A partition doesn\u0027t have to be equal pieces,"},{"Start":"00:26.040 ","End":"00:28.605","Text":"but in our example, it will be."},{"Start":"00:28.605 ","End":"00:31.955","Text":"The length of each partition will be 2,"},{"Start":"00:31.955 ","End":"00:36.290","Text":"which is this length divided by n and each"},{"Start":"00:36.290 ","End":"00:40.790","Text":"of the pieces will start from 0 and add 2 over n each time."},{"Start":"00:40.790 ","End":"00:47.490","Text":"The ith partition will be 2i over n and the ith interval be this."},{"Start":"00:47.490 ","End":"00:51.095","Text":"Now, let\u0027s compute the upper and lower sums for"},{"Start":"00:51.095 ","End":"00:55.220","Text":"this particular partition, and for 4 specific examples,"},{"Start":"00:55.220 ","End":"00:57.160","Text":"n equals 6, 8, 10, and 20."},{"Start":"00:57.160 ","End":"00:59.615","Text":"We want an actual numerical answers."},{"Start":"00:59.615 ","End":"01:03.440","Text":"Start with n equals 6 so 2 over 6 is 1/3,"},{"Start":"01:03.440 ","End":"01:07.720","Text":"so we go from 0-2 in jumps of 1/3,"},{"Start":"01:07.720 ","End":"01:10.890","Text":"and each pair is an interval."},{"Start":"01:10.890 ","End":"01:14.705","Text":"Because the function is increasing,"},{"Start":"01:14.705 ","End":"01:17.900","Text":"the infimum or minimum will be on"},{"Start":"01:17.900 ","End":"01:22.100","Text":"the left side of the interval and the maximum on the right."},{"Start":"01:22.100 ","End":"01:28.565","Text":"This is what the upper sum will look like graphically and here\u0027s the lower sum."},{"Start":"01:28.565 ","End":"01:32.205","Text":"On the right, we have x_i,"},{"Start":"01:32.205 ","End":"01:34.860","Text":"and on the left x_i minus 1,"},{"Start":"01:34.860 ","End":"01:37.260","Text":"and so this is x_i squared,"},{"Start":"01:37.260 ","End":"01:43.560","Text":"so it\u0027s i squared over 9 and here i minus 1 squared over 9."},{"Start":"01:43.560 ","End":"01:45.480","Text":"That\u0027s the height for each of these."},{"Start":"01:45.480 ","End":"01:47.625","Text":"Where i goes from 1-6,"},{"Start":"01:47.625 ","End":"01:51.230","Text":"and here\u0027s a formula we\u0027ll need in a moment."},{"Start":"01:51.230 ","End":"01:53.360","Text":"You can get by without it, but useful to have."},{"Start":"01:53.360 ","End":"01:59.060","Text":"The sum of 1 squared plus 2 squared up to n squared is given by this formula."},{"Start":"01:59.060 ","End":"02:03.020","Text":"The upper sum, it\u0027s the sum of M_i Delta x_i,"},{"Start":"02:03.020 ","End":"02:05.360","Text":"Delta x_i or 1/3,"},{"Start":"02:05.360 ","End":"02:08.554","Text":"that\u0027s from here for each of the i\u0027s."},{"Start":"02:08.554 ","End":"02:13.780","Text":"We can bring that out in front and M_i then will be i squared over 9,"},{"Start":"02:13.780 ","End":"02:20.220","Text":"and so using the formula where i goes from 1-6 or n"},{"Start":"02:20.220 ","End":"02:26.780","Text":"equals 6 and we get 6 times 7 times 13 over 6."},{"Start":"02:26.780 ","End":"02:31.850","Text":"Anyway, it comes out this numerator to be 91."},{"Start":"02:31.850 ","End":"02:37.420","Text":"We\u0027ve got 91 over 27 and numerically to 4 decimal places, is this."},{"Start":"02:37.420 ","End":"02:39.825","Text":"For the lowest sum, sort of i,"},{"Start":"02:39.825 ","End":"02:42.065","Text":"we have i minus 1 squared."},{"Start":"02:42.065 ","End":"02:44.405","Text":"If we do a substitution instead of i,"},{"Start":"02:44.405 ","End":"02:45.800","Text":"we put i minus 1."},{"Start":"02:45.800 ","End":"02:48.460","Text":"It goes from 0-5."},{"Start":"02:48.460 ","End":"02:50.640","Text":"When i is 0, it\u0027s nothing."},{"Start":"02:50.640 ","End":"02:53.615","Text":"It\u0027s just like this formula with n equals 5,"},{"Start":"02:53.615 ","End":"02:56.615","Text":"and that comes out to be 55."},{"Start":"02:56.615 ","End":"03:00.095","Text":"So 55 over 27 is this,"},{"Start":"03:00.095 ","End":"03:02.735","Text":"and that\u0027s for n equals 6."},{"Start":"03:02.735 ","End":"03:06.230","Text":"By the way, it means that the value of the integral,"},{"Start":"03:06.230 ","End":"03:09.500","Text":"which is the area under the curve is somewhere between this and this."},{"Start":"03:09.500 ","End":"03:11.555","Text":"But these 2 are still far apart."},{"Start":"03:11.555 ","End":"03:13.340","Text":"You\u0027ll see that as we increase n,"},{"Start":"03:13.340 ","End":"03:15.815","Text":"these 2 will get closer to each other."},{"Start":"03:15.815 ","End":"03:17.840","Text":"I\u0027ll just mention that by the way,"},{"Start":"03:17.840 ","End":"03:23.960","Text":"that we didn\u0027t really need this formula for n equals 5 or 6."},{"Start":"03:23.960 ","End":"03:25.760","Text":"We could just do it manually,"},{"Start":"03:25.760 ","End":"03:27.050","Text":"but when n is larger,"},{"Start":"03:27.050 ","End":"03:28.820","Text":"we will be using it."},{"Start":"03:28.820 ","End":"03:31.750","Text":"Next, let\u0027s move on to n equals 8."},{"Start":"03:31.750 ","End":"03:35.240","Text":"When n equals 8, from 0-2,"},{"Start":"03:35.240 ","End":"03:37.040","Text":"we divide it up into 8 pieces,"},{"Start":"03:37.040 ","End":"03:39.035","Text":"so each one of them is 1/4."},{"Start":"03:39.035 ","End":"03:42.365","Text":"We go from 0-2 in jumps of a 1/4."},{"Start":"03:42.365 ","End":"03:47.315","Text":"As before, the minimum is on the left and the maximum\u0027s on the right."},{"Start":"03:47.315 ","End":"03:49.175","Text":"Here\u0027s the picture."},{"Start":"03:49.175 ","End":"03:50.525","Text":"Just like before."},{"Start":"03:50.525 ","End":"03:51.680","Text":"It quicker this time,"},{"Start":"03:51.680 ","End":"04:00.515","Text":"we get i squared over 16 and the lower sum will be with i minus 1 squared over 16."},{"Start":"04:00.515 ","End":"04:04.955","Text":"The formula, we\u0027ll use it again with n equals 8."},{"Start":"04:04.955 ","End":"04:08.240","Text":"U is the sum of M_i Delta x_i."},{"Start":"04:08.240 ","End":"04:10.475","Text":"Each Delta x_i is a 1/4."},{"Start":"04:10.475 ","End":"04:13.765","Text":"Bring that out, i squared over 16,"},{"Start":"04:13.765 ","End":"04:16.635","Text":"4 with the 16 will give us 64,"},{"Start":"04:16.635 ","End":"04:21.580","Text":"and we need the sum of i squared from 1-8,"},{"Start":"04:21.580 ","End":"04:31.575","Text":"and this will give us using these formula where n is 8 and this comes out to be 204,"},{"Start":"04:31.575 ","End":"04:39.740","Text":"17 times 8 times 9 over 6 is 12 times 17,"},{"Start":"04:39.740 ","End":"04:42.170","Text":"which is 204 over 64."},{"Start":"04:42.170 ","End":"04:47.685","Text":"Calculator 3.1875, and the lower sum, you know what?"},{"Start":"04:47.685 ","End":"04:49.250","Text":"You can pause and take a look."},{"Start":"04:49.250 ","End":"04:54.900","Text":"It\u0027s very similar to this except that we go from 0-7,"},{"Start":"04:54.900 ","End":"05:02.825","Text":"which means 1-7, and just plug in 7 here and we get ultimately 2.1875."},{"Start":"05:02.825 ","End":"05:06.455","Text":"Now, let\u0027s go on to n equals 10 and 20."},{"Start":"05:06.455 ","End":"05:12.440","Text":"What I suggest is that before we do the case of n equals 10 and 20,"},{"Start":"05:12.440 ","End":"05:16.370","Text":"let\u0027s just solve it for a general n because this is quite a lot of work"},{"Start":"05:16.370 ","End":"05:20.915","Text":"to find the general formula for n and then plug in 10 and 20."},{"Start":"05:20.915 ","End":"05:24.560","Text":"We\u0027ll just do what we did above but in more general."},{"Start":"05:24.560 ","End":"05:27.050","Text":"The length of each sub-interval,"},{"Start":"05:27.050 ","End":"05:31.970","Text":"if we divide 02 into n equal pieces will be"},{"Start":"05:31.970 ","End":"05:38.005","Text":"2 over n and the points will be like so jumps of 2 over n,"},{"Start":"05:38.005 ","End":"05:39.740","Text":"and just like before,"},{"Start":"05:39.740 ","End":"05:43.955","Text":"we get the same formulas except instead of 6 or 8,"},{"Start":"05:43.955 ","End":"05:48.990","Text":"we leave this as n capital M_i is this,"},{"Start":"05:48.990 ","End":"05:54.250","Text":"and little m_i is the same but with i replaced by i minus 1."},{"Start":"05:54.250 ","End":"05:59.210","Text":"Now, the upper sum is the sum of big M_i Delta x_i."},{"Start":"05:59.210 ","End":"06:01.310","Text":"Delta x_i is 2 over n,"},{"Start":"06:01.310 ","End":"06:02.900","Text":"comes out in front."},{"Start":"06:02.900 ","End":"06:05.285","Text":"Plug this in here."},{"Start":"06:05.285 ","End":"06:08.300","Text":"Then we can take the n squared,"},{"Start":"06:08.300 ","End":"06:10.790","Text":"combine it with the n to form n cubed,"},{"Start":"06:10.790 ","End":"06:13.420","Text":"4 with 2 gives 8."},{"Start":"06:13.420 ","End":"06:16.740","Text":"Now, we have the sum of i squared from 1 to n,"},{"Start":"06:16.740 ","End":"06:18.155","Text":"and by the formula,"},{"Start":"06:18.155 ","End":"06:20.900","Text":"this is equal to this,"},{"Start":"06:20.900 ","End":"06:23.005","Text":"and we can simplify."},{"Start":"06:23.005 ","End":"06:25.060","Text":"The 8 of the 6 will take out."},{"Start":"06:25.060 ","End":"06:26.655","Text":"One of the n\u0027s here,"},{"Start":"06:26.655 ","End":"06:30.075","Text":"cancels with one of the n\u0027s here leaving n squared,"},{"Start":"06:30.075 ","End":"06:32.230","Text":"and then we have these 2 factors."},{"Start":"06:32.230 ","End":"06:36.320","Text":"Then we can divide each of these by n, one of the n is here,"},{"Start":"06:36.320 ","End":"06:41.175","Text":"one of the n is here and n plus 1 over n is 1 plus 1 over n,"},{"Start":"06:41.175 ","End":"06:46.085","Text":"2n plus 1 over n is 2 plus 1 over n. This is the formula we get"},{"Start":"06:46.085 ","End":"06:51.820","Text":"for the upper sum in terms of n. For the lower sum, it\u0027s very similar."},{"Start":"06:51.820 ","End":"06:56.155","Text":"Instead of big M, we have a little m. Instead of i,"},{"Start":"06:56.155 ","End":"06:58.555","Text":"we have i minus 1 here,"},{"Start":"06:58.555 ","End":"07:03.075","Text":"and then we do our usual substitution of index."},{"Start":"07:03.075 ","End":"07:05.745","Text":"Replace i by i minus 1."},{"Start":"07:05.745 ","End":"07:08.169","Text":"You can also take the n squared outside,"},{"Start":"07:08.169 ","End":"07:10.090","Text":"combine the 2 with the 4."},{"Start":"07:10.090 ","End":"07:12.640","Text":"Then instead of the sum of i minus 1 squared,"},{"Start":"07:12.640 ","End":"07:13.960","Text":"we have the sum of i squared,"},{"Start":"07:13.960 ","End":"07:16.520","Text":"but from 0 to n minus 1,"},{"Start":"07:16.520 ","End":"07:18.990","Text":"0 doesn\u0027t contribute anything."},{"Start":"07:18.990 ","End":"07:23.380","Text":"So we just use the formula that we can just look at this expression here,"},{"Start":"07:23.380 ","End":"07:26.620","Text":"but replace n with n minus 1."},{"Start":"07:26.620 ","End":"07:31.665","Text":"We get this, and after simplification,"},{"Start":"07:31.665 ","End":"07:35.810","Text":"we get very similar to here except instead of plus and plus,"},{"Start":"07:35.810 ","End":"07:37.685","Text":"we have minus and minus."},{"Start":"07:37.685 ","End":"07:41.390","Text":"This is a general formula for the upper and lower sums with the"},{"Start":"07:41.390 ","End":"07:45.530","Text":"general n. Now, we were asked for n equals 10 and 20."},{"Start":"07:45.530 ","End":"07:47.450","Text":"So n equals 10."},{"Start":"07:47.450 ","End":"07:51.975","Text":"We just plug in n equals 10 here,"},{"Start":"07:51.975 ","End":"07:54.500","Text":"and with the help of a calculator,"},{"Start":"07:54.500 ","End":"08:00.785","Text":"we get 3.08, and similarly here we get 2.28."},{"Start":"08:00.785 ","End":"08:03.880","Text":"For n equals 20,"},{"Start":"08:03.880 ","End":"08:06.735","Text":"these are the numbers we get,"},{"Start":"08:06.735 ","End":"08:12.065","Text":"and here are the graphs or the picture for the upper sum,"},{"Start":"08:12.065 ","End":"08:16.580","Text":"the sum of the areas of these blue rectangles on the lower sum,"},{"Start":"08:16.580 ","End":"08:19.645","Text":"the sum of the areas of the red rectangles."},{"Start":"08:19.645 ","End":"08:21.590","Text":"This will be above the integral,"},{"Start":"08:21.590 ","End":"08:23.000","Text":"will be below the integral."},{"Start":"08:23.000 ","End":"08:28.505","Text":"But you see that they\u0027re getting closer when n equals 20,"},{"Start":"08:28.505 ","End":"08:31.735","Text":"then we get still closer."},{"Start":"08:31.735 ","End":"08:33.930","Text":"This is what it looks like,"},{"Start":"08:33.930 ","End":"08:39.260","Text":"and let\u0027s put our results in a table that we had for 6,"},{"Start":"08:39.260 ","End":"08:42.790","Text":"8, 10, and 20."},{"Start":"08:42.790 ","End":"08:47.880","Text":"Notice that these are going down and down and down,"},{"Start":"08:47.880 ","End":"08:51.220","Text":"and these numbers are going up."},{"Start":"08:51.220 ","End":"08:53.380","Text":"There\u0027s still a distance apart."},{"Start":"08:53.380 ","End":"08:55.390","Text":"In fact in general,"},{"Start":"08:55.390 ","End":"08:58.885","Text":"if we just let n go to infinity,"},{"Start":"08:58.885 ","End":"09:00.210","Text":"this is what we\u0027ll get,"},{"Start":"09:00.210 ","End":"09:02.355","Text":"and I\u0027ll show you in a moment how I got this,"},{"Start":"09:02.355 ","End":"09:04.275","Text":"it\u0027s a 2 and 2/3."},{"Start":"09:04.275 ","End":"09:07.960","Text":"We note that we have the formula for L and U,"},{"Start":"09:07.960 ","End":"09:10.825","Text":"the lower and upper sum for the partition P_n."},{"Start":"09:10.825 ","End":"09:14.290","Text":"Now, if we take the upper bound,"},{"Start":"09:14.290 ","End":"09:17.470","Text":"the supremum, which is actually just the limit."},{"Start":"09:17.470 ","End":"09:25.620","Text":"When n goes to infinity this keeps increasing so the limit is 8/3 from below."},{"Start":"09:25.620 ","End":"09:27.559","Text":"Here the limit is 8/3,"},{"Start":"09:27.559 ","End":"09:30.360","Text":"but it approaches it from above."},{"Start":"09:30.360 ","End":"09:34.515","Text":"8/3 is in decimal to 4 places is this."},{"Start":"09:34.515 ","End":"09:37.280","Text":"Now, these 2 came out equal."},{"Start":"09:37.280 ","End":"09:40.865","Text":"What this means, and we\u0027ll see this in the next clip,"},{"Start":"09:40.865 ","End":"09:43.070","Text":"is that f is integrable."},{"Start":"09:43.070 ","End":"09:44.520","Text":"Just giving you a sneak preview."},{"Start":"09:44.520 ","End":"09:49.895","Text":"Whenever this supremum of the lowest sums is equal to the infimum of the upper sums."},{"Start":"09:49.895 ","End":"09:54.830","Text":"This means that f is integrable and the common value is called the integral."},{"Start":"09:54.830 ","End":"09:59.050","Text":"The area under the curve here is 8/3."},{"Start":"09:59.050 ","End":"10:02.610","Text":"That\u0027s enough for this clip."}],"ID":24683},{"Watched":false,"Name":"Upper,Lower Riemann integrals and Integrability","Duration":"6m 30s","ChapterTopicVideoID":23757,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23757.jpeg","UploadDate":"2021-01-08T10:38:18.6270000","DurationForVideoObject":"PT6M30S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.485","Text":"In the previous clip,"},{"Start":"00:01.485 ","End":"00:05.760","Text":"we talked about upper and lower Riemann sums and now"},{"Start":"00:05.760 ","End":"00:10.470","Text":"we\u0027re going to talk about upper and lower Riemann integrals."},{"Start":"00:10.470 ","End":"00:15.525","Text":"I\u0027m getting closer to the concepts of integrability and the Riemann integral."},{"Start":"00:15.525 ","End":"00:23.240","Text":"We defined the upper Riemann sum for a partition and a function as to be"},{"Start":"00:23.240 ","End":"00:27.335","Text":"this sum upper rectangles that enclose the area"},{"Start":"00:27.335 ","End":"00:33.295","Text":"and the lower Riemann sum was this and we even did an example."},{"Start":"00:33.295 ","End":"00:40.640","Text":"Also, we noted that we have the inequality that if little m and big M are bounds for f,"},{"Start":"00:40.640 ","End":"00:44.450","Text":"then for each interval little m_i is less than or equal to"},{"Start":"00:44.450 ","End":"00:49.730","Text":"big M_i where the definitions were as in the previous clip,"},{"Start":"00:49.730 ","End":"00:51.230","Text":"this is a continuation."},{"Start":"00:51.230 ","End":"00:54.105","Text":"Summing this up, these 4 things,"},{"Start":"00:54.105 ","End":"00:58.790","Text":"take a sum from i equals 1 to n of whatever this is,"},{"Start":"00:58.790 ","End":"01:01.135","Text":"times Delta x _i."},{"Start":"01:01.135 ","End":"01:07.260","Text":"Delta x_i appears in all places and then we sum i equals 1 to n for all 4 of them."},{"Start":"01:07.260 ","End":"01:09.295","Text":"It preserves the inequality."},{"Start":"01:09.295 ","End":"01:15.460","Text":"Now, recall that if we just take the sum of the lengths of the intervals,"},{"Start":"01:15.460 ","End":"01:18.970","Text":"it gives us the length in total of the interval a,"},{"Start":"01:18.970 ","End":"01:21.580","Text":"b, which is b minus a."},{"Start":"01:21.580 ","End":"01:24.280","Text":"This sum, we take m outside,"},{"Start":"01:24.280 ","End":"01:28.240","Text":"and then we have the sum here which is b minus"},{"Start":"01:28.240 ","End":"01:32.574","Text":"a and similarly at the right end of this chain,"},{"Start":"01:32.574 ","End":"01:35.005","Text":"we have big M b minus a."},{"Start":"01:35.005 ","End":"01:41.410","Text":"This by definition is the lower Riemann sum for P and f and this is"},{"Start":"01:41.410 ","End":"01:45.110","Text":"the upper Riemann sum for P and f. We have all these in"},{"Start":"01:45.110 ","End":"01:49.850","Text":"a row and that means that in shorthand,"},{"Start":"01:49.850 ","End":"01:52.880","Text":"we write just the integral from a to b of f,"},{"Start":"01:52.880 ","End":"01:55.610","Text":"but upper integral has a bar over the top."},{"Start":"01:55.610 ","End":"02:01.310","Text":"In longhand, we also include the x or whatever the independent variable is and"},{"Start":"02:01.310 ","End":"02:07.850","Text":"the definition of this is the infimum taken on all possible partitions of the interval a,"},{"Start":"02:07.850 ","End":"02:10.039","Text":"b of the upper sums."},{"Start":"02:10.039 ","End":"02:13.970","Text":"We take all the upper sums and take the least or the infimum."},{"Start":"02:13.970 ","End":"02:18.650","Text":"Similarly, the lower integral,"},{"Start":"02:18.650 ","End":"02:23.090","Text":"which is written with or without the x short way, the longer way,"},{"Start":"02:23.090 ","End":"02:28.795","Text":"is the supremum of all the lower partitions."},{"Start":"02:28.795 ","End":"02:31.970","Text":"The 1 is the infimum of the upper partitions and 1 is"},{"Start":"02:31.970 ","End":"02:35.240","Text":"the supremum of the lower partitions and these"},{"Start":"02:35.240 ","End":"02:36.890","Text":"2 are called the"},{"Start":"02:36.890 ","End":"02:42.380","Text":"upper and lower Riemann integrals of the function f over the interval a, b."},{"Start":"02:42.380 ","End":"02:45.185","Text":"Now, let\u0027s return to the example that we had."},{"Start":"02:45.185 ","End":"02:48.125","Text":"We had that the lower Riemann sum for"},{"Start":"02:48.125 ","End":"02:52.925","Text":"a particular partition when we break it up into n equal steps,"},{"Start":"02:52.925 ","End":"03:02.100","Text":"we had this expression and the upper for this partition was this expression."},{"Start":"03:02.510 ","End":"03:11.105","Text":"Lower Riemann integral, which is the supremum of all the lower sums,"},{"Start":"03:11.105 ","End":"03:19.145","Text":"is in particular bigger or equal to the supremum of part of the partitions, just the P_n."},{"Start":"03:19.145 ","End":"03:25.164","Text":"We take the supremum over a larger set it can get bigger, possibly."},{"Start":"03:25.164 ","End":"03:30.845","Text":"Now, this supremum we figured out was 8 over 3 with a simple limit."},{"Start":"03:30.845 ","End":"03:39.605","Text":"Similarly, the upper Riemann integral is the infimum of the upper Riemann sums."},{"Start":"03:39.605 ","End":"03:45.740","Text":"The infimum on all partitions is less than or equal to the infimum on"},{"Start":"03:45.740 ","End":"03:52.195","Text":"these special even spaced partitions and that was equal to 8 over 3 also."},{"Start":"03:52.195 ","End":"04:00.784","Text":"The lower Riemann integral and the upper Riemann integral both came out to be 8 over 3."},{"Start":"04:00.784 ","End":"04:05.390","Text":"The lower integral is always less than"},{"Start":"04:05.390 ","End":"04:09.590","Text":"or equal to the upper integral and we\u0027ll prove this in the next clip."},{"Start":"04:09.590 ","End":"04:13.730","Text":"From here, we have that 8/3 is less than or equal"},{"Start":"04:13.730 ","End":"04:18.540","Text":"to the lower Riemann integral of f on 0,"},{"Start":"04:18.540 ","End":"04:22.370","Text":"2 and that\u0027s less than or equal to the upper Riemann integral."},{"Start":"04:22.370 ","End":"04:25.250","Text":"That\u0027s always true. From here,"},{"Start":"04:25.250 ","End":"04:29.075","Text":"that\u0027s less than or equal to 8 over 3."},{"Start":"04:29.075 ","End":"04:31.730","Text":"Now, if you have a chain of inequalities that"},{"Start":"04:31.730 ","End":"04:35.180","Text":"begins and ends with the same thing throughout,"},{"Start":"04:35.180 ","End":"04:38.270","Text":"these must all be equalities."},{"Start":"04:38.270 ","End":"04:46.470","Text":"The lower integral of f is 8/3 and the upper integral of f is 8/3 and in particular,"},{"Start":"04:46.470 ","End":"04:48.470","Text":"these 2 are equal."},{"Start":"04:48.470 ","End":"04:51.115","Text":"When this happens, that\u0027s the good case,"},{"Start":"04:51.115 ","End":"04:55.425","Text":"then we know that the function is Riemann integrable."},{"Start":"04:55.425 ","End":"04:58.790","Text":"The Riemann integral and integrability."},{"Start":"04:58.790 ","End":"05:02.255","Text":"If it happens as it did in our example,"},{"Start":"05:02.255 ","End":"05:09.380","Text":"that the lower Riemann integral is equal to the upper Riemann integral then we say that"},{"Start":"05:09.380 ","End":"05:17.270","Text":"the function is integrable or Riemann integrable on that interval a, b."},{"Start":"05:17.270 ","End":"05:21.345","Text":"If this happens, the common value,"},{"Start":"05:21.345 ","End":"05:22.560","Text":"the 2 are equal,"},{"Start":"05:22.560 ","End":"05:27.790","Text":"so there\u0027s just 1 value there and we call that the Riemann integral of f on a,"},{"Start":"05:27.790 ","End":"05:32.930","Text":"b, and it\u0027s denoted like so without the bar above them,"},{"Start":"05:32.930 ","End":"05:34.490","Text":"without the bar below,"},{"Start":"05:34.490 ","End":"05:38.925","Text":"just the integral from a to b of f of x dx."},{"Start":"05:38.925 ","End":"05:48.195","Text":"In our example, lower and the upper integrals were both equal to 8/3 and so f of x,"},{"Start":"05:48.195 ","End":"05:50.975","Text":"which if you remember was x squared,"},{"Start":"05:50.975 ","End":"05:55.535","Text":"is integrable on the interval from 0-2,"},{"Start":"05:55.535 ","End":"05:59.900","Text":"and the integral is equal to 8/3."},{"Start":"05:59.900 ","End":"06:03.530","Text":"For those who\u0027ve learned basic integration,"},{"Start":"06:03.530 ","End":"06:06.220","Text":"simple definite integrals like that,"},{"Start":"06:06.220 ","End":"06:13.440","Text":"the way we do it is we find an indefinite integral of x squared as x cubed over 3,"},{"Start":"06:13.440 ","End":"06:14.755","Text":"is the simple formula,"},{"Start":"06:14.755 ","End":"06:16.640","Text":"and then we substitute 2,"},{"Start":"06:16.640 ","End":"06:18.550","Text":"we substitute 0 and subtract."},{"Start":"06:18.550 ","End":"06:23.600","Text":"We get 8 over 3 and that confirms what we had."},{"Start":"06:23.600 ","End":"06:26.750","Text":"It\u0027s time for another break and after that,"},{"Start":"06:26.750 ","End":"06:30.300","Text":"we\u0027ll do an example or exercise."}],"ID":24684},{"Watched":false,"Name":"Examples - Riemann Integrability","Duration":"6m 51s","ChapterTopicVideoID":23758,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23758.jpeg","UploadDate":"2021-01-08T10:41:53.6270000","DurationForVideoObject":"PT6M51S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.220","Text":"Continuing after the break,"},{"Start":"00:02.220 ","End":"00:07.035","Text":"another example, and this time, we\u0027ll take the function"},{"Start":"00:07.035 ","End":"00:13.634","Text":"defined on minus 1,1 to be f_(0) is 1,"},{"Start":"00:13.634 ","End":"00:19.200","Text":"but everywhere else f is 0 on this interval."},{"Start":"00:19.200 ","End":"00:24.165","Text":"Here\u0027s a picture, this is the x-axis,"},{"Start":"00:24.165 ","End":"00:27.749","Text":"and the function everywhere is on the x-axis"},{"Start":"00:27.749 ","End":"00:34.920","Text":"is 0 except for at the point 0 where the value of the function is 1."},{"Start":"00:34.920 ","End":"00:39.315","Text":"Let\u0027s think intuitively what the area under the curve might be."},{"Start":"00:39.315 ","End":"00:42.075","Text":"It\u0027s pretty much 0 everywhere,"},{"Start":"00:42.075 ","End":"00:46.020","Text":"and the only place it isn\u0027t 0 is like a single line,"},{"Start":"00:46.020 ","End":"00:47.960","Text":"but in terms of area,"},{"Start":"00:47.960 ","End":"00:50.870","Text":"a vertical line has no area so"},{"Start":"00:50.870 ","End":"00:53.810","Text":"we would think the answer will be 0."},{"Start":"00:53.810 ","End":"00:56.225","Text":"Let\u0027s see if that\u0027s what we get at the end."},{"Start":"00:56.225 ","End":"00:58.220","Text":"Now we\u0027ll show that f is integrable,"},{"Start":"00:58.220 ","End":"01:01.760","Text":"so let\u0027s take a general partition P,"},{"Start":"01:01.760 ","End":"01:04.070","Text":"which is x_0 to x_n,"},{"Start":"01:04.070 ","End":"01:08.215","Text":"where x_0 is minus 1 and x_n is 1."},{"Start":"01:08.215 ","End":"01:15.275","Text":"Let\u0027s assume that we have strict inequality in terms of the order of the points,"},{"Start":"01:15.275 ","End":"01:18.890","Text":"each point is strictly less than the previous."},{"Start":"01:18.890 ","End":"01:22.345","Text":"We said we can assume that if we throw out duplicates."},{"Start":"01:22.345 ","End":"01:26.265","Text":"Each x_i minus 1 is less than x_i,"},{"Start":"01:26.265 ","End":"01:28.760","Text":"x_1 is less than x_2, and so on."},{"Start":"01:28.760 ","End":"01:32.060","Text":"We can choose some point in this interval which is not 0,"},{"Start":"01:32.060 ","End":"01:36.750","Text":"since it\u0027s an interval of positive length we can obviously do that."},{"Start":"01:36.860 ","End":"01:40.050","Text":"Whenever x is not 0,"},{"Start":"01:40.050 ","End":"01:41.910","Text":"f of it is 0,"},{"Start":"01:41.910 ","End":"01:44.775","Text":"so f_(x) prime is 0."},{"Start":"01:44.775 ","End":"01:47.960","Text":"The minimum on this interval is 0,"},{"Start":"01:47.960 ","End":"01:50.420","Text":"the function everywhere is non-negative,"},{"Start":"01:50.420 ","End":"01:52.580","Text":"and if it\u0027s 0 somewhere,"},{"Start":"01:52.580 ","End":"01:54.110","Text":"then that\u0027s the minimum."},{"Start":"01:54.110 ","End":"02:00.770","Text":"The lower sum is the sum of 0 times Delta x_i, which is 0."},{"Start":"02:00.770 ","End":"02:03.870","Text":"Every lower sum is 0."},{"Start":"02:03.880 ","End":"02:10.370","Text":"Since each of these lower sums is 0 for each partition,"},{"Start":"02:10.370 ","End":"02:13.550","Text":"the supremum is also 0."},{"Start":"02:13.550 ","End":"02:18.114","Text":"The supremum of the set containing just 0 is 0."},{"Start":"02:18.114 ","End":"02:24.680","Text":"This is the definition of the lower Riemann sum for f on minus 1,"},{"Start":"02:24.680 ","End":"02:28.580","Text":"1, the lower sum. So that\u0027s 0."},{"Start":"02:28.580 ","End":"02:32.915","Text":"Note that each of these Delta x\u0027s"},{"Start":"02:32.915 ","End":"02:38.030","Text":"is less than or equal to the norm of P because P is the maximum of these,"},{"Start":"02:38.030 ","End":"02:40.115","Text":"each 1 is less than or equal to the maximum."},{"Start":"02:40.115 ","End":"02:43.420","Text":"Suppose some interval contains the 0,"},{"Start":"02:43.420 ","End":"02:48.844","Text":"in this case, the maximum will be 1 if the interval contains 0,"},{"Start":"02:48.844 ","End":"02:54.110","Text":"and therefore, M_i Delta x_i is less than or equal"},{"Start":"02:54.110 ","End":"03:00.480","Text":"to 1 times the norm of P. M_i is either 0 or 1,"},{"Start":"03:00.480 ","End":"03:04.505","Text":"0 or 1 times Delta x_i is less than or equal To Delta x_i,"},{"Start":"03:04.505 ","End":"03:06.215","Text":"which is less than or equal to P."},{"Start":"03:06.215 ","End":"03:07.955","Text":"So you have 2 cases:"},{"Start":"03:07.955 ","End":"03:13.790","Text":"It\u0027s either 0 or in the case that the interval contains 0,"},{"Start":"03:13.790 ","End":"03:23.235","Text":"then we can say it\u0027s between 0 and norm of P. The most 2 intervals that contain 0,"},{"Start":"03:23.235 ","End":"03:25.080","Text":"if 0 is 1 of these,"},{"Start":"03:25.080 ","End":"03:27.179","Text":"let\u0027s say 0 is x_2,"},{"Start":"03:27.179 ","End":"03:28.965","Text":"then it will belong to 2 intervals,"},{"Start":"03:28.965 ","End":"03:31.145","Text":"x_1, x_2, and x_2, x_3."},{"Start":"03:31.145 ","End":"03:34.120","Text":"Otherwise, it\u0027s going to belong to just 1 of the intervals,"},{"Start":"03:34.120 ","End":"03:37.875","Text":"it can belong to exactly 1 or 2 of the intervals."},{"Start":"03:37.875 ","End":"03:40.535","Text":"For at most 2 values of i,"},{"Start":"03:40.535 ","End":"03:42.695","Text":"0 is in this interval."},{"Start":"03:42.695 ","End":"03:45.325","Text":"Altogether, if we take the sum of these,"},{"Start":"03:45.325 ","End":"03:49.855","Text":"M_i Delta x_i\u0027s, all of them are 0 except 2 of them,"},{"Start":"03:49.855 ","End":"03:53.985","Text":"and those 2 are bounded by the norm of P,"},{"Start":"03:53.985 ","End":"03:58.510","Text":"so altogether we have at most twice the norm of P. Now"},{"Start":"03:58.510 ","End":"04:03.875","Text":"the upper integral is the infimum of all these upper sums,"},{"Start":"04:03.875 ","End":"04:09.390","Text":"each upper sum is between 0 and twice the norm of P. If"},{"Start":"04:09.390 ","End":"04:14.885","Text":"we let the norm of P 10^0 by making the partition finer and finer,"},{"Start":"04:14.885 ","End":"04:22.220","Text":"then what we get is that the upper Riemann integral is 0,"},{"Start":"04:22.220 ","End":"04:26.660","Text":"and it\u0027s between 0 and twice something which goes to 0."},{"Start":"04:26.660 ","End":"04:29.894","Text":"By the sandwich theorem this would be 0."},{"Start":"04:29.894 ","End":"04:36.060","Text":"On the other hand earlier, here we had that the lowest integral is equal to 0."},{"Start":"04:36.060 ","End":"04:38.690","Text":"If the upper Riemann integral is 0 and the lower"},{"Start":"04:38.690 ","End":"04:41.840","Text":"Riemann integral is 0, then by definition,"},{"Start":"04:41.840 ","End":"04:46.255","Text":"f is integrable and the integral is 0,"},{"Start":"04:46.255 ","End":"04:48.664","Text":"and that\u0027s what we expected anyway."},{"Start":"04:48.664 ","End":"04:50.270","Text":"That\u0027s this example."},{"Start":"04:50.270 ","End":"04:55.190","Text":"The next example will be 1 where the function is not integrable."},{"Start":"04:55.190 ","End":"04:59.825","Text":"This next example is about the Dirichlet function,"},{"Start":"04:59.825 ","End":"05:02.750","Text":"and it\u0027s not really possible to sketch it."},{"Start":"05:02.750 ","End":"05:06.035","Text":"It\u0027s defined on the interval 0,1,"},{"Start":"05:06.035 ","End":"05:10.865","Text":"and f_(x) is 1 if x is rational,"},{"Start":"05:10.865 ","End":"05:14.005","Text":"and 0 if x is irrational."},{"Start":"05:14.005 ","End":"05:17.870","Text":"Now, the Dirichlet function is certainly bounded,"},{"Start":"05:17.870 ","End":"05:20.255","Text":"it\u0027s only values are 1 and 0,"},{"Start":"05:20.255 ","End":"05:22.945","Text":"but we\u0027ll show that it\u0027s not integrable."},{"Start":"05:22.945 ","End":"05:28.930","Text":"All we have to do is show that the lower integral is not equal to the upper integral."},{"Start":"05:28.930 ","End":"05:31.310","Text":"Do this by showing that the lower integral is 0,"},{"Start":"05:31.310 ","End":"05:33.740","Text":"but the upper integral is 1."},{"Start":"05:33.740 ","End":"05:38.600","Text":"Now if we take any partition of the interval 0,1"},{"Start":"05:38.600 ","End":"05:43.085","Text":"where we assume that this is a strictly increasing sequence,"},{"Start":"05:43.085 ","End":"05:47.180","Text":"each interval has non-zero length and"},{"Start":"05:47.180 ","End":"05:51.950","Text":"any interval contains a rational number and an irrational number."},{"Start":"05:51.950 ","End":"05:58.594","Text":"The irrational number in the interval means that the infimum is 0,"},{"Start":"05:58.594 ","End":"06:00.935","Text":"can\u0027t be less than 0 certainly,"},{"Start":"06:00.935 ","End":"06:04.664","Text":"and because of the rational number in the interval,"},{"Start":"06:04.664 ","End":"06:06.560","Text":"the value 1 is attained,"},{"Start":"06:06.560 ","End":"06:07.760","Text":"but it can\u0027t be any higher,"},{"Start":"06:07.760 ","End":"06:10.370","Text":"so the supremum is 1."},{"Start":"06:10.370 ","End":"06:18.965","Text":"The lower sum for this partition is 0 and the upper sum is 1."},{"Start":"06:18.965 ","End":"06:20.720","Text":"If we plug in here,"},{"Start":"06:20.720 ","End":"06:25.550","Text":"m_i is 0 and here big M_i is 1 and we get that."},{"Start":"06:25.550 ","End":"06:28.790","Text":"Now, we take the supremum of this overall P,"},{"Start":"06:28.790 ","End":"06:31.610","Text":"the supremum of the constant 0 is 0."},{"Start":"06:31.610 ","End":"06:37.299","Text":"Similarly here, the infimum of the upper sums is going to be 1,"},{"Start":"06:37.299 ","End":"06:42.515","Text":"which means that the lower integral is 0 and the upper integral is 1."},{"Start":"06:42.515 ","End":"06:47.680","Text":"That concludes this example of a non-integrable function,"},{"Start":"06:47.680 ","End":"06:51.640","Text":"and it\u0027s time to take a break."}],"ID":24685},{"Watched":false,"Name":"Exercise 1a","Duration":"4m 54s","ChapterTopicVideoID":23759,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23759.jpeg","UploadDate":"2021-01-08T10:43:38.2870000","DurationForVideoObject":"PT4M54S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:08.190","Text":"This exercise is a 2 part exercise where Part 1 is a preparation for Part 2,"},{"Start":"00:08.190 ","End":"00:10.740","Text":"which is the real exercise,"},{"Start":"00:10.740 ","End":"00:14.195","Text":"and I\u0027ll just give you a preview."},{"Start":"00:14.195 ","End":"00:20.330","Text":"In Part 2, we\u0027re going to compute the integral of x^n from"},{"Start":"00:20.330 ","End":"00:27.530","Text":"0-1 using Riemann sums and not using the integration formulas that you are familiar with."},{"Start":"00:27.530 ","End":"00:30.320","Text":"Back to Part 1, which like I said,"},{"Start":"00:30.320 ","End":"00:32.960","Text":"is the preparation stage for Part 2,"},{"Start":"00:32.960 ","End":"00:34.880","Text":"which is the real exercise."},{"Start":"00:34.880 ","End":"00:38.885","Text":"Again, and Part 1 has 3 parts, sub-parts."},{"Start":"00:38.885 ","End":"00:41.060","Text":"We\u0027ll just read them as we come to them."},{"Start":"00:41.060 ","End":"00:43.295","Text":"Let\u0027s take Part 1 first."},{"Start":"00:43.295 ","End":"00:47.390","Text":"Show that b^n minus a^n equals b minus a,"},{"Start":"00:47.390 ","End":"00:48.860","Text":"well, I won\u0027t read it out,"},{"Start":"00:48.860 ","End":"00:51.560","Text":"where n is a positive integer"},{"Start":"00:51.560 ","End":"00:55.320","Text":"and a and b are real numbers."},{"Start":"00:55.350 ","End":"01:01.720","Text":"We\u0027ll start from the right-hand side and work our way through to the left-hand side."},{"Start":"01:01.720 ","End":"01:06.410","Text":"We\u0027ll expand, we\u0027ll take b times this minus a times this."},{"Start":"01:06.410 ","End":"01:09.180","Text":"B times this brackets is this,"},{"Start":"01:09.180 ","End":"01:12.640","Text":"and a times these brackets gives us this,"},{"Start":"01:12.640 ","End":"01:15.519","Text":"and we write it a little bit offset."},{"Start":"01:15.519 ","End":"01:20.375","Text":"That way, we\u0027ll get similar stuff above each other,"},{"Start":"01:20.375 ","End":"01:25.380","Text":"and you can see that everything comes in pairs and cancels out,"},{"Start":"01:25.380 ","End":"01:29.125","Text":"and all we\u0027re going to be left with is b^n minus"},{"Start":"01:29.125 ","End":"01:33.505","Text":"a^n, and that concludes the first part."},{"Start":"01:33.505 ","End":"01:38.260","Text":"Here it is, I scrolled back to show you what the second part is."},{"Start":"01:38.260 ","End":"01:44.910","Text":"I highlighted it and we\u0027re going to start by expanding the numerator here."},{"Start":"01:45.290 ","End":"01:53.059","Text":"Using Part 1, with b equals k plus 1 and a equals k,"},{"Start":"01:53.059 ","End":"01:58.880","Text":"we get that this equals this because the b minus a part is just 1,"},{"Start":"01:58.880 ","End":"02:02.650","Text":"so we can basically just throw this piece out."},{"Start":"02:02.650 ","End":"02:05.714","Text":"Now, each 1 of these,"},{"Start":"02:05.714 ","End":"02:07.680","Text":"the n plus 1 terms here,"},{"Start":"02:07.680 ","End":"02:12.590","Text":"each 1 of them is of the form k plus 1 to the power of something,"},{"Start":"02:12.590 ","End":"02:14.270","Text":"k to the power of something,"},{"Start":"02:14.270 ","End":"02:21.550","Text":"and these 2 add up to n. They have to be between k^n and k plus 1^n,"},{"Start":"02:21.550 ","End":"02:24.665","Text":"these are the 2 extreme possibilities."},{"Start":"02:24.665 ","End":"02:28.349","Text":"What we get, because they are are n plus 1 terms,"},{"Start":"02:28.349 ","End":"02:30.870","Text":"and each of the terms is between this and this,"},{"Start":"02:30.870 ","End":"02:37.610","Text":"so we get n plus 1 times k^n is less than what\u0027s written here,"},{"Start":"02:37.610 ","End":"02:43.340","Text":"which is less than n plus 1 times k plus 1^n, this here."},{"Start":"02:43.340 ","End":"02:48.515","Text":"Now all we need to do is divide everything by n plus 1,"},{"Start":"02:48.515 ","End":"02:50.105","Text":"and we get this,"},{"Start":"02:50.105 ","End":"02:51.800","Text":"and this is what we had to show,"},{"Start":"02:51.800 ","End":"02:53.765","Text":"so that\u0027s Part 2."},{"Start":"02:53.765 ","End":"02:58.790","Text":"Now, Part 3, you can pause and take a look at this."},{"Start":"02:58.790 ","End":"03:04.190","Text":"What we can do for Part 3 is to use Part 2 and"},{"Start":"03:04.190 ","End":"03:11.200","Text":"sum this double inequality from 0 to m minus 1."},{"Start":"03:11.330 ","End":"03:14.660","Text":"So k goes from 0 to m minus 1,"},{"Start":"03:14.660 ","End":"03:15.800","Text":"0 to m minus 1,"},{"Start":"03:15.800 ","End":"03:17.120","Text":"0 to m minus 1,"},{"Start":"03:17.120 ","End":"03:20.990","Text":"and we just have these things here as they are."},{"Start":"03:20.990 ","End":"03:22.635","Text":"I can, of course,"},{"Start":"03:22.635 ","End":"03:29.635","Text":"make this sum go from 1 to m minus 1 because when k is 0, it\u0027s just 0."},{"Start":"03:29.635 ","End":"03:33.535","Text":"We have the sum from 1 to m minus 1."},{"Start":"03:33.535 ","End":"03:39.885","Text":"For the middle bit, just take the n plus 1 outside the Sigma."},{"Start":"03:39.885 ","End":"03:41.525","Text":"For the last bit,"},{"Start":"03:41.525 ","End":"03:43.310","Text":"we do a substitution."},{"Start":"03:43.310 ","End":"03:46.445","Text":"Replace k plus 1 by k,"},{"Start":"03:46.445 ","End":"03:48.900","Text":"and then the sum will go"},{"Start":"03:48.900 ","End":"03:51.495","Text":"instead of from 0 to m minus 1,"},{"Start":"03:51.495 ","End":"03:55.730","Text":"it\u0027ll go from 1 to m. Instead of the k plus 1,"},{"Start":"03:55.730 ","End":"03:59.135","Text":"we have k. Here are some of the details."},{"Start":"03:59.135 ","End":"04:01.475","Text":"Now, the middle bit,"},{"Start":"04:01.475 ","End":"04:06.875","Text":"which I\u0027ve colored like this is actually a telescoping series."},{"Start":"04:06.875 ","End":"04:13.785","Text":"Because what we have if we write it out is 1^n plus 1 minus 0^n plus 1."},{"Start":"04:13.785 ","End":"04:16.215","Text":"Here we have a 2 minus a 1,"},{"Start":"04:16.215 ","End":"04:17.810","Text":"then a 3 minus a 2,"},{"Start":"04:17.810 ","End":"04:19.875","Text":"a 4 minus a 3."},{"Start":"04:19.875 ","End":"04:24.740","Text":"Everything\u0027s going to cancel out except for the first term which will stay with a minus,"},{"Start":"04:24.740 ","End":"04:26.870","Text":"and the last term which stays with a plus."},{"Start":"04:26.870 ","End":"04:30.395","Text":"We get m^n plus 1 minus 0^n plus 1"},{"Start":"04:30.395 ","End":"04:32.660","Text":"and this is just 0."},{"Start":"04:32.660 ","End":"04:36.260","Text":"We\u0027re just left with m^n plus 1."},{"Start":"04:36.260 ","End":"04:40.180","Text":"Now, substitute back in here,"},{"Start":"04:40.180 ","End":"04:47.305","Text":"so we get this less than m^n plus 1 over n plus 1 less than this,"},{"Start":"04:47.305 ","End":"04:50.475","Text":"and that concludes the third part."},{"Start":"04:50.475 ","End":"04:54.230","Text":"So we\u0027re done for Part 1 of 2."}],"ID":24686},{"Watched":false,"Name":"Exercise 1b","Duration":"7m 13s","ChapterTopicVideoID":23760,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23760.jpeg","UploadDate":"2021-01-08T10:46:26.6070000","DurationForVideoObject":"PT7M13S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.850","Text":"This is Part 2 of the exercise."},{"Start":"00:02.850 ","End":"00:07.605","Text":"In Part 1, we did some preparation work that we\u0027ll need in this part 2."},{"Start":"00:07.605 ","End":"00:10.740","Text":"Anyway, given the function f of x,"},{"Start":"00:10.740 ","End":"00:13.170","Text":"which is x^n on the interval 0,"},{"Start":"00:13.170 ","End":"00:16.185","Text":"1, where n is natural number."},{"Start":"00:16.185 ","End":"00:20.550","Text":"Using Riemann sums, we have to show that f is integrable on"},{"Start":"00:20.550 ","End":"00:26.130","Text":"the interval and to compute the value of the integral from 0 to 1."},{"Start":"00:26.130 ","End":"00:31.650","Text":"We\u0027re given a hint to partition the interval into m equal pieces,"},{"Start":"00:31.650 ","End":"00:34.020","Text":"meaning pieces of equal width or length."},{"Start":"00:34.020 ","End":"00:40.580","Text":"Then we\u0027re going to use Part 1 to help us evaluate the upper and lower sums."},{"Start":"00:40.580 ","End":"00:45.635","Text":"This is the partition we get when we divide into m pieces."},{"Start":"00:45.635 ","End":"00:51.080","Text":"The width of each piece is 1 over m. We"},{"Start":"00:51.080 ","End":"00:57.360","Text":"jump in steps of 1 over m. So x_k is going to be k over m,"},{"Start":"00:57.360 ","End":"00:59.910","Text":"where this is x_0, x_1, x_2."},{"Start":"00:59.910 ","End":"01:06.205","Text":"The kth interval, I_k is from k-1 over m to k over m."},{"Start":"01:06.205 ","End":"01:13.340","Text":"The function f of x equals x^n is increasing on all of the interval from 0 to 1,"},{"Start":"01:13.340 ","End":"01:16.995","Text":"in particular on the sub-interval I_k."},{"Start":"01:16.995 ","End":"01:23.090","Text":"It attains its minimum at the left endpoint and the maximum at the right endpoint."},{"Start":"01:23.090 ","End":"01:29.300","Text":"In other words, m_k small m is k-1 over m^n,"},{"Start":"01:29.300 ","End":"01:34.420","Text":"and M_k is k over m^n,"},{"Start":"01:34.420 ","End":"01:38.900","Text":"picking the right endpoint to the power of n. Here\u0027s a picture."},{"Start":"01:38.900 ","End":"01:44.180","Text":"In this case, m equals 5 divided up into 5 pieces."},{"Start":"01:44.180 ","End":"01:48.125","Text":"The upper picture is the upper sum."},{"Start":"01:48.125 ","End":"01:53.190","Text":"Notice that the height at each place is the right endpoint,"},{"Start":"01:53.190 ","End":"01:54.470","Text":"for the lower sum,"},{"Start":"01:54.470 ","End":"01:57.215","Text":"the height is the left endpoint."},{"Start":"01:57.215 ","End":"02:01.145","Text":"The upper sum, which is the sum of these rectangles,"},{"Start":"02:01.145 ","End":"02:07.910","Text":"is gotten by simply adding up the m_k times Delta x_k,"},{"Start":"02:07.910 ","End":"02:10.825","Text":"but the Delta x_k are all constant."},{"Start":"02:10.825 ","End":"02:15.430","Text":"We can bring it out of the brackets."},{"Start":"02:15.430 ","End":"02:21.170","Text":"Similarly, the lower Riemann sum is gotten by"},{"Start":"02:21.170 ","End":"02:27.860","Text":"taking the left-endpoints and we get 0^n plus 1 over m^n and so on."},{"Start":"02:27.860 ","End":"02:30.460","Text":"It\u0027s shifted along 1."},{"Start":"02:30.460 ","End":"02:35.195","Text":"It\u0027s almost the same except at the beginning and the end."},{"Start":"02:35.195 ","End":"02:39.140","Text":"In fact, if you look at these pictures,"},{"Start":"02:39.140 ","End":"02:41.825","Text":"this sum is the same as this sum,"},{"Start":"02:41.825 ","End":"02:44.270","Text":"except for 1 extra rectangle."},{"Start":"02:44.270 ","End":"02:46.835","Text":"If we added the area of this rectangle,"},{"Start":"02:46.835 ","End":"02:49.785","Text":"that will be the difference between this sum and this sum."},{"Start":"02:49.785 ","End":"02:52.850","Text":"The difference between the upper and the lower sums,"},{"Start":"02:52.850 ","End":"02:54.185","Text":"like I was saying,"},{"Start":"02:54.185 ","End":"02:56.060","Text":"just if we compute it,"},{"Start":"02:56.060 ","End":"03:01.145","Text":"we see that we have the 1 over m or we just have 1^n minus 0^n,"},{"Start":"03:01.145 ","End":"03:03.395","Text":"in other words, it\u0027s 1 over m,"},{"Start":"03:03.395 ","End":"03:06.830","Text":"which like I said, is the area of this rectangle."},{"Start":"03:06.830 ","End":"03:08.030","Text":"The width is 1 over m,"},{"Start":"03:08.030 ","End":"03:10.115","Text":"the height is 1. That\u0027s this."},{"Start":"03:10.115 ","End":"03:14.254","Text":"Note that this goes to 0 as m goes to infinity,"},{"Start":"03:14.254 ","End":"03:18.730","Text":"meaning that 1 over m will become as small as we like."},{"Start":"03:18.730 ","End":"03:22.140","Text":"If we\u0027re given some positive Epsilon,"},{"Start":"03:22.140 ","End":"03:26.485","Text":"we can choose m so that 1 over m is less than Epsilon."},{"Start":"03:26.485 ","End":"03:31.835","Text":"Just squeeze this narrower and narrower until the area is less than Epsilon."},{"Start":"03:31.835 ","End":"03:35.540","Text":"The difference between the upper and the lower sums is less than"},{"Start":"03:35.540 ","End":"03:40.820","Text":"Epsilon and that should ring a bell Riemann\u0027s criterion."},{"Start":"03:40.820 ","End":"03:47.000","Text":"If we can get the upper minus the lower sum to be smaller than any Epsilon,"},{"Start":"03:47.000 ","End":"03:50.330","Text":"we can find a partition for that for any epsilon,"},{"Start":"03:50.330 ","End":"03:52.165","Text":"then f is integrable."},{"Start":"03:52.165 ","End":"03:54.390","Text":"That\u0027s that part of the exercise,"},{"Start":"03:54.390 ","End":"03:59.330","Text":"now what remains is to compute the value of this integral."},{"Start":"03:59.330 ","End":"04:03.805","Text":"The upper sum here it is."},{"Start":"04:03.805 ","End":"04:09.450","Text":"We can take the m^n from the denominator of"},{"Start":"04:09.450 ","End":"04:16.830","Text":"everything out in front and m^n combines with the m so we get m^n plus 1."},{"Start":"04:16.830 ","End":"04:19.605","Text":"What\u0027s left is 1^n, 2^n,"},{"Start":"04:19.605 ","End":"04:23.400","Text":"and so on up to 1 which is m over m,"},{"Start":"04:23.400 ","End":"04:26.200","Text":"so this is the m^n."},{"Start":"04:28.090 ","End":"04:30.470","Text":"The first term is 0,"},{"Start":"04:30.470 ","End":"04:31.925","Text":"so we don\u0027t include that."},{"Start":"04:31.925 ","End":"04:35.290","Text":"It\u0027s basically the same as this except that we stop 1 term"},{"Start":"04:35.290 ","End":"04:41.420","Text":"short without this it\u0027s just the sum from 1 to m minus 1,"},{"Start":"04:41.420 ","End":"04:46.775","Text":"or to the power of n. Now if you go back and look at Part 1 of the exercise,"},{"Start":"04:46.775 ","End":"04:53.900","Text":"it had 3 sub-parts and the third sub-part was exactly to prove this."},{"Start":"04:53.900 ","End":"04:59.210","Text":"We have an estimate that the lower sum is less"},{"Start":"04:59.210 ","End":"05:04.625","Text":"than this expression and the upper sum is bigger than this expression."},{"Start":"05:04.625 ","End":"05:09.350","Text":"Next, we can divide everything by m^n plus 1,"},{"Start":"05:09.350 ","End":"05:17.230","Text":"and what we get is this over m^n plus 1 is exactly L,"},{"Start":"05:17.230 ","End":"05:20.035","Text":"lower sum of P_m."},{"Start":"05:20.035 ","End":"05:26.535","Text":"This divided by m^n plus 1 is the upper sum."},{"Start":"05:26.535 ","End":"05:32.795","Text":"We have 1 over n plus 1 sandwiched between the lower sum and the upper sum."},{"Start":"05:32.795 ","End":"05:36.005","Text":"On the other hand, we have another sandwich."},{"Start":"05:36.005 ","End":"05:38.420","Text":"Because f is integrable,"},{"Start":"05:38.420 ","End":"05:45.455","Text":"the integral is always between the lower sum and the upper sum for any partition."},{"Start":"05:45.455 ","End":"05:52.220","Text":"If we have 2 things that are trapped between the same 2 pieces of bread of the sandwich,"},{"Start":"05:52.220 ","End":"05:56.630","Text":"that means that the distance between these 2 in absolute value"},{"Start":"05:56.630 ","End":"06:01.415","Text":"is got to be less than or equal to this minus this."},{"Start":"06:01.415 ","End":"06:05.885","Text":"If you\u0027re not sure about that from algebra here,"},{"Start":"06:05.885 ","End":"06:10.250","Text":"I\u0027ve abstracted it, if you have 2 numbers, C_1 and C_2,"},{"Start":"06:10.250 ","End":"06:13.235","Text":"both between A and B,"},{"Start":"06:13.235 ","End":"06:20.545","Text":"it\u0027s clear that the distance from C_1 to C_2 has to be less than or equal to B minus A."},{"Start":"06:20.545 ","End":"06:22.445","Text":"That\u0027s what we\u0027re doing here."},{"Start":"06:22.445 ","End":"06:29.060","Text":"Now, this is less than Epsilon for all Epsilon because we took an arbitrary Epsilon,"},{"Start":"06:29.060 ","End":"06:31.045","Text":"it means that it\u0027s 0,"},{"Start":"06:31.045 ","End":"06:33.800","Text":"if this minus this is 0,"},{"Start":"06:33.800 ","End":"06:36.410","Text":"then this equals this, now,"},{"Start":"06:36.410 ","End":"06:39.560","Text":"we have that the value of the integral,"},{"Start":"06:39.560 ","End":"06:49.215","Text":"in other words, the integral of x^n from 0 to 1 is equal exactly to 1 over n plus 1."},{"Start":"06:49.215 ","End":"06:52.280","Text":"That\u0027s the end except that I\u0027d like to just do"},{"Start":"06:52.280 ","End":"06:55.850","Text":"a check since most of you know integration anyway,"},{"Start":"06:55.850 ","End":"06:58.505","Text":"using rules of integration,"},{"Start":"06:58.505 ","End":"07:03.290","Text":"we know that the integral of x^n is x^n plus 1 over n plus 1,"},{"Start":"07:03.290 ","End":"07:05.744","Text":"plug in 1 and 0, and subtract,"},{"Start":"07:05.744 ","End":"07:07.755","Text":"and we have 1 over n plus 1,"},{"Start":"07:07.755 ","End":"07:10.860","Text":"which comes out the same as this, so we\u0027re okay."},{"Start":"07:10.860 ","End":"07:13.720","Text":"That concludes this exercise."}],"ID":24687},{"Watched":false,"Name":"Exercise 2","Duration":"11m 37s","ChapterTopicVideoID":23761,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23761.jpeg","UploadDate":"2021-01-08T10:50:29.0230000","DurationForVideoObject":"PT11M37S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.560","Text":"In this exercise, the function f of x is cosine x on the interval from 0 to Pi over 2."},{"Start":"00:07.560 ","End":"00:09.675","Text":"That\u0027s like 90 degrees."},{"Start":"00:09.675 ","End":"00:17.925","Text":"We have to use Riemann sums to prove that f is integrable on the interval 0 Pi over 2,"},{"Start":"00:17.925 ","End":"00:22.635","Text":"and to compute the integral of f on this interval."},{"Start":"00:22.635 ","End":"00:24.900","Text":"Here\u0027s the sketch of the cosine."},{"Start":"00:24.900 ","End":"00:31.110","Text":"This bold part is the bit from 0 to Pi over 2."},{"Start":"00:31.110 ","End":"00:33.300","Text":"That is the only part that matters,"},{"Start":"00:33.300 ","End":"00:35.895","Text":"and note that the function is decreasing."},{"Start":"00:35.895 ","End":"00:37.440","Text":"We\u0027re given 2 hints."},{"Start":"00:37.440 ","End":"00:43.985","Text":"The first hint is to partition the interval into n equal pieces,"},{"Start":"00:43.985 ","End":"00:46.825","Text":"meaning pieces of equal length or width."},{"Start":"00:46.825 ","End":"00:51.215","Text":"Then ultimately to let n go to infinity."},{"Start":"00:51.215 ","End":"00:53.300","Text":"We did that in a previous exercise."},{"Start":"00:53.300 ","End":"00:55.130","Text":"This is a common thing to do."},{"Start":"00:55.130 ","End":"00:58.490","Text":"The second hint is about trigonometry."},{"Start":"00:58.490 ","End":"01:02.720","Text":"We\u0027re given the trigonometric identity,"},{"Start":"01:02.720 ","End":"01:04.250","Text":"the one that\u0027s written here,"},{"Start":"01:04.250 ","End":"01:05.660","Text":"I won\u0027t read it out,"},{"Start":"01:05.660 ","End":"01:10.520","Text":"and using this to evaluate this expression,"},{"Start":"01:10.520 ","End":"01:13.880","Text":"which should come out to be like so."},{"Start":"01:13.880 ","End":"01:20.900","Text":"It\u0027s for notation, z plus is the positive integers,"},{"Start":"01:20.900 ","End":"01:23.770","Text":"like the natural numbers but without 0."},{"Start":"01:23.770 ","End":"01:26.585","Text":"Let\u0027s figure out what is this partition."},{"Start":"01:26.585 ","End":"01:28.880","Text":"Now we said n equal pieces."},{"Start":"01:28.880 ","End":"01:33.380","Text":"We can compute the length of each piece first."},{"Start":"01:33.380 ","End":"01:36.075","Text":"We have pi over 2 is the total length."},{"Start":"01:36.075 ","End":"01:38.430","Text":"If we divide pi over 2 by n,"},{"Start":"01:38.430 ","End":"01:40.870","Text":"then we have pi over 2n."},{"Start":"01:40.870 ","End":"01:44.900","Text":"Which means that we go in jumps of pi over 2n,"},{"Start":"01:44.900 ","End":"01:47.485","Text":"multiples of Pi over 2n."},{"Start":"01:47.485 ","End":"01:51.075","Text":"In n steps we get to pi over 2."},{"Start":"01:51.075 ","End":"01:53.280","Text":"There are n plus 1 points,"},{"Start":"01:53.280 ","End":"01:56.085","Text":"but only n intervals."},{"Start":"01:56.085 ","End":"02:00.420","Text":"Typical endpoint xk is k Pi over 2n."},{"Start":"02:00.420 ","End":"02:07.260","Text":"It\u0027s Pi over 2n, but multiplied by k from 0 to n. The kth interval,"},{"Start":"02:07.260 ","End":"02:13.950","Text":"the k only goes from 1 to n is from k minus 1 to k. Now like I pointed out, f,"},{"Start":"02:13.950 ","End":"02:19.230","Text":"which is cosine is decreasing on the whole interval from 0 to pi over 2,"},{"Start":"02:19.230 ","End":"02:22.070","Text":"and if I split it up into sub-intervals,"},{"Start":"02:22.070 ","End":"02:24.805","Text":"it\u0027s still going to be decreasing on each of them,"},{"Start":"02:24.805 ","End":"02:30.535","Text":"so the minimum is obtained on the right and the maximum on the left."},{"Start":"02:30.535 ","End":"02:34.900","Text":"It\u0027s the cosine of the left end point, which is this,"},{"Start":"02:34.900 ","End":"02:40.535","Text":"and the minimum is obtained by taking the right end point which is this."},{"Start":"02:40.535 ","End":"02:46.475","Text":"The upper Riemann sum is just the sum of mk times delta xk,"},{"Start":"02:46.475 ","End":"02:52.550","Text":"the product of these 2 where k goes from 1 to n. But since this doesn\u0027t depend on k,"},{"Start":"02:52.550 ","End":"02:55.600","Text":"we can pull that in front of the sum, like so."},{"Start":"02:55.600 ","End":"02:59.015","Text":"The lower sum is very similar."},{"Start":"02:59.015 ","End":"03:03.005","Text":"It\u0027s just that we take little mk,"},{"Start":"03:03.005 ","End":"03:06.770","Text":"which means that it\u0027s without the k minus 1 here."},{"Start":"03:06.770 ","End":"03:08.090","Text":"I mean it\u0027s without the minus 1,"},{"Start":"03:08.090 ","End":"03:09.575","Text":"it\u0027s cosine k Pi,"},{"Start":"03:09.575 ","End":"03:11.980","Text":"so it like offset by 1."},{"Start":"03:11.980 ","End":"03:16.160","Text":"I\u0027ve written it in such a way that the corresponding ones line up."},{"Start":"03:16.160 ","End":"03:19.500","Text":"Now, if we subtract the 2 sums,"},{"Start":"03:19.500 ","End":"03:21.165","Text":"the upper minus the lower,"},{"Start":"03:21.165 ","End":"03:22.685","Text":"by the way, I\u0027ve written it,"},{"Start":"03:22.685 ","End":"03:30.220","Text":"you can see that it\u0027s just this minus this with the Pi over 2n in front."},{"Start":"03:30.220 ","End":"03:32.385","Text":"Cosine of 0 is 1."},{"Start":"03:32.385 ","End":"03:35.145","Text":"Cosine of pi over 2 is 0,"},{"Start":"03:35.145 ","End":"03:36.480","Text":"1 minus 0 is 1,"},{"Start":"03:36.480 ","End":"03:38.760","Text":"so it\u0027s just pi over 2n."},{"Start":"03:38.760 ","End":"03:43.310","Text":"This can be as small as we like because it goes to 0."},{"Start":"03:43.310 ","End":"03:45.680","Text":"If we\u0027re given any positive epsilon,"},{"Start":"03:45.680 ","End":"03:53.225","Text":"we can find a big enough n so that this pi over 2n is less than the given epsilon."},{"Start":"03:53.225 ","End":"04:02.115","Text":"Let the partition P be the Pn for chosen n. Then we have the u of Pf minus l of Pf,"},{"Start":"04:02.115 ","End":"04:05.570","Text":"the upper sum minus the lower sum for the partition and"},{"Start":"04:05.570 ","End":"04:09.785","Text":"the function is going to be less than epsilon."},{"Start":"04:09.785 ","End":"04:14.180","Text":"By Riemann\u0027s criterion, if this is the case,"},{"Start":"04:14.180 ","End":"04:15.800","Text":"assuming that the function is bounded,"},{"Start":"04:15.800 ","End":"04:17.285","Text":"which of course it is,"},{"Start":"04:17.285 ","End":"04:20.675","Text":"then this condition means that f is integrable."},{"Start":"04:20.675 ","End":"04:23.040","Text":"That\u0027s part of the question,"},{"Start":"04:23.040 ","End":"04:25.640","Text":"and the other part is to do the actual computation of"},{"Start":"04:25.640 ","End":"04:28.940","Text":"the value of the integral by using Riemann sums,"},{"Start":"04:28.940 ","End":"04:31.705","Text":"not using the Table of Integrals."},{"Start":"04:31.705 ","End":"04:34.975","Text":"Note 2 things; number 1,"},{"Start":"04:34.975 ","End":"04:39.080","Text":"that the integral, when it exists,"},{"Start":"04:39.080 ","End":"04:41.375","Text":"which in our case it does,"},{"Start":"04:41.375 ","End":"04:43.130","Text":"it\u0027s always sandwiched between"},{"Start":"04:43.130 ","End":"04:47.480","Text":"the lower Riemann sum and the upper Riemann sum for any partition,"},{"Start":"04:47.480 ","End":"04:49.565","Text":"be the same thing here and here."},{"Start":"04:49.565 ","End":"04:54.915","Text":"Secondly, the difference between these 2 tends to 0."},{"Start":"04:54.915 ","End":"04:58.315","Text":"From this, this minus this goes to 0."},{"Start":"04:58.315 ","End":"05:04.120","Text":"Now, we can conclude from this that both the lowest sum and the upper sum,"},{"Start":"05:04.120 ","End":"05:06.655","Text":"they both tend in the limit to the integral."},{"Start":"05:06.655 ","End":"05:09.309","Text":"Meaning if this minus this goes to 0,"},{"Start":"05:09.309 ","End":"05:12.730","Text":"then the integral minus the lower sum goes to 0,"},{"Start":"05:12.730 ","End":"05:14.780","Text":"which means that this tends to this,"},{"Start":"05:14.780 ","End":"05:19.900","Text":"and similarly, the upper sum tends to the integral."},{"Start":"05:20.090 ","End":"05:24.390","Text":"We can get the integral by taking the limit of either 1."},{"Start":"05:24.390 ","End":"05:25.690","Text":"Let\u0027s take the lower sum."},{"Start":"05:25.690 ","End":"05:30.099","Text":"The lower sum in the limit is the integral."},{"Start":"05:30.099 ","End":"05:32.065","Text":"Yeah, we could have chosen the upper,"},{"Start":"05:32.065 ","End":"05:33.700","Text":"had to choose 1."},{"Start":"05:33.700 ","End":"05:37.735","Text":"Now recall, we have the formula somewhere. Yeah, here."},{"Start":"05:37.735 ","End":"05:44.320","Text":"That the lower sum for Pn is equal to this expression, this finite series."},{"Start":"05:44.320 ","End":"05:49.235","Text":"For a given n, let theta equals Pi over 2n."},{"Start":"05:49.235 ","End":"05:51.595","Text":"This now simplifies to,"},{"Start":"05:51.595 ","End":"05:53.755","Text":"this is the term and we have theta,"},{"Start":"05:53.755 ","End":"05:59.090","Text":"2 theta n minus 1 theta, n theta."},{"Start":"05:59.090 ","End":"06:06.905","Text":"In sigma notation it\u0027s just the sum k goes from 1 to n of theta cosine k theta."},{"Start":"06:06.905 ","End":"06:09.104","Text":"Now if you look back,"},{"Start":"06:09.104 ","End":"06:12.895","Text":"hint number 1 was to prove this."},{"Start":"06:12.895 ","End":"06:15.670","Text":"Well, not exactly."},{"Start":"06:15.670 ","End":"06:18.040","Text":"The hint was written this way,"},{"Start":"06:18.040 ","End":"06:20.050","Text":"but I slightly rewrote it."},{"Start":"06:20.050 ","End":"06:24.290","Text":"It was 2k plus 1 over 2 is k plus 0.5."},{"Start":"06:24.290 ","End":"06:28.010","Text":"This identity just follows from 1 of"},{"Start":"06:28.010 ","End":"06:33.965","Text":"the standard trigonometric identities for sine Alpha cosine Beta,"},{"Start":"06:33.965 ","End":"06:35.765","Text":"which is this, well,"},{"Start":"06:35.765 ","End":"06:43.009","Text":"alpha is k plus 0.5 times theta and beta is k minus 0.5 times theta,"},{"Start":"06:43.009 ","End":"06:49.680","Text":"and I\u0027ll leave you to think about that to show that this does indeed imply this."},{"Start":"06:50.990 ","End":"06:54.665","Text":"From this, we can deduce this."},{"Start":"06:54.665 ","End":"06:56.675","Text":"Now, why am I concerned with this?"},{"Start":"06:56.675 ","End":"06:59.570","Text":"Because that is the typical term in this sum,"},{"Start":"06:59.570 ","End":"07:02.290","Text":"theta cosine k theta."},{"Start":"07:02.290 ","End":"07:06.990","Text":"How do I have to modify this to get this?"},{"Start":"07:06.990 ","End":"07:15.020","Text":"We have to throw in a theta in front and we have to divide by sine theta over 2."},{"Start":"07:15.020 ","End":"07:18.395","Text":"The sine theta over 2 goes here."},{"Start":"07:18.395 ","End":"07:21.070","Text":"The theta goes here,"},{"Start":"07:21.070 ","End":"07:25.130","Text":"it combines with the 0.5 and gives us theta over 2."},{"Start":"07:25.130 ","End":"07:29.625","Text":"This as is already you probably thinking,"},{"Start":"07:29.625 ","End":"07:32.795","Text":"this is like sine alpha over alpha,"},{"Start":"07:32.795 ","End":"07:36.575","Text":"but upside down because we know its limit and alpha goes to 0."},{"Start":"07:36.575 ","End":"07:40.550","Text":"Anyway, the lower sum, this here,"},{"Start":"07:40.550 ","End":"07:45.710","Text":"which is the sigma of theta cosine k theta means"},{"Start":"07:45.710 ","End":"07:51.400","Text":"the sum k equals 1 to n of this expression."},{"Start":"07:51.400 ","End":"07:53.815","Text":"Now, whatever doesn\u0027t have k in it,"},{"Start":"07:53.815 ","End":"07:56.645","Text":"like this, I can pull it in front of this sigma."},{"Start":"07:56.645 ","End":"07:59.755","Text":"I have sigma of this."},{"Start":"07:59.755 ","End":"08:02.025","Text":"This is where we got 2,"},{"Start":"08:02.025 ","End":"08:09.620","Text":"the lowest sum for Pn is this expression which expands to this."},{"Start":"08:09.620 ","End":"08:15.710","Text":"The reason this is better for us is this is what is called a telescoping series."},{"Start":"08:15.710 ","End":"08:17.690","Text":"I\u0027ll develop it 2 ways."},{"Start":"08:17.690 ","End":"08:21.065","Text":"First, I\u0027ll show you the telescoping series way."},{"Start":"08:21.065 ","End":"08:26.405","Text":"Just start writing out individual members of this series."},{"Start":"08:26.405 ","End":"08:29.315","Text":"If k is 1,"},{"Start":"08:29.315 ","End":"08:32.555","Text":"then we have 1.5 and 0.5."},{"Start":"08:32.555 ","End":"08:35.780","Text":"If k is 2, we have 2.5,"},{"Start":"08:35.780 ","End":"08:39.500","Text":"1.5, then 3.5, 2.5."},{"Start":"08:39.500 ","End":"08:43.715","Text":"Next to last is n minus 0.5 n minus 1.5,"},{"Start":"08:43.715 ","End":"08:45.410","Text":"where k is n minus 1."},{"Start":"08:45.410 ","End":"08:49.550","Text":"Then when k is n, we get n plus 0.5, n minus 0.5."},{"Start":"08:49.550 ","End":"08:54.259","Text":"Now I\u0027ve colored it in such a way that everything in green cancels."},{"Start":"08:54.259 ","End":"08:57.860","Text":"The 1.5 cancels with 1.5 because it\u0027s plus and minus,"},{"Start":"08:57.860 ","End":"09:03.105","Text":"2.5 with 2.5, 3.5, well, with dot-dot-dot."},{"Start":"09:03.105 ","End":"09:07.880","Text":"But everything cancels, and the only things that are left are sine of n"},{"Start":"09:07.880 ","End":"09:13.160","Text":"plus 0.5 theta and minus sine of 0.5 theta."},{"Start":"09:13.160 ","End":"09:15.750","Text":"Let\u0027s do it more formally now."},{"Start":"09:15.750 ","End":"09:21.320","Text":"What we\u0027re going to do is a substitution in the second part."},{"Start":"09:21.320 ","End":"09:24.780","Text":"Instead of k, going from 1 to n,"},{"Start":"09:24.780 ","End":"09:33.650","Text":"we can replace k by k plus 1 and then start 1 lower from 0 to n minus 1,"},{"Start":"09:33.650 ","End":"09:36.465","Text":"k plus 1 minus0.5 is k plus 0.5."},{"Start":"09:36.465 ","End":"09:40.400","Text":"Now we see that these 2 are almost the same."},{"Start":"09:40.400 ","End":"09:43.040","Text":"The difference is the start and the end point."},{"Start":"09:43.040 ","End":"09:48.230","Text":"All the middle bit is common from 1 to n minus 1 it\u0027s in common."},{"Start":"09:48.230 ","End":"09:50.930","Text":"We just have 1 extra term here for n,"},{"Start":"09:50.930 ","End":"09:53.830","Text":"and 1 extra term here for the 0."},{"Start":"09:53.830 ","End":"09:56.990","Text":"What we\u0027re left with is k equals n here,"},{"Start":"09:56.990 ","End":"10:01.295","Text":"which is this, and k equals 0 here, which is this."},{"Start":"10:01.295 ","End":"10:04.880","Text":"This bit is exactly the same as this bit here,"},{"Start":"10:04.880 ","End":"10:07.160","Text":"so we know we\u0027re okay."},{"Start":"10:07.160 ","End":"10:12.185","Text":"I remember theta was just a shortcut to tidy up,"},{"Start":"10:12.185 ","End":"10:14.690","Text":"if theta is Pi over n,"},{"Start":"10:14.690 ","End":"10:18.530","Text":"and also note that when n goes to infinity,"},{"Start":"10:18.530 ","End":"10:21.480","Text":"then theta goes to 0."},{"Start":"10:22.220 ","End":"10:30.385","Text":"The limit of this is the limit as n goes to infinity of this, and what is this?"},{"Start":"10:30.385 ","End":"10:33.405","Text":"Replace theta by Pi over 2n."},{"Start":"10:33.405 ","End":"10:37.515","Text":"Pi over 2n times n is pi over 2,"},{"Start":"10:37.515 ","End":"10:40.970","Text":"pi over 2n times 0.5 is Pi over 4n."},{"Start":"10:40.970 ","End":"10:44.140","Text":"Similarly here, pi over 4n."},{"Start":"10:44.140 ","End":"10:48.310","Text":"Now, when n goes to infinity, theta goes to 0,"},{"Start":"10:48.310 ","End":"10:50.020","Text":"and this is a famous limit,"},{"Start":"10:50.020 ","End":"10:51.130","Text":"is just upside down,"},{"Start":"10:51.130 ","End":"10:54.640","Text":"doesn\u0027t matter, so 1 over 1 is still 1."},{"Start":"10:54.640 ","End":"10:56.500","Text":"This bit goes to 0,"},{"Start":"10:56.500 ","End":"10:58.570","Text":"so we get sine pi over 2,"},{"Start":"10:58.570 ","End":"11:02.675","Text":"and here also sine 0."},{"Start":"11:02.675 ","End":"11:06.400","Text":"Sine of pi over 2 is 1."},{"Start":"11:06.400 ","End":"11:07.970","Text":"It\u0027s like sine of 90 degrees,"},{"Start":"11:07.970 ","End":"11:10.805","Text":"sine of 0 is 0."},{"Start":"11:10.805 ","End":"11:13.940","Text":"This just comes out to be 1."},{"Start":"11:13.940 ","End":"11:16.835","Text":"Which means that the integral,"},{"Start":"11:16.835 ","End":"11:20.430","Text":"because we said that this limit is the integral."},{"Start":"11:20.430 ","End":"11:22.170","Text":"The integral is 1,"},{"Start":"11:22.170 ","End":"11:23.700","Text":"and let\u0027s just spell it out,"},{"Start":"11:23.700 ","End":"11:26.595","Text":"f of x was cosine x,"},{"Start":"11:26.595 ","End":"11:29.810","Text":"so we computed the integral from 0 to pi over 2 of"},{"Start":"11:29.810 ","End":"11:34.940","Text":"cosine x to be equal to 1 using Riemann sums."},{"Start":"11:34.940 ","End":"11:37.770","Text":"That concludes this exercise."}],"ID":24688},{"Watched":false,"Name":"Partition and Refinements","Duration":"10m 40s","ChapterTopicVideoID":23762,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23762.jpeg","UploadDate":"2021-01-08T10:53:42.0030000","DurationForVideoObject":"PT10M40S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.615","Text":"At this point, we\u0027ve already defined the Riemann integral."},{"Start":"00:03.615 ","End":"00:06.390","Text":"We\u0027ve defined what it means for a function to be"},{"Start":"00:06.390 ","End":"00:09.890","Text":"integrable and what the actual value of the Riemann integral is."},{"Start":"00:09.890 ","End":"00:13.710","Text":"What we have is not powerful enough to do any real work,"},{"Start":"00:13.710 ","End":"00:14.790","Text":"it\u0027s not useful enough,"},{"Start":"00:14.790 ","End":"00:18.945","Text":"we need to develop some tools and that\u0027s what we\u0027re going to do next."},{"Start":"00:18.945 ","End":"00:20.400","Text":"To start with that,"},{"Start":"00:20.400 ","End":"00:24.870","Text":"we\u0027ll need some concepts and partition we\u0027ve seen already,"},{"Start":"00:24.870 ","End":"00:28.425","Text":"but we need to know what a refinement of a partition is."},{"Start":"00:28.425 ","End":"00:32.820","Text":"Suppose we have 2 partitions: P_1 and P_2 of the same interval a,"},{"Start":"00:32.820 ","End":"00:37.930","Text":"b, then sometimes one is said to be finer than the other."},{"Start":"00:37.930 ","End":"00:41.280","Text":"If P_2 has more points in it than P_1,"},{"Start":"00:41.280 ","End":"00:44.430","Text":"it has P_1 plus some extra partition points,"},{"Start":"00:44.430 ","End":"00:47.660","Text":"then we say that the partition is fine or P_2 is finer than"},{"Start":"00:47.660 ","End":"00:52.465","Text":"P_1 and we also say that P_2 is a refinement of P_1."},{"Start":"00:52.465 ","End":"00:55.430","Text":"In a simple example,"},{"Start":"00:55.430 ","End":"00:57.380","Text":"let\u0027s take the interval 0,"},{"Start":"00:57.380 ","End":"00:59.570","Text":"1 and look at 3 partitions, P,"},{"Start":"00:59.570 ","End":"01:04.735","Text":"Q and R, and see which is a refinement and which isn\u0027t."},{"Start":"01:04.735 ","End":"01:07.230","Text":"0, 1/2, 1 here,"},{"Start":"01:07.230 ","End":"01:10.005","Text":"we jump every 1/3 and here every 1/4."},{"Start":"01:10.005 ","End":"01:13.880","Text":"Now, Q is not a refinement of P,"},{"Start":"01:13.880 ","End":"01:17.210","Text":"because 1/2 is in here and it\u0027s not in here."},{"Start":"01:17.210 ","End":"01:20.845","Text":"A refinement has to contain these points and more."},{"Start":"01:20.845 ","End":"01:24.075","Text":"R is not a refinement of Q,"},{"Start":"01:24.075 ","End":"01:27.240","Text":"because 1/3 and 2/3 are not in here,"},{"Start":"01:27.240 ","End":"01:31.130","Text":"but R is a refinement of P. It\u0027s got all these points,"},{"Start":"01:31.130 ","End":"01:34.825","Text":"0, 1/2, 1 and extra stuff."},{"Start":"01:34.825 ","End":"01:39.860","Text":"Next a theorem and I\u0027ll bring the picture with it already."},{"Start":"01:39.860 ","End":"01:45.500","Text":"Suppose that f is a bounded function on the interval a,"},{"Start":"01:45.500 ","End":"01:47.269","Text":"b, and that we have 2 partitions,"},{"Start":"01:47.269 ","End":"01:54.785","Text":"P and Q of this interval and that Q is a refinement of P. There\u0027s 2 things we can say,"},{"Start":"01:54.785 ","End":"02:02.600","Text":"the lowest sum of Q and F is bigger or equal to the lower sum of P and F. In other words,"},{"Start":"02:02.600 ","End":"02:10.520","Text":"the finer partition makes the lower sum possibly grow and likewise for the upper sum,"},{"Start":"02:10.520 ","End":"02:16.465","Text":"the one which is finer has a smaller or equal to upper sum."},{"Start":"02:16.465 ","End":"02:19.200","Text":"For the picture that shows this,"},{"Start":"02:19.200 ","End":"02:25.535","Text":"the top 2 show the partition with 4 points,"},{"Start":"02:25.535 ","End":"02:28.465","Text":"a, another point, another point, and b,"},{"Start":"02:28.465 ","End":"02:32.405","Text":"and this is the upper sum and this is the lower sum."},{"Start":"02:32.405 ","End":"02:36.005","Text":"Then this next partition would be Q,"},{"Start":"02:36.005 ","End":"02:38.615","Text":"which has the same points as in here,"},{"Start":"02:38.615 ","End":"02:41.390","Text":"but extra points thrown in."},{"Start":"02:41.390 ","End":"02:46.245","Text":"Then we have the upper and lower sums for the finer partition."},{"Start":"02:46.245 ","End":"02:52.130","Text":"You notice that the upper sum goes down and the lower sum from here to here it goes up,"},{"Start":"02:52.130 ","End":"02:55.825","Text":"it gets closer and closer to the actual area."},{"Start":"02:55.825 ","End":"02:58.010","Text":"The proof for this is not hard,"},{"Start":"02:58.010 ","End":"02:59.450","Text":"so let\u0027s do it."},{"Start":"02:59.450 ","End":"03:03.620","Text":"What we\u0027ll do is we first consider a simple case where Q"},{"Start":"03:03.620 ","End":"03:08.195","Text":"just has one more point than P and then we do an induction."},{"Start":"03:08.195 ","End":"03:10.180","Text":"Let\u0027s start with this."},{"Start":"03:10.180 ","End":"03:16.530","Text":"P is x naught to x_n and Q is P union with single-point,"},{"Start":"03:16.530 ","End":"03:17.955","Text":"we call it x-star,"},{"Start":"03:17.955 ","End":"03:19.730","Text":"where x-star is not one of these,"},{"Start":"03:19.730 ","End":"03:23.644","Text":"it\u0027s strictly between some pair of these,"},{"Start":"03:23.644 ","End":"03:26.420","Text":"say between x_i minus 1 and x_i."},{"Start":"03:26.420 ","End":"03:29.285","Text":"Now, the point x-star actually splits"},{"Start":"03:29.285 ","End":"03:33.785","Text":"this interval x_i minus 1 x_i into 2 pieces from here to here,"},{"Start":"03:33.785 ","End":"03:35.830","Text":"and from here to here."},{"Start":"03:35.830 ","End":"03:44.240","Text":"Let\u0027s define, m prime is the infimum of f on the left piece"},{"Start":"03:44.240 ","End":"03:52.805","Text":"from x_i minus 1 to x-star and m double prime will be the infimum on the right-hand part."},{"Start":"03:52.805 ","End":"03:59.600","Text":"Now, each of these infimum is bigger or equal to the infimum on the whole."},{"Start":"03:59.600 ","End":"04:02.270","Text":"Because when you take the infimum of less,"},{"Start":"04:02.270 ","End":"04:04.665","Text":"it could possibly grow."},{"Start":"04:04.665 ","End":"04:06.890","Text":"The more you take the infimum of,"},{"Start":"04:06.890 ","End":"04:09.575","Text":"the more likely it is you\u0027ll get a lower candidate,"},{"Start":"04:09.575 ","End":"04:11.375","Text":"so this should be clear."},{"Start":"04:11.375 ","End":"04:21.049","Text":"What we have is that L of Q minus L of P is each of these is a sum, it\u0027s a Sigma."},{"Start":"04:21.049 ","End":"04:30.230","Text":"But the only difference is over the interval that was from x_i minus 1 to x_i,"},{"Start":"04:30.230 ","End":"04:34.145","Text":"what was 1 piece in P is 2 pieces in Q,"},{"Start":"04:34.145 ","End":"04:36.425","Text":"so only that term changes."},{"Start":"04:36.425 ","End":"04:40.490","Text":"Instead of m_i x_i minus x_i minus 1,"},{"Start":"04:40.490 ","End":"04:44.870","Text":"it breaks up into 2: the infimum of the left piece times"},{"Start":"04:44.870 ","End":"04:50.660","Text":"the Delta for the left bit and same for the right bit."},{"Start":"04:50.660 ","End":"04:52.565","Text":"We can rearrange it,"},{"Start":"04:52.565 ","End":"04:57.185","Text":"this part stays, but x_i minus x_i minus 1,"},{"Start":"04:57.185 ","End":"05:05.835","Text":"we can split as x_i minus x-star plus x-star minus x_i minus 1,"},{"Start":"05:05.835 ","End":"05:08.985","Text":"and the same m_i here and here."},{"Start":"05:08.985 ","End":"05:11.800","Text":"Now we can regroup."},{"Start":"05:11.800 ","End":"05:13.430","Text":"From here to here,"},{"Start":"05:13.430 ","End":"05:16.055","Text":"we just collect some like terms."},{"Start":"05:16.055 ","End":"05:22.475","Text":"Notice that we have x-star minus x_i minus 1 here,"},{"Start":"05:22.475 ","End":"05:27.145","Text":"and also x-star minus x_i minus 1 here,"},{"Start":"05:27.145 ","End":"05:30.525","Text":"and so collecting those 2,"},{"Start":"05:30.525 ","End":"05:36.795","Text":"we have m-prime, there\u0027s a minus and m_i."},{"Start":"05:36.795 ","End":"05:44.000","Text":"Then we can collect x_i minus x-star from here and here and that gives us"},{"Start":"05:44.000 ","End":"05:52.540","Text":"m double prime minus m_i times the same x_i minus x-star."},{"Start":"05:52.760 ","End":"05:57.350","Text":"Now, this is bigger or equal to 0 because"},{"Start":"05:57.350 ","End":"06:01.685","Text":"we see here that m-prime is bigger or equal to m_i."},{"Start":"06:01.685 ","End":"06:04.580","Text":"Similarly, this is bigger than 0,"},{"Start":"06:04.580 ","End":"06:07.910","Text":"which we see because this is bigger or equal to this."},{"Start":"06:07.910 ","End":"06:11.690","Text":"This is strictly positive because we have"},{"Start":"06:11.690 ","End":"06:15.890","Text":"this with strict inequalities and this is also positive."},{"Start":"06:15.890 ","End":"06:18.290","Text":"Anyway, altogether, we get something that\u0027s"},{"Start":"06:18.290 ","End":"06:23.510","Text":"non-negative and if this minus this is bigger or equal to 0,"},{"Start":"06:23.510 ","End":"06:27.450","Text":"then we can say that this is bigger or equal to this."},{"Start":"06:27.450 ","End":"06:32.580","Text":"That\u0027s the first half of these 2 and the other one is very similar,"},{"Start":"06:32.580 ","End":"06:35.120","Text":"so I\u0027ll go through it more quickly."},{"Start":"06:35.120 ","End":"06:39.050","Text":"We let m prime and m double prime be"},{"Start":"06:39.050 ","End":"06:44.725","Text":"the supremum of the left-half and the right-half, that interval."},{"Start":"06:44.725 ","End":"06:52.415","Text":"Actually it is so similar to what\u0027s above that I\u0027ll just expose it."},{"Start":"06:52.415 ","End":"06:56.005","Text":"You can pause and take a look at it. It\u0027s so similar."},{"Start":"06:56.005 ","End":"07:00.440","Text":"That takes care of the case where Q has one more point than P naught."},{"Start":"07:00.440 ","End":"07:02.615","Text":"If Q has n more points than P,"},{"Start":"07:02.615 ","End":"07:06.860","Text":"then we can apply induction on n. Basically what you do is you"},{"Start":"07:06.860 ","End":"07:11.089","Text":"add the points 1 at a time and the inequality keeps holding,"},{"Start":"07:11.089 ","End":"07:14.930","Text":"so the lower sum keeps"},{"Start":"07:14.930 ","End":"07:19.430","Text":"getting bigger or equal to bigger or equal to and in the end bigger or equal to."},{"Start":"07:19.430 ","End":"07:21.710","Text":"Similarly, the upper sum,"},{"Start":"07:21.710 ","End":"07:23.780","Text":"we get something that\u0027s still less than or equal to,"},{"Start":"07:23.780 ","End":"07:25.220","Text":"still less than or equal to,"},{"Start":"07:25.220 ","End":"07:30.335","Text":"so it can extend the result where n is more than 1 by induction."},{"Start":"07:30.335 ","End":"07:34.880","Text":"Still on the topic of refinement of partitions."},{"Start":"07:34.880 ","End":"07:37.700","Text":"Note that if we have any 2 partitions,"},{"Start":"07:37.700 ","End":"07:40.670","Text":"P and Q of an interval a, b,"},{"Start":"07:40.670 ","End":"07:43.490","Text":"they have a common refinement,"},{"Start":"07:43.490 ","End":"07:45.770","Text":"meaning a third partition,"},{"Start":"07:45.770 ","End":"07:49.025","Text":"which is a refinement of each one of them."},{"Start":"07:49.025 ","End":"07:52.010","Text":"The obvious thing to do is just to take"},{"Start":"07:52.010 ","End":"07:54.770","Text":"the union of the points in the first partition and"},{"Start":"07:54.770 ","End":"07:58.845","Text":"the second partition and then we get a partition which"},{"Start":"07:58.845 ","End":"08:03.170","Text":"contains P and it contains Q and therefore,"},{"Start":"08:03.170 ","End":"08:05.525","Text":"it\u0027s a common refinement."},{"Start":"08:05.525 ","End":"08:08.510","Text":"In fact, it\u0027s the smallest common refinement."},{"Start":"08:08.510 ","End":"08:11.180","Text":"Any other common refinement contains"},{"Start":"08:11.180 ","End":"08:16.615","Text":"R. Because if something contains P and something contains Q,"},{"Start":"08:16.615 ","End":"08:19.085","Text":"then it has to contain their union."},{"Start":"08:19.085 ","End":"08:23.450","Text":"An important corollary from this observation about"},{"Start":"08:23.450 ","End":"08:28.910","Text":"the common refinement is that if we have a function that\u0027s bounded on a,"},{"Start":"08:28.910 ","End":"08:33.170","Text":"b, then the lower Riemann integral is always"},{"Start":"08:33.170 ","End":"08:38.610","Text":"less than or equal to the upper Riemann integral of f on a, b."},{"Start":"08:38.610 ","End":"08:43.490","Text":"Here\u0027s the proof. Let P and Q be partitions of a, b."},{"Start":"08:43.490 ","End":"08:50.750","Text":"Now, the claim is that the lower sum for P is less than or equal to the upper sum of Q."},{"Start":"08:50.750 ","End":"08:56.395","Text":"In other words, every lower sum is less than or equal to every upper sum."},{"Start":"08:56.395 ","End":"09:01.430","Text":"The proof of this claim is if P and Q are given,"},{"Start":"09:01.430 ","End":"09:04.730","Text":"we take R to be a common partition of P and Q,"},{"Start":"09:04.730 ","End":"09:07.580","Text":"just like here and then we have"},{"Start":"09:07.580 ","End":"09:12.185","Text":"the following string of inequalities and I\u0027ll explain each one."},{"Start":"09:12.185 ","End":"09:17.690","Text":"This is less than or equal to this from the theorem,"},{"Start":"09:17.690 ","End":"09:19.370","Text":"that if you refine a partition,"},{"Start":"09:19.370 ","End":"09:22.205","Text":"the lower sum can only get larger."},{"Start":"09:22.205 ","End":"09:25.100","Text":"We also know that for a given partition,"},{"Start":"09:25.100 ","End":"09:27.920","Text":"the lower sum is less than or equal to the upper sum."},{"Start":"09:27.920 ","End":"09:30.485","Text":"Then again, using the theorem,"},{"Start":"09:30.485 ","End":"09:34.250","Text":"if you take a refinement then the upper sum for"},{"Start":"09:34.250 ","End":"09:36.710","Text":"the refinement is always less than or equal to"},{"Start":"09:36.710 ","End":"09:40.775","Text":"the upper sum of the one that we take the refinement of."},{"Start":"09:40.775 ","End":"09:42.860","Text":"We have a string of inequalities,"},{"Start":"09:42.860 ","End":"09:47.525","Text":"so this is less than or equal to this. We\u0027re almost there."},{"Start":"09:47.525 ","End":"09:55.280","Text":"Fix Q certain partition and take the supremum over all P. What we get is that"},{"Start":"09:55.280 ","End":"10:04.085","Text":"the lower Riemann integral is the supremum over all P of the lower sum for P,"},{"Start":"10:04.085 ","End":"10:07.880","Text":"this is also going to be less than or equal to the upper sum for"},{"Start":"10:07.880 ","End":"10:11.970","Text":"Q and F. Because each one of these is less than or equal to,"},{"Start":"10:11.970 ","End":"10:15.125","Text":"so the supremum is less than or equal to."},{"Start":"10:15.125 ","End":"10:20.165","Text":"Now, we can take the infimum over all of Q."},{"Start":"10:20.165 ","End":"10:23.120","Text":"We have that this is less than or equal to this for"},{"Start":"10:23.120 ","End":"10:28.280","Text":"each Q and so if we take the infimum over all Q of this,"},{"Start":"10:28.280 ","End":"10:30.290","Text":"and it\u0027s also bigger or equal to."},{"Start":"10:30.290 ","End":"10:33.544","Text":"But this infimum is the upper Riemann integral,"},{"Start":"10:33.544 ","End":"10:36.885","Text":"so this is less than or equal to this."},{"Start":"10:36.885 ","End":"10:40.580","Text":"This proves our corollary."}],"ID":24689},{"Watched":false,"Name":"Riemann s criterion for Integrability","Duration":"12m 41s","ChapterTopicVideoID":23763,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23763.jpeg","UploadDate":"2021-01-08T10:58:02.6130000","DurationForVideoObject":"PT12M41S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.860","Text":"In the previous clip,"},{"Start":"00:01.860 ","End":"00:04.605","Text":"we proved this corollary."},{"Start":"00:04.605 ","End":"00:06.690","Text":"Now we\u0027re going to use this for"},{"Start":"00:06.690 ","End":"00:13.275","Text":"a very important theorem called Riemann\u0027s criterion for integrability."},{"Start":"00:13.275 ","End":"00:19.380","Text":"Riemann\u0027s criteria is an if and only if condition for integrability."},{"Start":"00:19.380 ","End":"00:21.900","Text":"Now we\u0027re only talking about bounded functions here."},{"Start":"00:21.900 ","End":"00:25.890","Text":"We start off with a bounded function and then it is integrable on a,"},{"Start":"00:25.890 ","End":"00:28.950","Text":"b if and only if this condition."},{"Start":"00:28.950 ","End":"00:30.150","Text":"What is this condition?"},{"Start":"00:30.150 ","End":"00:33.740","Text":"For each Epsilon, there is a partition of,"},{"Start":"00:33.740 ","End":"00:35.190","Text":"a, b of course,"},{"Start":"00:35.190 ","End":"00:42.755","Text":"such that the upper sum minus the lower sum is less Epsilon."},{"Start":"00:42.755 ","End":"00:45.695","Text":"For the proof, we have to do both directions."},{"Start":"00:45.695 ","End":"00:47.210","Text":"Let\u0027s do this direction,"},{"Start":"00:47.210 ","End":"00:50.900","Text":"meaning that this condition holds about for each Epsilon"},{"Start":"00:50.900 ","End":"00:55.070","Text":"there is a P. We have to show that f is integrable."},{"Start":"00:55.070 ","End":"00:57.990","Text":"Let\u0027s say we\u0027re given Epsilon bigger than 0,"},{"Start":"00:57.990 ","End":"01:04.340","Text":"we choose P as in the conditions such that this minus this is less than Epsilon."},{"Start":"01:04.340 ","End":"01:07.310","Text":"Then by the corollary above,"},{"Start":"01:07.310 ","End":"01:10.130","Text":"this is less than or equal to this."},{"Start":"01:10.130 ","End":"01:15.080","Text":"The lower integral is less than or equal to the upper integral,"},{"Start":"01:15.080 ","End":"01:19.370","Text":"but this is the supremum of L of P,"},{"Start":"01:19.370 ","End":"01:22.950","Text":"f and it\u0027s the infimum of U of P,"},{"Start":"01:22.950 ","End":"01:28.170","Text":"f, so it\u0027s less than or equal to here and bigger or equal to here."},{"Start":"01:28.220 ","End":"01:33.125","Text":"Then this minus this, well it\u0027s non-negative,"},{"Start":"01:33.125 ","End":"01:37.205","Text":"but it\u0027s also less than or equal to this minus this,"},{"Start":"01:37.205 ","End":"01:39.560","Text":"because these are all ordered."},{"Start":"01:39.560 ","End":"01:42.635","Text":"This difference is less than this difference."},{"Start":"01:42.635 ","End":"01:46.430","Text":"This is less than Epsilon from here."},{"Start":"01:46.430 ","End":"01:50.705","Text":"This difference of the upper"},{"Start":"01:50.705 ","End":"01:55.160","Text":"integral minus the lower integral is sandwiched between 0 and Epsilon."},{"Start":"01:55.160 ","End":"01:57.215","Text":"This is true for all Epsilons."},{"Start":"01:57.215 ","End":"01:59.600","Text":"Letting Epsilon go to 0,"},{"Start":"01:59.600 ","End":"02:03.230","Text":"what we get is that this minus this is 0,"},{"Start":"02:03.230 ","End":"02:05.880","Text":"which means that these 2 are equal."},{"Start":"02:05.880 ","End":"02:09.130","Text":"By definition of integrable,"},{"Start":"02:09.130 ","End":"02:12.370","Text":"this means that f is integrable."},{"Start":"02:12.370 ","End":"02:15.200","Text":"That takes care of 1 direction."},{"Start":"02:15.200 ","End":"02:17.930","Text":"Now let\u0027s do the other direction."},{"Start":"02:17.930 ","End":"02:20.380","Text":"Suppose f is integrable,"},{"Start":"02:20.380 ","End":"02:24.995","Text":"we have to show that for each Epsilon there is P such that blah, blah, blah."},{"Start":"02:24.995 ","End":"02:28.000","Text":"Now because of this equality,"},{"Start":"02:28.000 ","End":"02:30.940","Text":"this is just the supremum of the lower sums"},{"Start":"02:30.940 ","End":"02:33.985","Text":"and this is the infimum of the upper sum so we get this."},{"Start":"02:33.985 ","End":"02:36.905","Text":"By definition of supremum and infimum,"},{"Start":"02:36.905 ","End":"02:38.760","Text":"we can find partitions."},{"Start":"02:38.760 ","End":"02:40.750","Text":"We can find 1 here and 1 here,"},{"Start":"02:40.750 ","End":"02:44.650","Text":"but we have to use a different letter, not the same 1,"},{"Start":"02:44.650 ","End":"02:48.305","Text":"such that for this partition P_2, the U,"},{"Start":"02:48.305 ","End":"02:52.350","Text":"it is less than this plus a 1/2 Epsilon,"},{"Start":"02:52.350 ","End":"02:54.740","Text":"which means that this minus this is less than 1/2 Epsilon."},{"Start":"02:54.740 ","End":"03:00.065","Text":"Similarly, on the other side we have this minus this is less than 1/2 Epsilon."},{"Start":"03:00.065 ","End":"03:02.975","Text":"You\u0027ll see why we chose 1/2 Epsilon in a moment."},{"Start":"03:02.975 ","End":"03:05.585","Text":"Now if we add these up,"},{"Start":"03:05.585 ","End":"03:12.065","Text":"what we get is that this minus this is less than Epsilon."},{"Start":"03:12.065 ","End":"03:14.960","Text":"When we add these 2 inequalities,"},{"Start":"03:14.960 ","End":"03:16.820","Text":"the 1/2 Epsilon plus 1/2 Epsilon gives us"},{"Start":"03:16.820 ","End":"03:19.490","Text":"the Epsilon and this integral is 1 to the minus,"},{"Start":"03:19.490 ","End":"03:21.140","Text":"and 1 to the plus cancels out,"},{"Start":"03:21.140 ","End":"03:23.110","Text":"so we get this."},{"Start":"03:23.110 ","End":"03:28.670","Text":"Now let\u0027s choose P to be a common refinement of P_1 and P_2."},{"Start":"03:28.670 ","End":"03:31.970","Text":"If you remember, a common refinement,"},{"Start":"03:31.970 ","End":"03:38.315","Text":"can possibly increase the lowest sum and possibly decrease the upper sum."},{"Start":"03:38.315 ","End":"03:41.450","Text":"What we get is the value U for P is less than or"},{"Start":"03:41.450 ","End":"03:44.750","Text":"equal to U for P_2 and similarly for the L,"},{"Start":"03:44.750 ","End":"03:46.540","Text":"or the lower sum."},{"Start":"03:46.540 ","End":"03:52.745","Text":"If we take something bigger or equal to here and something less than or equal to here,"},{"Start":"03:52.745 ","End":"03:55.534","Text":"the inequality can only get sharper."},{"Start":"03:55.534 ","End":"03:58.645","Text":"This will still be less than Epsilon."},{"Start":"03:58.645 ","End":"04:02.555","Text":"This means that this P is exactly what we\u0027re looking for."},{"Start":"04:02.555 ","End":"04:04.835","Text":"This is the P that we wanted."},{"Start":"04:04.835 ","End":"04:07.550","Text":"We\u0027re done with the proof."},{"Start":"04:07.550 ","End":"04:12.965","Text":"A corollary to this theorem or criterion."},{"Start":"04:12.965 ","End":"04:17.300","Text":"Suppose that f is bounded on a, b,"},{"Start":"04:17.300 ","End":"04:20.449","Text":"and suppose we have a sequence of partitions P_n,"},{"Start":"04:20.449 ","End":"04:23.139","Text":"where n runs over the natural numbers,"},{"Start":"04:23.139 ","End":"04:24.690","Text":"sequence of partitions of a,"},{"Start":"04:24.690 ","End":"04:32.230","Text":"b such that the limit of U of P_n minus L of P_n is 0,"},{"Start":"04:32.230 ","End":"04:36.380","Text":"well, as written here, then f is integrable."},{"Start":"04:36.380 ","End":"04:38.240","Text":"We are going to prove that;"},{"Start":"04:38.240 ","End":"04:40.325","Text":"by the above theorem,"},{"Start":"04:40.325 ","End":"04:44.570","Text":"I\u0027m using the Riemann criterion all we have to do is show that for"},{"Start":"04:44.570 ","End":"04:50.120","Text":"each Epsilon there exists a partition P of ab such that U of P,"},{"Start":"04:50.120 ","End":"04:53.080","Text":"f minus L of P, f is less than Epsilon."},{"Start":"04:53.080 ","End":"05:00.200","Text":"Now this expression, the U of P_n minus L of P_n goes to 0 as n goes to infinity."},{"Start":"05:00.200 ","End":"05:03.080","Text":"By definition of the limit of a sequence,"},{"Start":"05:03.080 ","End":"05:05.855","Text":"it means that for sufficiently large n,"},{"Start":"05:05.855 ","End":"05:09.244","Text":"this is less than any positive Epsilon we choose,"},{"Start":"05:09.244 ","End":"05:11.740","Text":"in particular for this Epsilon."},{"Start":"05:11.740 ","End":"05:18.290","Text":"Choose n and take P to be P_n for this particular n. Then U of P,"},{"Start":"05:18.290 ","End":"05:19.520","Text":"f minus L of P,"},{"Start":"05:19.520 ","End":"05:22.235","Text":"f will be less than Epsilon."},{"Start":"05:22.235 ","End":"05:27.590","Text":"By Riemann\u0027s criterion, f is integrable on a, b."},{"Start":"05:27.590 ","End":"05:33.440","Text":"Next, we\u0027ll do an example exercise using Riemann\u0027s criterion."},{"Start":"05:33.440 ","End":"05:36.560","Text":"We have the function f on the interval 0,"},{"Start":"05:36.560 ","End":"05:44.480","Text":"1 defined by f of x is equal to x for x equals 1 over n and it\u0027s equals 1,"},{"Start":"05:44.480 ","End":"05:46.790","Text":"2, 3, 4, 5, and so on."},{"Start":"05:46.790 ","End":"05:51.730","Text":"We\u0027ll let f be 0 except for these points."},{"Start":"05:51.730 ","End":"05:55.550","Text":"Our task is to show that f is integrable"},{"Start":"05:55.550 ","End":"05:59.680","Text":"on this interval and to find the value of the integral."},{"Start":"05:59.680 ","End":"06:03.320","Text":"Like I said, we\u0027ll use the Riemann criterion."},{"Start":"06:03.320 ","End":"06:06.790","Text":"Obviously, f is bounded on this interval."},{"Start":"06:06.790 ","End":"06:09.295","Text":"It\u0027s always between 0 and 1."},{"Start":"06:09.295 ","End":"06:14.310","Text":"Now that Epsilon bigger than 0 be given arbitrary."},{"Start":"06:14.310 ","End":"06:17.560","Text":"We can assume it\u0027s also less than 1 for convenience,"},{"Start":"06:17.560 ","End":"06:20.000","Text":"I mean, it goes to 0."},{"Start":"06:20.360 ","End":"06:24.820","Text":"We need to find a partition of 0,"},{"Start":"06:24.820 ","End":"06:26.740","Text":"1, call it P,"},{"Start":"06:26.740 ","End":"06:32.980","Text":"such that the upper sum minus the lower sum is less than Epsilon."},{"Start":"06:32.980 ","End":"06:39.715","Text":"Only finite number of 1 over n can lie in the interval from Epsilon to 1;"},{"Start":"06:39.715 ","End":"06:42.025","Text":"1 over 1, 1/2,"},{"Start":"06:42.025 ","End":"06:44.300","Text":"1/3, 1/4 and so on."},{"Start":"06:44.300 ","End":"06:49.705","Text":"At some point, 1 over n is going to be less than Epsilon."},{"Start":"06:49.705 ","End":"06:54.410","Text":"This is the formula for n. It\u0027s the floor function of 1 over Epsilon,"},{"Start":"06:54.410 ","End":"06:56.870","Text":"but that\u0027s less important."},{"Start":"06:56.870 ","End":"07:00.140","Text":"Now we can cover these 1 over n\u0027s,"},{"Start":"07:00.140 ","End":"07:01.880","Text":"the finite number of them,"},{"Start":"07:01.880 ","End":"07:07.050","Text":"by closed intervals, can label them like so."},{"Start":"07:07.050 ","End":"07:10.270","Text":"You can make them so that none are overlapping."},{"Start":"07:10.270 ","End":"07:13.160","Text":"We have these inequalities,"},{"Start":"07:13.160 ","End":"07:17.000","Text":"they\u0027re all increasing and the first one\u0027s bigger or equal to"},{"Start":"07:17.000 ","End":"07:21.105","Text":"Epsilon and the last one has to be 1."},{"Start":"07:21.105 ","End":"07:22.565","Text":"That\u0027s the endpoint."},{"Start":"07:22.565 ","End":"07:24.424","Text":"When I say cover,"},{"Start":"07:24.424 ","End":"07:32.870","Text":"it means that each of the 1 over n\u0027s that are in this set lies in 1 of these intervals."},{"Start":"07:32.870 ","End":"07:37.925","Text":"Furthermore, we can choose these intervals by shrinking them if necessary,"},{"Start":"07:37.925 ","End":"07:40.735","Text":"so that the sum of the lengths is less than Epsilon."},{"Start":"07:40.735 ","End":"07:43.790","Text":"We cover each 1 of these with a very thin,"},{"Start":"07:43.790 ","End":"07:48.095","Text":"narrow interval, that the sum is less than Epsilon."},{"Start":"07:48.095 ","End":"07:54.535","Text":"That, means that the sum from 1-N of x_2i minus x_2i minus 1 is less than Epsilon."},{"Start":"07:54.535 ","End":"07:59.270","Text":"Now let\u0027s consider the partition which is all these points,"},{"Start":"07:59.270 ","End":"08:03.425","Text":"x_1 up to x_2N and also Epsilon and 0."},{"Start":"08:03.425 ","End":"08:07.030","Text":"This is a partition of 0, 1."},{"Start":"08:07.030 ","End":"08:12.120","Text":"Altogether there are 2N plus 1 intervals, I."},{"Start":"08:12.120 ","End":"08:15.360","Text":"There are altogether 2N plus 2 points in here,"},{"Start":"08:15.360 ","End":"08:17.460","Text":"so 2N plus 1 intervals,"},{"Start":"08:17.460 ","End":"08:19.950","Text":"each one will be some I."},{"Start":"08:19.950 ","End":"08:23.069","Text":"For each such I that denote,"},{"Start":"08:23.069 ","End":"08:27.650","Text":"m_I is the infimum of the function on the interval big M_I,"},{"Start":"08:27.650 ","End":"08:30.125","Text":"the supremum of the function,"},{"Start":"08:30.125 ","End":"08:34.510","Text":"and Delta I is the width of I."},{"Start":"08:34.510 ","End":"08:38.750","Text":"We can sort these intervals into 3 kinds;"},{"Start":"08:38.750 ","End":"08:43.340","Text":"there\u0027s going to be 1 plus N plus N of these."},{"Start":"08:43.340 ","End":"08:45.295","Text":"Well, I\u0027ll show you what I mean."},{"Start":"08:45.295 ","End":"08:50.205","Text":"There\u0027s 1 interval which is just the 0 Epsilon."},{"Start":"08:50.205 ","End":"08:55.040","Text":"For this interval, the infimum is 0 because there"},{"Start":"08:55.040 ","End":"08:59.960","Text":"are points in this interval which are not of the form 1 over N,"},{"Start":"08:59.960 ","End":"09:04.105","Text":"so f is 0 on those points."},{"Start":"09:04.105 ","End":"09:11.455","Text":"There are N intervals which are exactly these that are on the line here,"},{"Start":"09:11.455 ","End":"09:13.790","Text":"x_2i minus 1, x_2i."},{"Start":"09:13.790 ","End":"09:16.084","Text":"On these intervals also,"},{"Start":"09:16.084 ","End":"09:18.980","Text":"the infimum is 0 because it\u0027s"},{"Start":"09:18.980 ","End":"09:22.730","Text":"an interval of positive length and it contains just 1 point,"},{"Start":"09:22.730 ","End":"09:26.270","Text":"where f is not 0 everywhere else is."},{"Start":"09:26.270 ","End":"09:30.500","Text":"The supremum is less than or equal to 1,"},{"Start":"09:30.500 ","End":"09:32.360","Text":"because on the whole interval 0,"},{"Start":"09:32.360 ","End":"09:34.610","Text":"1, it\u0027s less than or equal to 1."},{"Start":"09:34.610 ","End":"09:38.660","Text":"The width of this interval is just this."},{"Start":"09:38.660 ","End":"09:41.779","Text":"Note that each point,"},{"Start":"09:41.779 ","End":"09:46.090","Text":"1 over n is in 1 of these 2."},{"Start":"09:46.090 ","End":"09:50.255","Text":"It\u0027s in here if n is bigger than N,"},{"Start":"09:50.255 ","End":"09:55.100","Text":"or here if n is less than or equal to N. That\u0027s what we said here."},{"Start":"09:55.100 ","End":"09:57.590","Text":"All these points are in 1 of these."},{"Start":"09:57.590 ","End":"10:01.410","Text":"All the remainder are the last group,"},{"Start":"10:01.410 ","End":"10:04.385","Text":"the last kind, is N intervals also,"},{"Start":"10:04.385 ","End":"10:13.365","Text":"which contain no 1 over n. That would be from Epsilon to x_1 and from x_2 to x_3,"},{"Start":"10:13.365 ","End":"10:15.170","Text":"as the remaining ones."},{"Start":"10:15.170 ","End":"10:18.230","Text":"The infimum is 0,"},{"Start":"10:18.230 ","End":"10:22.040","Text":"supremum is also 0 because there are no 1"},{"Start":"10:22.040 ","End":"10:26.300","Text":"over n. The function is 0 everywhere in this third type of interval."},{"Start":"10:26.300 ","End":"10:28.320","Text":"The width, well, we don\u0027t care."},{"Start":"10:28.320 ","End":"10:31.360","Text":"We\u0027re going to be multiplying it by zeros, so don\u0027t care."},{"Start":"10:31.360 ","End":"10:36.140","Text":"Now the lowest sum for this partition is the sum"},{"Start":"10:36.140 ","End":"10:41.430","Text":"of little m_I Delta I for the intervals I."},{"Start":"10:41.430 ","End":"10:45.875","Text":"Because little m_i is 0 for all the kinds,"},{"Start":"10:45.875 ","End":"10:48.040","Text":"then this sum is going to be 0,"},{"Start":"10:48.040 ","End":"10:51.440","Text":"so the lowest sum is 0."},{"Start":"10:51.440 ","End":"10:57.110","Text":"Now the upper sum is the sum of big M times Delta,"},{"Start":"10:57.110 ","End":"10:59.540","Text":"and it depends on the interval."},{"Start":"10:59.540 ","End":"11:03.530","Text":"For type 3, the supremum is 0,"},{"Start":"11:03.530 ","End":"11:08.590","Text":"so I only have to sum these over the intervals in 1 and 2."},{"Start":"11:08.590 ","End":"11:12.860","Text":"For the second kind where I is this,"},{"Start":"11:12.860 ","End":"11:17.015","Text":"the supremum is less than or equal to 1."},{"Start":"11:17.015 ","End":"11:22.070","Text":"Delta I is x_2i minus x_2i minus 1."},{"Start":"11:22.070 ","End":"11:25.880","Text":"The sum for i equals 1 to n of these is"},{"Start":"11:25.880 ","End":"11:32.220","Text":"less than or equal to the sum of x_2i minus x_2 minus 1."},{"Start":"11:32.220 ","End":"11:37.950","Text":"This is less than Epsilon, still written here."},{"Start":"11:37.950 ","End":"11:41.040","Text":"For the first kind,"},{"Start":"11:41.040 ","End":"11:42.570","Text":"which is just one of them, the 0,"},{"Start":"11:42.570 ","End":"11:50.265","Text":"Epsilon big M_I Delta I is M_I is less than or equal to Epsilon."},{"Start":"11:50.265 ","End":"11:52.500","Text":"That\u0027s from here."},{"Start":"11:52.500 ","End":"11:55.920","Text":"The width is Epsilon."},{"Start":"11:55.920 ","End":"11:57.335","Text":"Epsilon is less than 1,"},{"Start":"11:57.335 ","End":"12:03.680","Text":"so it\u0027s less than or equal to or less than 1 times Epsilon, less than Epsilon."},{"Start":"12:03.680 ","End":"12:12.175","Text":"Altogether, the upper sum for the partition P is less than Epsilon plus Epsilon,"},{"Start":"12:12.175 ","End":"12:14.280","Text":"which is 2 Epsilon."},{"Start":"12:14.280 ","End":"12:18.450","Text":"Now that gives us this minus this is less than 2 Epsilon."},{"Start":"12:18.450 ","End":"12:20.309","Text":"Now we wanted Epsilon,"},{"Start":"12:20.309 ","End":"12:25.280","Text":"so all we have to do is start with Epsilon over 2 instead of Epsilon."},{"Start":"12:25.280 ","End":"12:32.040","Text":"In this computation, we can make this Epsilon over 2,"},{"Start":"12:32.040 ","End":"12:37.410","Text":"so we get twice Epsilon over 2 and that would give us Epsilon."},{"Start":"12:37.410 ","End":"12:41.050","Text":"That complete the exercise."}],"ID":24690},{"Watched":false,"Name":"Applications of Riemann s criterion for Integrability","Duration":"12m 15s","ChapterTopicVideoID":23764,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23764.jpeg","UploadDate":"2021-01-08T11:02:02.2770000","DurationForVideoObject":"PT12M15S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.110","Text":"In this clip, we\u0027ll talk about some of the applications or"},{"Start":"00:04.110 ","End":"00:07.935","Text":"uses of Riemann\u0027s criterion for integrability."},{"Start":"00:07.935 ","End":"00:11.420","Text":"We can prove a lot of interesting results,"},{"Start":"00:11.420 ","End":"00:16.380","Text":"for example that continuous functions are integrable on a closed interval."},{"Start":"00:16.380 ","End":"00:20.370","Text":"Anyway, let\u0027s first recall what is Riemann\u0027s criterion."},{"Start":"00:20.370 ","End":"00:23.040","Text":"It talks about a bounded function."},{"Start":"00:23.040 ","End":"00:24.840","Text":"It\u0027s an if and only if condition."},{"Start":"00:24.840 ","End":"00:29.130","Text":"F is integrable if and only if for each Epsilon there exists"},{"Start":"00:29.130 ","End":"00:36.040","Text":"a partition such that its upper sum minus lower sum is less than Epsilon."},{"Start":"00:36.190 ","End":"00:40.145","Text":"We can use this criterion to show that"},{"Start":"00:40.145 ","End":"00:44.870","Text":"continuous functions and also monotone functions are integrable."},{"Start":"00:44.870 ","End":"00:48.055","Text":"Let\u0027s start with the continuous ones."},{"Start":"00:48.055 ","End":"00:50.040","Text":"If f is continuous on a,"},{"Start":"00:50.040 ","End":"00:52.295","Text":"b then it\u0027s Riemann integrable."},{"Start":"00:52.295 ","End":"00:56.090","Text":"Now the proof, we first of all have to show that f is bounded,"},{"Start":"00:56.090 ","End":"00:58.670","Text":"but there\u0027s a boundedness theorem that says that"},{"Start":"00:58.670 ","End":"01:01.690","Text":"a continuous function on a closed interval is bounded."},{"Start":"01:01.690 ","End":"01:06.035","Text":"There\u0027s another theorem that says that on a closed interval or any compact set,"},{"Start":"01:06.035 ","End":"01:10.505","Text":"a continuous function is uniformly continuous."},{"Start":"01:10.505 ","End":"01:15.895","Text":"That Epsilon bigger than 0 be given by uniform continuity,"},{"Start":"01:15.895 ","End":"01:24.370","Text":"there is a Delta such that the distance between f of x and f of y is less than Epsilon,"},{"Start":"01:24.370 ","End":"01:27.310","Text":"but instead of Epsilon we can take Epsilon over b minus a."},{"Start":"01:27.310 ","End":"01:28.795","Text":"That\u0027s our new Epsilon,"},{"Start":"01:28.795 ","End":"01:35.065","Text":"and this is for all x and y in the interval whose distance is less than Delta."},{"Start":"01:35.065 ","End":"01:40.475","Text":"Now let\u0027s choose a partition whose mesh or norm is less than Delta."},{"Start":"01:40.475 ","End":"01:42.860","Text":"Such partitions exist, for example,"},{"Start":"01:42.860 ","End":"01:47.180","Text":"you could divide the interval into n intervals of equal length as"},{"Start":"01:47.180 ","End":"01:51.770","Text":"long as b minus a over n is less than Delta,"},{"Start":"01:51.770 ","End":"01:55.660","Text":"which will be true if n is bigger than b minus a over Delta."},{"Start":"01:55.660 ","End":"02:03.320","Text":"Consider an arbitrary interval in the partition x_i minus 1 x_i equal to big I."},{"Start":"02:03.320 ","End":"02:09.110","Text":"We\u0027ll let little m and big M with a subscript I denote the infimum,"},{"Start":"02:09.110 ","End":"02:14.060","Text":"supremum respectively of f of I on the interval."},{"Start":"02:14.060 ","End":"02:18.485","Text":"Because f is continuous by the extreme value theorem,"},{"Start":"02:18.485 ","End":"02:22.880","Text":"big M_i is attained at some point x and"},{"Start":"02:22.880 ","End":"02:27.985","Text":"little m_i is attained at some other point y in the interval."},{"Start":"02:27.985 ","End":"02:35.570","Text":"Recall that normal mesh if p is the maximum of the Delta x_i."},{"Start":"02:35.570 ","End":"02:39.065","Text":"Since x and y are in the interval,"},{"Start":"02:39.065 ","End":"02:43.820","Text":"the difference has to be less than x_i minus x_i minus 1,"},{"Start":"02:43.820 ","End":"02:45.560","Text":"which is Delta x_i,"},{"Start":"02:45.560 ","End":"02:48.830","Text":"which is less than or equal to the norm,"},{"Start":"02:48.830 ","End":"02:51.660","Text":"which is less than Delta."},{"Start":"02:51.660 ","End":"02:57.190","Text":"By the uniform continuity or what we said here,"},{"Start":"02:57.190 ","End":"03:00.770","Text":"we get that big M_i minus little m_i,"},{"Start":"03:00.770 ","End":"03:06.355","Text":"which is f of x minus f of y is less than Epsilon over b minus a."},{"Start":"03:06.355 ","End":"03:10.210","Text":"Now the upper sum minus the lower sum is the sum"},{"Start":"03:10.210 ","End":"03:14.915","Text":"of big M_i Delta x_i minus the sum of little m_i Delta x_i,"},{"Start":"03:14.915 ","End":"03:16.500","Text":"and that\u0027s equal to,"},{"Start":"03:16.500 ","End":"03:19.605","Text":"we can take the Delta x_i outside the brackets,"},{"Start":"03:19.605 ","End":"03:22.270","Text":"and also because this is bigger than this,"},{"Start":"03:22.270 ","End":"03:25.725","Text":"we can just stick in absolute value here."},{"Start":"03:25.725 ","End":"03:30.405","Text":"Here we noted that this is less than Epsilon over b minus a,"},{"Start":"03:30.405 ","End":"03:35.635","Text":"so this is going to be less than Epsilon over b minus a times the sum of the Delta x_i."},{"Start":"03:35.635 ","End":"03:40.490","Text":"Since this is equal to b minus a,"},{"Start":"03:40.490 ","End":"03:43.805","Text":"the b minus a cancels and we\u0027re left with Epsilon,"},{"Start":"03:43.805 ","End":"03:47.465","Text":"so U minus L is less than Epsilon."},{"Start":"03:47.465 ","End":"03:50.515","Text":"That concludes the proof of this theorem."},{"Start":"03:50.515 ","End":"03:54.845","Text":"Now let\u0027s show that a monotonic function is integrable."},{"Start":"03:54.845 ","End":"03:58.955","Text":"Again, we\u0027ll be using Riemann\u0027s criterion,"},{"Start":"03:58.955 ","End":"04:01.160","Text":"which I\u0027ve put here as a reminder,"},{"Start":"04:01.160 ","End":"04:02.645","Text":"I won\u0027t read it out."},{"Start":"04:02.645 ","End":"04:05.480","Text":"Let\u0027s assume that f is increasing because"},{"Start":"04:05.480 ","End":"04:08.060","Text":"monotonic means either increasing or decreasing,"},{"Start":"04:08.060 ","End":"04:11.270","Text":"and the case where it\u0027s decreasing will be very similar."},{"Start":"04:11.270 ","End":"04:13.010","Text":"F is bounded."},{"Start":"04:13.010 ","End":"04:14.300","Text":"If it is increasing,"},{"Start":"04:14.300 ","End":"04:17.360","Text":"then it\u0027s bigger or equal to f at a and less than or"},{"Start":"04:17.360 ","End":"04:20.825","Text":"equal to the value of f at b, so yeah, bounded."},{"Start":"04:20.825 ","End":"04:23.075","Text":"Given Epsilon bigger than 0,"},{"Start":"04:23.075 ","End":"04:26.390","Text":"we need to find P as in the theorem."},{"Start":"04:26.390 ","End":"04:33.410","Text":"Divide a, b into n equal pieces and you get the partition P_n, depends on n,"},{"Start":"04:33.410 ","End":"04:38.270","Text":"where the width of each interval is b minus"},{"Start":"04:38.270 ","End":"04:41.630","Text":"a over n. It\u0027s a constant for all i from 1 to"},{"Start":"04:41.630 ","End":"04:45.514","Text":"n. We\u0027re going to show that if n is big enough,"},{"Start":"04:45.514 ","End":"04:51.245","Text":"then the upper sum minus the lower sum for this partition is less than Epsilon."},{"Start":"04:51.245 ","End":"04:54.080","Text":"The difference is equal to the sum of"},{"Start":"04:54.080 ","End":"04:58.780","Text":"big M_i Delta x_i minus the sum of little m_i Delta x_i."},{"Start":"04:58.780 ","End":"05:01.440","Text":"Now each of the Delta x_i is b minus a over n,"},{"Start":"05:01.440 ","End":"05:03.779","Text":"so we can bring it in front of the Sigma,"},{"Start":"05:03.779 ","End":"05:07.035","Text":"and we have the sum of M_i minus m_i."},{"Start":"05:07.035 ","End":"05:11.600","Text":"If you think about it, little m_i is simply the value of the function at"},{"Start":"05:11.600 ","End":"05:14.510","Text":"the left end point of the interval and"},{"Start":"05:14.510 ","End":"05:18.815","Text":"big M_i is the value of f at the right end point of the interval."},{"Start":"05:18.815 ","End":"05:22.235","Text":"Talking about the interval x_i minus 1 x_i."},{"Start":"05:22.235 ","End":"05:26.435","Text":"Just have to add n of these for each i."},{"Start":"05:26.435 ","End":"05:29.340","Text":"The right minus the left,"},{"Start":"05:29.440 ","End":"05:34.270","Text":"and this becomes a telescopic series and"},{"Start":"05:34.270 ","End":"05:39.350","Text":"all the middle terms cancel out and we\u0027re just left with the last minus, the first,"},{"Start":"05:39.350 ","End":"05:42.485","Text":"because everything else appears in a minus and then in a plus,"},{"Start":"05:42.485 ","End":"05:47.815","Text":"and this is f of b minus f of a. Spelling it out here,"},{"Start":"05:47.815 ","End":"05:50.000","Text":"if we just reverse the order,"},{"Start":"05:50.000 ","End":"05:53.240","Text":"it\u0027s easier to take it as minus f of x_i minus 1 plus f of x_i,"},{"Start":"05:53.240 ","End":"05:54.980","Text":"so we have minus this plus this,"},{"Start":"05:54.980 ","End":"05:56.525","Text":"and then minus this plus this,"},{"Start":"05:56.525 ","End":"06:01.135","Text":"and all the middle ones cancel out and here this is what we have."},{"Start":"06:01.135 ","End":"06:06.020","Text":"This upper sum minus the lower sum is b minus a over n,"},{"Start":"06:06.020 ","End":"06:08.015","Text":"f of b minus f of a."},{"Start":"06:08.015 ","End":"06:12.515","Text":"Everything\u0027s a constant except n. If we take n big enough,"},{"Start":"06:12.515 ","End":"06:15.490","Text":"surely we\u0027ll get less than any given Epsilon,"},{"Start":"06:15.490 ","End":"06:18.570","Text":"and that concludes the proof of this theorem,"},{"Start":"06:18.570 ","End":"06:22.805","Text":"so monotonic functions are integrable."},{"Start":"06:22.805 ","End":"06:27.080","Text":"Now a small topic that doesn\u0027t really fit in anywhere,"},{"Start":"06:27.080 ","End":"06:29.240","Text":"it doesn\u0027t relate to the Riemann criterion,"},{"Start":"06:29.240 ","End":"06:30.920","Text":"but we\u0027ll put it here."},{"Start":"06:30.920 ","End":"06:34.385","Text":"It\u0027s a question that might come up."},{"Start":"06:34.385 ","End":"06:39.650","Text":"Can you interchange the limit and the integral operations?"},{"Start":"06:39.650 ","End":"06:43.280","Text":"To be specific, suppose we have a sequence of"},{"Start":"06:43.280 ","End":"06:47.390","Text":"functions f_n that converge to f point-wise on the interval a,"},{"Start":"06:47.390 ","End":"06:52.280","Text":"b and suppose that all the functions here are integrable,"},{"Start":"06:52.280 ","End":"06:59.880","Text":"is it always true that the integral of f_n turns to the integral of f?"},{"Start":"06:59.880 ","End":"07:04.535","Text":"Rephrasing that, is the limit of the integral of"},{"Start":"07:04.535 ","End":"07:09.470","Text":"f_n equal to the integral of the limit of fn?"},{"Start":"07:09.470 ","End":"07:12.860","Text":"In other words, can you exchange limit an integral?"},{"Start":"07:12.860 ","End":"07:15.230","Text":"The general answer is no,"},{"Start":"07:15.230 ","End":"07:18.500","Text":"and we just need 1 counterexample for that."},{"Start":"07:18.500 ","End":"07:21.080","Text":"That\u0027ll be this example exercise."},{"Start":"07:21.080 ","End":"07:26.360","Text":"We have to define a sequence f_n of functions on 0,"},{"Start":"07:26.360 ","End":"07:31.904","Text":"1 by f_n of 1 is 0 and f_n for"},{"Start":"07:31.904 ","End":"07:38.040","Text":"everything except 1 is nx^n minus 1 over 1 plus x."},{"Start":"07:38.040 ","End":"07:43.100","Text":"We have to show that the limit of the integral is 1/2,"},{"Start":"07:43.100 ","End":"07:46.175","Text":"whereas the integral of the limit is 0,"},{"Start":"07:46.175 ","End":"07:49.379","Text":"and certainly 1/2 is not equal to 0."},{"Start":"07:49.379 ","End":"07:54.800","Text":"Let\u0027s define f to be the limit of f_n."},{"Start":"07:54.800 ","End":"07:56.540","Text":"What is this function f?"},{"Start":"07:56.540 ","End":"07:58.550","Text":"In other words, what\u0027s the limit of these?"},{"Start":"07:58.550 ","End":"08:07.550","Text":"My claim is that f of x is equal to 0 for x bigger or equal to 0 and less than 1."},{"Start":"08:07.550 ","End":"08:11.990","Text":"Then we\u0027ll use the ratio test for limits of sequences."},{"Start":"08:11.990 ","End":"08:20.075","Text":"The ratio test says that if you take the ratio of 2 consecutive terms,"},{"Start":"08:20.075 ","End":"08:24.065","Text":"assuming everything\u0027s positive, we\u0027re not dividing by 0,"},{"Start":"08:24.065 ","End":"08:27.500","Text":"suppose this ratio turns to L,"},{"Start":"08:27.500 ","End":"08:30.470","Text":"and suppose that this L is less than 1,"},{"Start":"08:30.470 ","End":"08:36.510","Text":"then the sequence converges and in fact the limit is 0."},{"Start":"08:36.510 ","End":"08:40.365","Text":"In our case, f_n plus 1 over f_n of x,"},{"Start":"08:40.365 ","End":"08:45.265","Text":"where x is between 0 and 1 excluding 1,"},{"Start":"08:45.265 ","End":"08:50.185","Text":"is equal to this divided by this."},{"Start":"08:50.185 ","End":"08:52.125","Text":"Things cancel."},{"Start":"08:52.125 ","End":"08:54.420","Text":"The 1 plus x cancels,"},{"Start":"08:54.420 ","End":"08:58.860","Text":"and x^n over x^n minus 1 comes out x,"},{"Start":"08:58.860 ","End":"09:01.905","Text":"and n plus 1 over n is 1 plus 1 over n,"},{"Start":"09:01.905 ","End":"09:06.210","Text":"and this turns to x, and so f of x,"},{"Start":"09:06.210 ","End":"09:09.859","Text":"which is the point y is limit of the f_n is 0,"},{"Start":"09:09.859 ","End":"09:11.690","Text":"provided x is less than 1."},{"Start":"09:11.690 ","End":"09:12.980","Text":"As a matter of fact,"},{"Start":"09:12.980 ","End":"09:15.125","Text":"f_n of 1 is always 0,"},{"Start":"09:15.125 ","End":"09:19.940","Text":"so this limit is also 0. Anyway, we don\u0027t need that."},{"Start":"09:19.940 ","End":"09:25.310","Text":"We don\u0027t care what f of 1 is because at a single point it doesn\u0027t matter."},{"Start":"09:25.310 ","End":"09:29.360","Text":"Recall the result that if we have an integrable function and"},{"Start":"09:29.360 ","End":"09:34.430","Text":"another function differs from it at a finite number of points,"},{"Start":"09:34.430 ","End":"09:39.710","Text":"then that function is also integrable and they have the same integral."},{"Start":"09:39.710 ","End":"09:45.020","Text":"In our case, f is equal to 0 except perhaps at x equals 1,"},{"Start":"09:45.020 ","End":"09:48.304","Text":"it\u0027s actually equal everywhere, but never mind."},{"Start":"09:48.304 ","End":"09:52.215","Text":"I put that in brackets since we don\u0027t need to know that."},{"Start":"09:52.215 ","End":"09:56.460","Text":"The limit of f_n,"},{"Start":"09:56.460 ","End":"10:00.110","Text":"which is the integral of f of x dx is the same as the integral"},{"Start":"10:00.110 ","End":"10:04.024","Text":"of the 0 function and that is equal to 0."},{"Start":"10:04.024 ","End":"10:07.100","Text":"That\u0027s the second half."},{"Start":"10:07.100 ","End":"10:09.185","Text":"That\u0027s this 0 part."},{"Start":"10:09.185 ","End":"10:12.020","Text":"Now we need to do the integral and then the limit."},{"Start":"10:12.020 ","End":"10:17.025","Text":"So f_n is this except at x equals 1."},{"Start":"10:17.025 ","End":"10:23.755","Text":"The integral of f_n from 0 to 1 is the integral of this."},{"Start":"10:23.755 ","End":"10:26.300","Text":"F_n is equal to this except at one point,"},{"Start":"10:26.300 ","End":"10:27.380","Text":"so it doesn\u0027t matter,"},{"Start":"10:27.380 ","End":"10:31.250","Text":"so we can assume that f_n is everywhere equal to this."},{"Start":"10:31.250 ","End":"10:34.310","Text":"Then we\u0027ll do an integration by parts."},{"Start":"10:34.310 ","End":"10:35.869","Text":"This is fairly routine,"},{"Start":"10:35.869 ","End":"10:38.930","Text":"so I won\u0027t go into this in detail,"},{"Start":"10:38.930 ","End":"10:42.770","Text":"reminding you of the formula for integration by parts."},{"Start":"10:42.770 ","End":"10:46.475","Text":"Using this formula, we end up with this result,"},{"Start":"10:46.475 ","End":"10:49.885","Text":"but we still have an unknown integral here."},{"Start":"10:49.885 ","End":"10:52.670","Text":"We\u0027re not actually going to evaluate this."},{"Start":"10:52.670 ","End":"10:57.215","Text":"Just have to show that this turns to 0 as n goes to infinity."},{"Start":"10:57.215 ","End":"11:01.840","Text":"Since 1 over 1 plus x squared is less than or equal to 1,"},{"Start":"11:01.840 ","End":"11:07.780","Text":"and that\u0027s because 1 plus x is bigger than 1,"},{"Start":"11:07.780 ","End":"11:13.850","Text":"then we get that x to the n over 1 plus x squared is less than or equal to x^n,"},{"Start":"11:13.850 ","End":"11:17.705","Text":"and so the integral is less than or equal to the integral."},{"Start":"11:17.705 ","End":"11:21.860","Text":"But this integral, which is x^n plus 1"},{"Start":"11:21.860 ","End":"11:26.345","Text":"over n plus 1 between 0 and 1 comes out to be 1 over n plus 1,"},{"Start":"11:26.345 ","End":"11:30.200","Text":"and it goes to 0 as n goes to infinity."},{"Start":"11:30.200 ","End":"11:34.500","Text":"The limit of f_n,"},{"Start":"11:34.500 ","End":"11:38.325","Text":"the integral as n goes to infinity,"},{"Start":"11:38.325 ","End":"11:48.905","Text":"is this limit which is 1/2 plus the limit of this which we just showed is equal to 0,"},{"Start":"11:48.905 ","End":"11:51.715","Text":"so this is equal to 1/2."},{"Start":"11:51.715 ","End":"12:00.230","Text":"The limit of the integral is 1/2 as opposed to the integral of the limit which is 0."},{"Start":"12:00.230 ","End":"12:04.835","Text":"The question we asked has the answer,"},{"Start":"12:04.835 ","End":"12:09.260","Text":"no, we cannot exchange limit and integral operations in general."},{"Start":"12:09.260 ","End":"12:12.395","Text":"In a specific situation it might be true."},{"Start":"12:12.395 ","End":"12:15.690","Text":"Okay, That\u0027s enough for this clip."}],"ID":24691},{"Watched":false,"Name":"Exercise 3","Duration":"4m 54s","ChapterTopicVideoID":23765,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23765.jpeg","UploadDate":"2021-01-08T11:03:25.6170000","DurationForVideoObject":"PT4M54S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.670","Text":"When you paraphrase this exercise,"},{"Start":"00:02.670 ","End":"00:05.790","Text":"basically we have to show that the square root of"},{"Start":"00:05.790 ","End":"00:10.305","Text":"an integrable function is also integrable provided of course,"},{"Start":"00:10.305 ","End":"00:13.380","Text":"that that function is non-negative otherwise how we\u0027re going to take"},{"Start":"00:13.380 ","End":"00:17.550","Text":"the square root and more precisely is how it\u0027s written here."},{"Start":"00:17.550 ","End":"00:20.640","Text":"Our main tool will be the Riemann criterion for"},{"Start":"00:20.640 ","End":"00:26.010","Text":"integrability and if and only if criterion for bounded functions to be integrable."},{"Start":"00:26.010 ","End":"00:30.980","Text":"What it says is that a bounded function f is integrable on a,"},{"Start":"00:30.980 ","End":"00:37.970","Text":"b if and only if every Epsilon has a partition subject to certain conditions"},{"Start":"00:37.970 ","End":"00:41.150","Text":"and that condition is that the upper sum for"},{"Start":"00:41.150 ","End":"00:46.285","Text":"this partition minus the lower sum for this partition is less than Epsilon."},{"Start":"00:46.285 ","End":"00:49.995","Text":"Because f is integrable,"},{"Start":"00:49.995 ","End":"00:53.535","Text":"it\u0027s also bounded on a, b."},{"Start":"00:53.535 ","End":"00:57.500","Text":"There are constant c and d such that f of"},{"Start":"00:57.500 ","End":"01:01.710","Text":"x is sandwiched between these 2 constants on the interval."},{"Start":"01:01.710 ","End":"01:08.645","Text":"We can take the square root of this inequality so square root of f is bounded."},{"Start":"01:08.645 ","End":"01:11.180","Text":"Now that we know it\u0027s bounded,"},{"Start":"01:11.180 ","End":"01:14.960","Text":"we\u0027re going to use this criterion here but instead of f,"},{"Start":"01:14.960 ","End":"01:17.720","Text":"we\u0027re going to use it with square root of f. In other word,"},{"Start":"01:17.720 ","End":"01:19.955","Text":"we have to show that given Epsilon,"},{"Start":"01:19.955 ","End":"01:25.910","Text":"we can produce the partition P such that this is true except with square root of"},{"Start":"01:25.910 ","End":"01:30.185","Text":"f. Let Epsilon bigger than 0 be given"},{"Start":"01:30.185 ","End":"01:35.160","Text":"and we need to find p such that you have P and square root of f minus L,"},{"Start":"01:35.160 ","End":"01:37.810","Text":"P and square root of f less than Epsilon."},{"Start":"01:37.810 ","End":"01:44.105","Text":"Now, f is integrable so by the Riemann criterion,"},{"Start":"01:44.105 ","End":"01:46.820","Text":"we can use it in this direction."},{"Start":"01:46.820 ","End":"01:51.620","Text":"We can choose a partition P like so such that the upper sum"},{"Start":"01:51.620 ","End":"01:56.000","Text":"minus the lower sum for f is less than Epsilon 1."},{"Start":"01:56.000 ","End":"01:58.250","Text":"You don\u0027t have to use the same Epsilon."},{"Start":"01:58.250 ","End":"02:03.685","Text":"In fact, going to let Epsilon 1 be 2 square root of c Epsilon."},{"Start":"02:03.685 ","End":"02:06.080","Text":"You\u0027ll see why this choice when we get to the end,"},{"Start":"02:06.080 ","End":"02:08.465","Text":"is like reverse engineered."},{"Start":"02:08.465 ","End":"02:10.645","Text":"That\u0027s the thing that will make it work."},{"Start":"02:10.645 ","End":"02:14.750","Text":"Label m_i and M_i to be"},{"Start":"02:14.750 ","End":"02:20.065","Text":"the supremum and infimum respectively of the function f on the sub-interval"},{"Start":"02:20.065 ","End":"02:26.580","Text":"X_i minus 1 x_i and choose another letter L. Let l_i"},{"Start":"02:26.580 ","End":"02:29.540","Text":"and L_i be the corresponding supremum and"},{"Start":"02:29.540 ","End":"02:34.445","Text":"infimum for square root of f on the same sub-interval."},{"Start":"02:34.445 ","End":"02:39.080","Text":"Now, because the square root is a monotonically increasing function,"},{"Start":"02:39.080 ","End":"02:41.569","Text":"which preserves the order,"},{"Start":"02:41.569 ","End":"02:46.160","Text":"the square root of the infimum is the infimum"},{"Start":"02:46.160 ","End":"02:48.110","Text":"of the square root and the square root of"},{"Start":"02:48.110 ","End":"02:51.485","Text":"the supremum is the supremum of the square root."},{"Start":"02:51.485 ","End":"02:59.675","Text":"Now we want to evaluate this expression and see if it comes out to be less than Epsilon."},{"Start":"02:59.675 ","End":"03:04.340","Text":"The usual formula, the sum of the supremum minus"},{"Start":"03:04.340 ","End":"03:10.430","Text":"the infimum on each interval times its width but we use the Ls not the Ms of course,"},{"Start":"03:10.430 ","End":"03:12.990","Text":"because it\u0027s for the square root."},{"Start":"03:13.220 ","End":"03:23.380","Text":"This comes out to be the square root of M_i minus square root of m_i as we said here."},{"Start":"03:23.380 ","End":"03:25.685","Text":"There\u0027s this formula."},{"Start":"03:25.685 ","End":"03:28.145","Text":"Well, if you multiply this out, you\u0027ll get this."},{"Start":"03:28.145 ","End":"03:34.310","Text":"It\u0027s a difference of squares basically and we can use this expression to say"},{"Start":"03:34.310 ","End":"03:37.760","Text":"that this minus this is what we get"},{"Start":"03:37.760 ","End":"03:41.750","Text":"if we bring this to the other side on the denominator."},{"Start":"03:41.750 ","End":"03:43.970","Text":"Now, we can simplify this."},{"Start":"03:43.970 ","End":"03:48.860","Text":"The square root of m_i and the square root of M_i,"},{"Start":"03:48.860 ","End":"03:53.180","Text":"each of them is bigger or equal to the square root of c,"},{"Start":"03:53.180 ","End":"03:56.455","Text":"because square root of c is a lower bound."},{"Start":"03:56.455 ","End":"04:06.950","Text":"Yes, the square root of f is bounded below by the square root of c and so this plus this"},{"Start":"04:06.950 ","End":"04:10.610","Text":"is bigger or equal to twice the square root of c but it\u0027s on"},{"Start":"04:10.610 ","End":"04:14.270","Text":"the denominator so we flip the direction of"},{"Start":"04:14.270 ","End":"04:18.020","Text":"the inequality and say this is less than or equal to this when we"},{"Start":"04:18.020 ","End":"04:21.920","Text":"replace this by twice the square root of c. Now,"},{"Start":"04:21.920 ","End":"04:25.670","Text":"this numerator is exactly the expression for the upper sum"},{"Start":"04:25.670 ","End":"04:30.080","Text":"minus the lower sum for f. We know that this minus"},{"Start":"04:30.080 ","End":"04:34.200","Text":"this is less than Epsilon 1 so"},{"Start":"04:34.200 ","End":"04:38.520","Text":"we get here Epsilon 1 over twice the square root of c. Now,"},{"Start":"04:38.520 ","End":"04:42.800","Text":"you see why we chose Epsilon 1 to be the way it is because now"},{"Start":"04:42.800 ","End":"04:49.655","Text":"Epsilon 1 over twice the square root of c is exactly equal to Epsilon."},{"Start":"04:49.655 ","End":"04:55.140","Text":"That\u0027s the missing piece that we needed and so we are done."}],"ID":24692},{"Watched":false,"Name":"Exercise 4","Duration":"6m 47s","ChapterTopicVideoID":23766,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23766.jpeg","UploadDate":"2021-01-08T11:05:16.9500000","DurationForVideoObject":"PT6M47S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.465","Text":"This exercise is really a foreign 1."},{"Start":"00:03.465 ","End":"00:07.830","Text":"In each of them f is a function from the interval a,"},{"Start":"00:07.830 ","End":"00:09.900","Text":"b to the reals."},{"Start":"00:09.900 ","End":"00:14.565","Text":"In each case we have to prove or disprove the assertion."},{"Start":"00:14.565 ","End":"00:19.380","Text":"Disproof typically means finding a counter example."},{"Start":"00:19.380 ","End":"00:22.650","Text":"Part a, if f is integrable,"},{"Start":"00:22.650 ","End":"00:26.610","Text":"non-negative and not identically zero,"},{"Start":"00:26.610 ","End":"00:33.580","Text":"then the integral of f on the interval a, b is positive."},{"Start":"00:33.580 ","End":"00:36.120","Text":"The answer is no,"},{"Start":"00:36.120 ","End":"00:40.865","Text":"it\u0027s not true and we\u0027ll disprove it with a counter example."},{"Start":"00:40.865 ","End":"00:43.355","Text":"Take the function on the interval,"},{"Start":"00:43.355 ","End":"00:45.605","Text":"this is a and this is b."},{"Start":"00:45.605 ","End":"00:54.155","Text":"Make it 1 at a and 0 everywhere else then let\u0027s see if it meets the conditions."},{"Start":"00:54.155 ","End":"00:59.840","Text":"It\u0027s integrable because it differs from an integrable function by 1,"},{"Start":"00:59.840 ","End":"01:02.780","Text":"or you could say it\u0027s decreasing,"},{"Start":"01:02.780 ","End":"01:05.330","Text":"it\u0027s monotone, it\u0027s non-negative."},{"Start":"01:05.330 ","End":"01:10.005","Text":"That\u0027s for sure. It\u0027s either zero or 1 and not identically zero."},{"Start":"01:10.005 ","End":"01:12.645","Text":"Here it\u0027s not zero at a."},{"Start":"01:12.645 ","End":"01:15.570","Text":"But is the integral positive?"},{"Start":"01:15.570 ","End":"01:18.110","Text":"No. I guess I should have written a few more words."},{"Start":"01:18.110 ","End":"01:20.480","Text":"The integral is obviously 0 because it"},{"Start":"01:20.480 ","End":"01:24.475","Text":"differs at a finite number of points from a function"},{"Start":"01:24.475 ","End":"01:27.755","Text":"zero whose integral is zero and changing"},{"Start":"01:27.755 ","End":"01:31.490","Text":"a function at a finite number of points doesn\u0027t change the integral,"},{"Start":"01:31.490 ","End":"01:35.020","Text":"so zero is not positive, so no."},{"Start":"01:35.020 ","End":"01:38.540","Text":"Now part b, it\u0027s the same as the above,"},{"Start":"01:38.540 ","End":"01:42.980","Text":"but change the word integrable to the word"},{"Start":"01:42.980 ","End":"01:50.075","Text":"continuous and all ready excludes this example because this example is not continuous."},{"Start":"01:50.075 ","End":"01:53.840","Text":"Turns out that in this case it\u0027s actually true."},{"Start":"01:53.840 ","End":"01:57.695","Text":"Suppose x naught belongs to the interval a, b,"},{"Start":"01:57.695 ","End":"02:01.760","Text":"and f at that point is strictly positive,"},{"Start":"02:01.760 ","End":"02:03.260","Text":"say it\u0027s y naught."},{"Start":"02:03.260 ","End":"02:08.455","Text":"Now we can find an interval of positive length around x naught"},{"Start":"02:08.455 ","End":"02:14.330","Text":"such that on this whole interval f is bigger than a half of y naught."},{"Start":"02:14.330 ","End":"02:17.960","Text":"The way to do this is to use the Epsilon Delta."},{"Start":"02:17.960 ","End":"02:23.345","Text":"Take Epsilon is a half y then f will be close to y naught"},{"Start":"02:23.345 ","End":"02:29.479","Text":"within a half y whenever x is close to x naught within Delta,"},{"Start":"02:29.479 ","End":"02:31.280","Text":"so that gives us an interval."},{"Start":"02:31.280 ","End":"02:39.180","Text":"Now, F is non-negative so the integral from a to b is bigger or"},{"Start":"02:39.180 ","End":"02:42.945","Text":"equal to the integral on this interval I"},{"Start":"02:42.945 ","End":"02:47.570","Text":"because on the rest of it this might be 2 other intervals to the left and right of I,"},{"Start":"02:47.570 ","End":"02:52.250","Text":"and on each of those it\u0027s non-negative so we can take those pieces away."},{"Start":"02:52.250 ","End":"02:58.070","Text":"This inequality comes from a property of the integrals."},{"Start":"02:58.070 ","End":"03:01.100","Text":"It appeared in an earlier exercise and it"},{"Start":"03:01.100 ","End":"03:05.410","Text":"follows directly from the monotonicity property."},{"Start":"03:05.410 ","End":"03:11.840","Text":"What we\u0027re interested in is that if f of x is bigger or equal to M on an interval,"},{"Start":"03:11.840 ","End":"03:18.860","Text":"then the integral is bigger or equal to that constant times b minus a."},{"Start":"03:18.860 ","End":"03:22.620","Text":"That\u0027s basically what we have here,"},{"Start":"03:22.620 ","End":"03:28.805","Text":"m is a half y naught and b minus a is the length of the interval I."},{"Start":"03:28.805 ","End":"03:30.799","Text":"This is positive, this is positive,"},{"Start":"03:30.799 ","End":"03:35.730","Text":"so positive and that takes care of part b."},{"Start":"03:35.730 ","End":"03:38.240","Text":"Part c is also true."},{"Start":"03:38.240 ","End":"03:43.760","Text":"We have to show that if f is integrable then so is f squared."},{"Start":"03:43.760 ","End":"03:46.055","Text":"I\u0027ll just give you a sketch."},{"Start":"03:46.055 ","End":"03:49.190","Text":"We\u0027ll use the Riemann criterion for integrability but"},{"Start":"03:49.190 ","End":"03:52.745","Text":"first we have to check that f squared is bounded."},{"Start":"03:52.745 ","End":"03:55.475","Text":"Well, since f is bounded,"},{"Start":"03:55.475 ","End":"03:58.835","Text":"absolute value less than or equal to M, square both sides."},{"Start":"03:58.835 ","End":"04:05.470","Text":"The square can come inside and we have that f squared is bounded by M squared."},{"Start":"04:05.470 ","End":"04:13.725","Text":"Let Epsilon bigger than zero be given and choose Epsilon 1 to be Epsilon over 2M."},{"Start":"04:13.725 ","End":"04:17.660","Text":"This Epsilon for the integrability of f squared and this Epsilon for"},{"Start":"04:17.660 ","End":"04:21.915","Text":"the integrability of f. Now f we know is integrable,"},{"Start":"04:21.915 ","End":"04:25.830","Text":"so there exists a partition P of the interval a,"},{"Start":"04:25.830 ","End":"04:30.770","Text":"b such that the upper sum minus the lower sum is less than Epsilon 1."},{"Start":"04:30.770 ","End":"04:34.339","Text":"Now, with algebra, for each i,"},{"Start":"04:34.339 ","End":"04:40.340","Text":"mi squared minus mi squared is less than or equal to 2 difference of squares."},{"Start":"04:40.340 ","End":"04:43.490","Text":"We can put the absolute value around this here."},{"Start":"04:43.490 ","End":"04:47.000","Text":"We don\u0027t need absolute value because this is bigger than this."},{"Start":"04:47.000 ","End":"04:50.135","Text":"Anyway, this plus this is less than or equal to 2M,"},{"Start":"04:50.135 ","End":"04:52.345","Text":"so this is what we have."},{"Start":"04:52.345 ","End":"04:57.360","Text":"Then take the sum over all the i. I didn\u0027t write that that"},{"Start":"04:57.360 ","End":"05:02.320","Text":"i equals 1 to n of mi squared minus mi squared."},{"Start":"05:02.320 ","End":"05:07.370","Text":"By this we can take this out as 2M."},{"Start":"05:07.370 ","End":"05:09.110","Text":"This should be familiar."},{"Start":"05:09.110 ","End":"05:13.400","Text":"This is exactly the sum for the upper minus"},{"Start":"05:13.400 ","End":"05:19.475","Text":"the lower sum of f. This we know is less than Epsilon 1,"},{"Start":"05:19.475 ","End":"05:22.100","Text":"so all together 2M Epsilon 1,"},{"Start":"05:22.100 ","End":"05:25.180","Text":"and because of the way we chose Epsilon 1,"},{"Start":"05:25.180 ","End":"05:27.740","Text":"2M Epsilon 1 is Epsilon."},{"Start":"05:27.740 ","End":"05:31.310","Text":"That takes care of part c. Now what was d?"},{"Start":"05:31.310 ","End":"05:37.115","Text":"To show that if absolute value of f is integrable then so is f,"},{"Start":"05:37.115 ","End":"05:39.440","Text":"and prove or disprove."},{"Start":"05:39.440 ","End":"05:42.679","Text":"Turns out that it\u0027s false,"},{"Start":"05:42.679 ","End":"05:48.400","Text":"and we\u0027ll give an example where the absolute value of f is integrable but f itself isn\u0027t."},{"Start":"05:48.400 ","End":"05:50.300","Text":"Similar to a function we had."},{"Start":"05:50.300 ","End":"05:55.565","Text":"We had the Dirichlet function which was 1 for rationals and 0 for irrationals."},{"Start":"05:55.565 ","End":"05:58.550","Text":"Let\u0027s modify it slightly so that we have a minus"},{"Start":"05:58.550 ","End":"06:03.590","Text":"1 for irrational number and 1 for a rational number."},{"Start":"06:03.590 ","End":"06:06.260","Text":"Oh yeah, I forgot to write on 01."},{"Start":"06:06.260 ","End":"06:08.345","Text":"Well, this is not integrable,"},{"Start":"06:08.345 ","End":"06:12.320","Text":"and precisely the same way that the Dirichlet function isn\u0027t."},{"Start":"06:12.320 ","End":"06:15.110","Text":"We can also use 1 to prove the other."},{"Start":"06:15.110 ","End":"06:20.030","Text":"For example, if I add 1 to f of x and then divide by 2,"},{"Start":"06:20.030 ","End":"06:21.950","Text":"I\u0027ll get the Dirichlet function."},{"Start":"06:21.950 ","End":"06:28.370","Text":"Each 1 is a linear function of the other and linearity preserves integrability,"},{"Start":"06:28.370 ","End":"06:29.885","Text":"so we could do it that way."},{"Start":"06:29.885 ","End":"06:36.920","Text":"Anyway, the absolute value of f is integrable because it\u0027s a constant, the constant 1."},{"Start":"06:36.920 ","End":"06:40.085","Text":"That\u0027s why we altered the Dirichlet function."},{"Start":"06:40.085 ","End":"06:42.470","Text":"This is integrable, but like we say,"},{"Start":"06:42.470 ","End":"06:44.350","Text":"f itself is not,"},{"Start":"06:44.350 ","End":"06:47.650","Text":"and that concludes this exercise."}],"ID":24693},{"Watched":false,"Name":"Exercise 5","Duration":"2m ","ChapterTopicVideoID":23767,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23767.jpeg","UploadDate":"2021-01-08T11:05:52.0100000","DurationForVideoObject":"PT2M","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:09.690","Text":"In this exercise, we have to compute the integral from 0.25 to 4.3 of the floor function."},{"Start":"00:09.690 ","End":"00:13.965","Text":"The floor function of x is defined like this,"},{"Start":"00:13.965 ","End":"00:20.415","Text":"it\u0027s the greatest integer that\u0027s still less than or equal to the number."},{"Start":"00:20.415 ","End":"00:30.105","Text":"For example, the floor function of 3.3 is just 3 but the floor function of 3 is still 3."},{"Start":"00:30.105 ","End":"00:37.265","Text":"Note that if we take a half-open interval from n to n plus 1,"},{"Start":"00:37.265 ","End":"00:40.405","Text":"from 3 to 4, not including the 4,"},{"Start":"00:40.405 ","End":"00:46.920","Text":"then the floor function will be n or in our example 3 constant."},{"Start":"00:46.920 ","End":"00:49.970","Text":"Except for the right endpoint,"},{"Start":"00:49.970 ","End":"00:54.260","Text":"we can extend this to the closed interval and say it\u0027s true except for"},{"Start":"00:54.260 ","End":"00:59.640","Text":"1 point because the floor function of n plus 1 is n plus 1 but in all of the rest of it,"},{"Start":"00:59.640 ","End":"01:04.774","Text":"it\u0027s just n. The integral of the floor function of x"},{"Start":"01:04.774 ","End":"01:11.225","Text":"is the same as the integral of n on the interval from n to n plus 1."},{"Start":"01:11.225 ","End":"01:13.415","Text":"By the way, here\u0027s a picture,"},{"Start":"01:13.415 ","End":"01:15.605","Text":"the graph of the floor function."},{"Start":"01:15.605 ","End":"01:17.990","Text":"Anyway, let\u0027s do the computation,"},{"Start":"01:17.990 ","End":"01:23.390","Text":"we\u0027ll break the integral into pieces using the additive property."},{"Start":"01:23.390 ","End":"01:26.210","Text":"We go from 0.25-1, from 1-2,"},{"Start":"01:26.210 ","End":"01:30.125","Text":"from 2-3, from 3-4, and from 4-4.3."},{"Start":"01:30.125 ","End":"01:33.680","Text":"Now we\u0027ll use this formula that we developed"},{"Start":"01:33.680 ","End":"01:38.000","Text":"and we get that on this interval it\u0027s equal to 0."},{"Start":"01:38.000 ","End":"01:40.670","Text":"Here the function is 1, here it\u0027s 2, here it\u0027s 3,"},{"Start":"01:40.670 ","End":"01:46.455","Text":"here it\u0027s 4 and that\u0027s equal to 0 times,"},{"Start":"01:46.455 ","End":"01:49.350","Text":"the width of this interval is 0.75,"},{"Start":"01:49.350 ","End":"01:50.670","Text":"here it\u0027s 1, here it\u0027s 1,"},{"Start":"01:50.670 ","End":"01:53.945","Text":"here it\u0027s 1, and here it\u0027s 0.3."},{"Start":"01:53.945 ","End":"02:00.780","Text":"Add this all together and we get 7.2 and that\u0027s the answer and we\u0027re done."}],"ID":24694},{"Watched":false,"Name":"Exercise 6 - Properties of Definite Integrals","Duration":"11m 20s","ChapterTopicVideoID":23768,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23768.jpeg","UploadDate":"2021-01-08T11:09:17.4030000","DurationForVideoObject":"PT11M20S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.490","Text":"In this exercise, we\u0027ll prove part of the linearity property of integrability,"},{"Start":"00:05.490 ","End":"00:08.415","Text":"the multiplication by a constant part."},{"Start":"00:08.415 ","End":"00:11.985","Text":"The other part is addition of 2 integrable functions."},{"Start":"00:11.985 ","End":"00:19.140","Text":"Anyway, if f is integrable on ab and Alpha of some constant,"},{"Start":"00:19.140 ","End":"00:23.610","Text":"then Alpha f is also integrable on ab and what\u0027s more,"},{"Start":"00:23.610 ","End":"00:30.420","Text":"the integral of Alpha f is Alpha times the integral of f and we\u0027re given a hint."},{"Start":"00:30.420 ","End":"00:33.360","Text":"First, we\u0027ll deal with the case where Alpha\u0027s"},{"Start":"00:33.360 ","End":"00:37.040","Text":"non-negative and for the case where Alpha is negative,"},{"Start":"00:37.040 ","End":"00:41.240","Text":"we\u0027ll consider the function minus f. I want to start by"},{"Start":"00:41.240 ","End":"00:45.995","Text":"giving a tool that we can use and we\u0027ll use it more than once in this proof."},{"Start":"00:45.995 ","End":"00:50.825","Text":"Formally, if F is a function on any set,"},{"Start":"00:50.825 ","End":"00:52.565","Text":"doesn\u0027t have to be numbers,"},{"Start":"00:52.565 ","End":"00:55.760","Text":"could be functions, partitions, numbers,"},{"Start":"00:55.760 ","End":"00:59.420","Text":"whatever, as long as it has real values and assume it\u0027s"},{"Start":"00:59.420 ","End":"01:04.730","Text":"bounded and Alpha is a non-negative constant,"},{"Start":"01:04.730 ","End":"01:07.220","Text":"then Alpha\u0027s interchangeable with"},{"Start":"01:07.220 ","End":"01:10.669","Text":"the supremum operation as well as with the infimum operation."},{"Start":"01:10.669 ","End":"01:16.505","Text":"In other words, the supremum of Alpha f is Alpha times the supremum of f"},{"Start":"01:16.505 ","End":"01:19.720","Text":"and the infimum of Alpha f is Alpha times"},{"Start":"01:19.720 ","End":"01:23.330","Text":"the infimum of f. Let me try and give you some intuition."},{"Start":"01:23.330 ","End":"01:26.420","Text":"If you have function on a set, let\u0027s say,"},{"Start":"01:26.420 ","End":"01:32.060","Text":"the set of people and f is the height function of a person and Alpha\u0027s 2."},{"Start":"01:32.060 ","End":"01:36.485","Text":"If you first take the maximum height and then double it,"},{"Start":"01:36.485 ","End":"01:42.155","Text":"or measure all the double heights of people and then take the maximum,"},{"Start":"01:42.155 ","End":"01:44.330","Text":"the maximum of the double is the same as double of"},{"Start":"01:44.330 ","End":"01:47.990","Text":"the maximum and similarly with the minimum."},{"Start":"01:47.990 ","End":"01:50.870","Text":"But it wouldn\u0027t work if we had a negative number."},{"Start":"01:50.870 ","End":"01:52.805","Text":"If we multiply by a negative number,"},{"Start":"01:52.805 ","End":"01:54.190","Text":"actually you can\u0027t get a theorem,"},{"Start":"01:54.190 ","End":"01:58.190","Text":"you can just have to reverse supremum and infimum and then it will work."},{"Start":"01:58.190 ","End":"02:01.339","Text":"But anyway, This is not hard to prove,"},{"Start":"02:01.339 ","End":"02:03.125","Text":"but I\u0027ll skip that part."},{"Start":"02:03.125 ","End":"02:07.325","Text":"First, we will take the case Alpha bigger or equal to 0."},{"Start":"02:07.325 ","End":"02:09.810","Text":"Forgot to write that."},{"Start":"02:11.240 ","End":"02:17.250","Text":"Let p be any partition of ab, xi,"},{"Start":"02:17.250 ","End":"02:22.680","Text":"i goes from 0-n and the usual notations,"},{"Start":"02:22.680 ","End":"02:28.935","Text":"Delta xi., i_i and start with the uppersum."},{"Start":"02:28.935 ","End":"02:34.160","Text":"The uppersum for the partition and Alpha f. Let me just take you all the way to the end."},{"Start":"02:34.160 ","End":"02:40.955","Text":"At first, my goal is to bring the Alpha outside of this uppersum operation."},{"Start":"02:40.955 ","End":"02:43.194","Text":"But we have to do it in steps."},{"Start":"02:43.194 ","End":"02:49.190","Text":"The uppersum is the sum of what we call big MI, Delta x i."},{"Start":"02:49.190 ","End":"02:57.185","Text":"But big MI spelling it out is the supremum of the functional for f over this interval."},{"Start":"02:57.185 ","End":"03:04.009","Text":"Applying this with i_i in the role of S,"},{"Start":"03:04.009 ","End":"03:08.300","Text":"we see that we can take the Alpha out of the supremum."},{"Start":"03:08.300 ","End":"03:10.070","Text":"That\u0027s basically what this is saying."},{"Start":"03:10.070 ","End":"03:14.320","Text":"You can interchange the Alpha and the superman provide Alphas non-negative."},{"Start":"03:14.320 ","End":"03:18.470","Text":"Then, this is equal to by the linearity of Sigma,"},{"Start":"03:18.470 ","End":"03:22.040","Text":"we can bring the Alpha in front and what we have left is"},{"Start":"03:22.040 ","End":"03:27.665","Text":"just the expression for the upper sum of f for this partition."},{"Start":"03:27.665 ","End":"03:33.395","Text":"Alpha\u0027s been brought all the way out to the front and very similarly with the Nth,"},{"Start":"03:33.395 ","End":"03:35.450","Text":"I won\u0027t go through all the details,"},{"Start":"03:35.450 ","End":"03:41.540","Text":"but we bring the Alpha in front and the important 1 is this equality,"},{"Start":"03:41.540 ","End":"03:43.835","Text":"which is based on this."},{"Start":"03:43.835 ","End":"03:50.670","Text":"Again with i_i replacing s in this general claim."},{"Start":"03:51.110 ","End":"03:54.550","Text":"Next, we\u0027ll define another set,"},{"Start":"03:54.550 ","End":"04:00.470","Text":"capital Pi to be the set of all partitions p of ab."},{"Start":"04:00.470 ","End":"04:04.575","Text":"This will take the place of s in a moment, you\u0027ll see."},{"Start":"04:04.575 ","End":"04:08.110","Text":"Now we\u0027ll evaluate the upper Riemann integral of"},{"Start":"04:08.110 ","End":"04:13.300","Text":"Alpha f. That\u0027s the infimum on all partitions of"},{"Start":"04:13.300 ","End":"04:17.580","Text":"the upper sum for Alpha F and we can bring"},{"Start":"04:17.580 ","End":"04:23.140","Text":"the Alpha out of the upper sum because of this here,"},{"Start":"04:23.140 ","End":"04:26.905","Text":"I\u0027m just taking this as equal to this so the Alpha comes out in front."},{"Start":"04:26.905 ","End":"04:29.575","Text":"Then for this equality,"},{"Start":"04:29.575 ","End":"04:34.095","Text":"we\u0027re going to use this principle again with the infimum,"},{"Start":"04:34.095 ","End":"04:39.620","Text":"but this time with the set s being capital Pi and wanted to be formal,"},{"Start":"04:39.620 ","End":"04:44.970","Text":"this big F on a partition p is"},{"Start":"04:44.970 ","End":"04:51.410","Text":"the function upper sum of p and f. Each p gives us a real number."},{"Start":"04:51.410 ","End":"04:55.550","Text":"Anyway, the point is that we can take Alpha outside of"},{"Start":"04:55.550 ","End":"05:00.950","Text":"an infimum even when it\u0027s an infimum over all partitions and finally,"},{"Start":"05:00.950 ","End":"05:05.420","Text":"this is equal to Alpha times this is the expression for the upper"},{"Start":"05:05.420 ","End":"05:10.460","Text":"integral for f. If we look at the beginning and the end,"},{"Start":"05:10.460 ","End":"05:14.480","Text":"anyway, upper integral of Alpha f is Alpha times upper"},{"Start":"05:14.480 ","End":"05:20.090","Text":"integral of f if Alpha\u0027s non-negative, and similarly,"},{"Start":"05:20.090 ","End":"05:23.420","Text":"the lower Riemann integral of Alpha f is"},{"Start":"05:23.420 ","End":"05:27.335","Text":"Alpha times the lower Riemann integral of f. Now,"},{"Start":"05:27.335 ","End":"05:30.700","Text":"we\u0027re given that f is integrable."},{"Start":"05:30.700 ","End":"05:32.630","Text":"Since f is integrable,"},{"Start":"05:32.630 ","End":"05:38.735","Text":"the upper and lower Riemann integrals are equal and they\u0027re equal to the integral."},{"Start":"05:38.735 ","End":"05:42.770","Text":"What we get is that the upper integral of"},{"Start":"05:42.770 ","End":"05:47.450","Text":"Alpha f is equal to Alpha times the upper integral of f,"},{"Start":"05:47.450 ","End":"05:50.944","Text":"that\u0027s from here and this is equal to"},{"Start":"05:50.944 ","End":"05:55.370","Text":"Alpha times the lower Riemann integral of f. I mean this equals,"},{"Start":"05:55.370 ","End":"06:00.380","Text":"this is what we just said because of the integrability and this is equal to,"},{"Start":"06:00.380 ","End":"06:02.720","Text":"we can put the Alpha back inside,"},{"Start":"06:02.720 ","End":"06:05.000","Text":"which is what we said here,"},{"Start":"06:05.000 ","End":"06:08.380","Text":"that this is equal to this."},{"Start":"06:08.380 ","End":"06:13.010","Text":"If we just look at the first and the last parts of this equality,"},{"Start":"06:13.010 ","End":"06:18.410","Text":"we have that the upper integral of Alpha f is equal to the lower"},{"Start":"06:18.410 ","End":"06:25.110","Text":"integral of Alpha f and that means that Alpha f is integrable."},{"Start":"06:25.110 ","End":"06:30.410","Text":"Furthermore, that the integral of Alpha f is Alpha times the integral of"},{"Start":"06:30.410 ","End":"06:35.840","Text":"f. We can see this by saying that this is equal to either this or this."},{"Start":"06:35.840 ","End":"06:41.680","Text":"It doesn\u0027t matter because each of these that\u0027s in a box is the integral of f,"},{"Start":"06:41.680 ","End":"06:42.840","Text":"They\u0027re both equal to it,"},{"Start":"06:42.840 ","End":"06:43.970","Text":"so when this equals this,"},{"Start":"06:43.970 ","End":"06:48.200","Text":"we can just drop this under bar and get this."},{"Start":"06:48.200 ","End":"06:51.380","Text":"But still it was for Alpha bigger or equal to 0."},{"Start":"06:51.380 ","End":"06:55.835","Text":"Continuing, we want to generalize to all Alpha,"},{"Start":"06:55.835 ","End":"07:02.240","Text":"but we\u0027ll follow the hint and do the case of minus f. In other words,"},{"Start":"07:02.240 ","End":"07:03.950","Text":"that if f is integrable,"},{"Start":"07:03.950 ","End":"07:05.780","Text":"then minus f is integrable."},{"Start":"07:05.780 ","End":"07:09.290","Text":"Well, let\u0027s look. The upper integral of"},{"Start":"07:09.290 ","End":"07:15.290","Text":"minus f is the infimum of all partitions of the upper sum for p and"},{"Start":"07:15.290 ","End":"07:20.360","Text":"minus f and perhaps I should have made a small proposition or"},{"Start":"07:20.360 ","End":"07:26.625","Text":"claim earlier that when you change f with minus f,"},{"Start":"07:26.625 ","End":"07:30.540","Text":"upper becomes lower and vice versa."},{"Start":"07:30.540 ","End":"07:35.240","Text":"The minus comes out but we also flip up or some with lower sum."},{"Start":"07:35.240 ","End":"07:39.710","Text":"Basically, because MI and mi,"},{"Start":"07:39.710 ","End":"07:42.305","Text":"when you take the negative function,"},{"Start":"07:42.305 ","End":"07:47.260","Text":"they become minus mi and minus MI."},{"Start":"07:47.260 ","End":"07:49.389","Text":"But in the reverse order,"},{"Start":"07:49.389 ","End":"07:51.855","Text":"M is greater than m,"},{"Start":"07:51.855 ","End":"07:56.525","Text":"then minus m is greater than minus M. Anyway,"},{"Start":"07:56.525 ","End":"07:58.955","Text":"when we have a minus comes in front,"},{"Start":"07:58.955 ","End":"08:02.070","Text":"we change u to l and vice versa."},{"Start":"08:03.740 ","End":"08:07.970","Text":"Again, we\u0027re using a property of infimum and"},{"Start":"08:07.970 ","End":"08:13.504","Text":"supremum that if you make all the values negative and take the least,"},{"Start":"08:13.504 ","End":"08:18.050","Text":"It\u0027s actually the same as taking the greatest and then putting the minus,"},{"Start":"08:18.050 ","End":"08:20.470","Text":"it\u0027s also intuitively clear."},{"Start":"08:20.470 ","End":"08:24.470","Text":"In these cases, bringing a minus out causes"},{"Start":"08:24.470 ","End":"08:31.950","Text":"a flipping 1 time a u flips to an l and the other time an infimum flips to a supremum."},{"Start":"08:32.330 ","End":"08:38.499","Text":"Similarly, if we take the lower Riemann integral with the minus,"},{"Start":"08:38.499 ","End":"08:40.780","Text":"again, we have 2 flips."},{"Start":"08:40.780 ","End":"08:46.030","Text":"We have the supremum of the lower sums,"},{"Start":"08:46.030 ","End":"08:47.650","Text":"but with minus f,"},{"Start":"08:47.650 ","End":"08:50.620","Text":"when we bring the minus in front here,"},{"Start":"08:50.620 ","End":"08:57.790","Text":"it becomes an upper sum and then when we bring the minus in front of the supremum,"},{"Start":"08:57.790 ","End":"09:06.225","Text":"it becomes an infimum and this is equal to minus the upper Riemann integral."},{"Start":"09:06.225 ","End":"09:11.480","Text":"Now note that these 2 are"},{"Start":"09:11.480 ","End":"09:17.845","Text":"equal because f is integrable and if these 2 are equal,"},{"Start":"09:17.845 ","End":"09:20.644","Text":"that means that these 2 are equal."},{"Start":"09:20.644 ","End":"09:24.260","Text":"I mean, still equal when you add a minus and then we go all the way back here so"},{"Start":"09:24.260 ","End":"09:28.760","Text":"this equals this and not only is this equal to this,"},{"Start":"09:28.760 ","End":"09:31.130","Text":"but it\u0027s equal to this expression here,"},{"Start":"09:31.130 ","End":"09:35.120","Text":"which is minus the integral of f and this tells"},{"Start":"09:35.120 ","End":"09:39.350","Text":"us essentially the summary that minus f is integrable and the"},{"Start":"09:39.350 ","End":"09:43.070","Text":"integral of minus f is minus the integral of"},{"Start":"09:43.070 ","End":"09:47.870","Text":"f. We\u0027ve taken care of the case for Alphas non-negative,"},{"Start":"09:47.870 ","End":"09:50.420","Text":"and we\u0027ve also taken care of the case of"},{"Start":"09:50.420 ","End":"09:56.425","Text":"minus f. Now we still need the case of Alpha f with a negative Alpha."},{"Start":"09:56.425 ","End":"10:01.880","Text":"The way we tackle this is we let Beta equal minus Alpha,"},{"Start":"10:01.880 ","End":"10:05.050","Text":"and then Beta is positive."},{"Start":"10:05.050 ","End":"10:06.990","Text":"If f is integrable,"},{"Start":"10:06.990 ","End":"10:11.990","Text":"then what we had above shows that minus f is also integrable"},{"Start":"10:11.990 ","End":"10:18.520","Text":"and an integrable function can be multiplied by a positive constant like Beta."},{"Start":"10:18.520 ","End":"10:21.965","Text":"Beta times minus f is integrable,"},{"Start":"10:21.965 ","End":"10:27.860","Text":"but this is equal to minus Beta times f and minus Beta is Alpha."},{"Start":"10:27.860 ","End":"10:34.625","Text":"Alpha f is integrable and all we\u0027re missing is the formula that Alpha f,"},{"Start":"10:34.625 ","End":"10:38.465","Text":"the integral is Alpha times the integral of f. Let\u0027s see."},{"Start":"10:38.465 ","End":"10:47.825","Text":"The integral of Alpha f is the integral of Beta times minus f and this is equal to"},{"Start":"10:47.825 ","End":"10:51.285","Text":"Beta times the integral of minus f"},{"Start":"10:51.285 ","End":"10:55.190","Text":"because Beta\u0027s a positive constant so it can be pulled out of the"},{"Start":"10:55.190 ","End":"11:02.540","Text":"integral and we also showed that you can take a minus in front of the integral."},{"Start":"11:02.540 ","End":"11:06.470","Text":"Minus f is integrable and the integral is minus the"},{"Start":"11:06.470 ","End":"11:10.400","Text":"integral of f. Then when we combine Beta with minus,"},{"Start":"11:10.400 ","End":"11:12.920","Text":"we have minus Beta, which is Alpha,"},{"Start":"11:12.920 ","End":"11:21.220","Text":"so this equals this as required and we are done."}],"ID":24695},{"Watched":false,"Name":"Exercise 7 - Properties of Definite Integrals","Duration":"7m 6s","ChapterTopicVideoID":23769,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23769.jpeg","UploadDate":"2021-01-08T11:11:36.2270000","DurationForVideoObject":"PT7M6S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.210","Text":"In this exercise, we have to show that if we have"},{"Start":"00:03.210 ","End":"00:06.600","Text":"2 functions f and g that are both integrable on a,"},{"Start":"00:06.600 ","End":"00:10.095","Text":"b, then the third function, f plus g,"},{"Start":"00:10.095 ","End":"00:13.620","Text":"is also integrable on the interval a,"},{"Start":"00:13.620 ","End":"00:16.185","Text":"b and not just that,"},{"Start":"00:16.185 ","End":"00:21.360","Text":"but the value of the integral of f plus g is the integral of f plus the integral of"},{"Start":"00:21.360 ","End":"00:27.480","Text":"g. This is part of what is called the linearity property of the integral."},{"Start":"00:27.480 ","End":"00:31.500","Text":"The other half of it is that the integral of Alpha times"},{"Start":"00:31.500 ","End":"00:35.640","Text":"f is Alpha times the integral of f and that\u0027s done in another exercise."},{"Start":"00:35.640 ","End":"00:40.945","Text":"We\u0027re also given a hint of how to go about it, prove 2 things,"},{"Start":"00:40.945 ","End":"00:46.595","Text":"that the upper integral of f plus g is less than or equal to the upper integral of f"},{"Start":"00:46.595 ","End":"00:48.920","Text":"plus the upper integral of g."},{"Start":"00:48.920 ","End":"00:52.650","Text":"Similarly for the lower integral only it\u0027s bigger than or equal to."},{"Start":"00:52.650 ","End":"00:58.810","Text":"So let\u0027s start with a partition P of the interval a, b."},{"Start":"00:58.810 ","End":"01:01.970","Text":"The usual notations apply,"},{"Start":"01:01.970 ","End":"01:05.420","Text":"and let\u0027s see what the upper sum is for f plus g,"},{"Start":"01:05.420 ","End":"01:13.310","Text":"it\u0027s the sum of the supremum on each sub-interval times the width."},{"Start":"01:13.310 ","End":"01:16.175","Text":"This is what we call big MI."},{"Start":"01:16.175 ","End":"01:23.150","Text":"Now the supremum of a sum is always less than or equal to the sum of the suprema."},{"Start":"01:23.150 ","End":"01:27.410","Text":"If you fix any x in this interval,"},{"Start":"01:27.410 ","End":"01:30.320","Text":"then f of x plus g of x is less than or equal to"},{"Start":"01:30.320 ","End":"01:34.385","Text":"the supremum of f of x plus the supremum of g of x."},{"Start":"01:34.385 ","End":"01:36.530","Text":"If it\u0027s true for a given x,"},{"Start":"01:36.530 ","End":"01:38.770","Text":"we can take the supremum over all x,"},{"Start":"01:38.770 ","End":"01:43.635","Text":"and then this will still be less than or equal to this plus this."},{"Start":"01:43.635 ","End":"01:48.350","Text":"Delta x_i is just a multiplier and as long as it\u0027s non-negative,"},{"Start":"01:48.350 ","End":"01:50.680","Text":"this will continue to hold."},{"Start":"01:50.680 ","End":"01:53.540","Text":"Now this expression if you interpret it,"},{"Start":"01:53.540 ","End":"01:59.030","Text":"this is the upper sum of f and this is the upper sum of"},{"Start":"01:59.030 ","End":"02:05.390","Text":"g for the partition P. Similarly for the lower sum,"},{"Start":"02:05.390 ","End":"02:08.615","Text":"except that we have reverse inequality"},{"Start":"02:08.615 ","End":"02:12.635","Text":"and I won\u0027t go through this it\u0027s almost the same as this."},{"Start":"02:12.635 ","End":"02:16.835","Text":"I\u0027ve written something here and I realized the need to explain a bit more."},{"Start":"02:16.835 ","End":"02:23.750","Text":"I want to remind you that the upper integral is the greatest lower bound of"},{"Start":"02:23.750 ","End":"02:31.190","Text":"the upper sums for the function in this case f and in this case g. Let\u0027s take this 1."},{"Start":"02:31.190 ","End":"02:35.040","Text":"If this is the greatest lower bound,"},{"Start":"02:35.040 ","End":"02:37.080","Text":"if I add up sum over 2,"},{"Start":"02:37.080 ","End":"02:39.515","Text":"it\u0027s no longer a lower bound."},{"Start":"02:39.515 ","End":"02:42.050","Text":"Otherwise there would be a greater 1."},{"Start":"02:42.050 ","End":"02:43.400","Text":"If it\u0027s not a lower bound,"},{"Start":"02:43.400 ","End":"02:45.590","Text":"that means that for some partition,"},{"Start":"02:45.590 ","End":"02:47.990","Text":"the upper sum is less than it."},{"Start":"02:47.990 ","End":"02:52.820","Text":"Similarly for g, This is the greatest lower"},{"Start":"02:52.820 ","End":"02:58.235","Text":"bound for the upper sums on G. So if I add up sim over 2,"},{"Start":"02:58.235 ","End":"03:00.560","Text":"there\u0027s got to be something less than it."},{"Start":"03:00.560 ","End":"03:04.750","Text":"Otherwise it would be a greater lower bound."},{"Start":"03:04.750 ","End":"03:06.750","Text":"Now that we have Q and R,"},{"Start":"03:06.750 ","End":"03:10.205","Text":"remember that any 2 partitions have"},{"Start":"03:10.205 ","End":"03:19.220","Text":"a common refinement of each of them and call that P. When you take a refinement,"},{"Start":"03:19.220 ","End":"03:22.730","Text":"the upper sum can only get smaller."},{"Start":"03:22.730 ","End":"03:24.415","Text":"It\u0027ll be less than or equal to."},{"Start":"03:24.415 ","End":"03:28.505","Text":"In both cases, it\u0027s going to be less than or equal to."},{"Start":"03:28.505 ","End":"03:34.175","Text":"In this case we have a Q, in this case we have an R. The upper sum of"},{"Start":"03:34.175 ","End":"03:40.220","Text":"P and F is also less than this expression that we had here,"},{"Start":"03:40.220 ","End":"03:44.335","Text":"and the upper sum of P and G is less than this expression."},{"Start":"03:44.335 ","End":"03:47.030","Text":"The thing I was aiming for was to get the same partition"},{"Start":"03:47.030 ","End":"03:50.660","Text":"here and here and not to have a Q here and an R here."},{"Start":"03:50.660 ","End":"03:57.290","Text":"Now what follows is that the upper integral of f plus g is less than or"},{"Start":"03:57.290 ","End":"04:03.845","Text":"equal to any upper sum for f plus g. By what we had here,"},{"Start":"04:03.845 ","End":"04:05.900","Text":"this is less than or equal to the sum of"},{"Start":"04:05.900 ","End":"04:10.795","Text":"the upper sums for f and g separately with this partition P,"},{"Start":"04:10.795 ","End":"04:12.485","Text":"and this plus this,"},{"Start":"04:12.485 ","End":"04:14.870","Text":"if we add these 2 inequalities,"},{"Start":"04:14.870 ","End":"04:19.585","Text":"is less than this plus this plus Epsilon."},{"Start":"04:19.585 ","End":"04:22.909","Text":"This will be true for any Epsilon bigger than 0."},{"Start":"04:22.909 ","End":"04:26.195","Text":"I mean, the beginning is less than the end"},{"Start":"04:26.195 ","End":"04:30.340","Text":"if you ignore the P for all Epsilon bigger than 0."},{"Start":"04:30.340 ","End":"04:38.765","Text":"The upper integral of f plus g is less than or equal to the sum of the upper integrals."},{"Start":"04:38.765 ","End":"04:45.684","Text":"If it\u0027s less than this sum plus Epsilon for any Epsilon we can have Epsilon go to 0,"},{"Start":"04:45.684 ","End":"04:50.530","Text":"and the less than could become less than or equal to."},{"Start":"04:50.530 ","End":"04:53.915","Text":"Similarly won\u0027t go through all the details."},{"Start":"04:53.915 ","End":"04:55.610","Text":"The symmetrical part of this,"},{"Start":"04:55.610 ","End":"05:01.340","Text":"the lower sum of f plus g is time bigger or equal to lower sum of f plus lower sum"},{"Start":"05:01.340 ","End":"05:08.120","Text":"of g. This if you know it\u0027s what the hint asked us to do this part here."},{"Start":"05:08.120 ","End":"05:14.900","Text":"What we get is that the upper integral of f plus g is less than"},{"Start":"05:14.900 ","End":"05:17.240","Text":"or equal to the upper integral of f plus the upper integral of"},{"Start":"05:17.240 ","End":"05:22.220","Text":"g. This is equal to because of the integrability of f and g,"},{"Start":"05:22.220 ","End":"05:25.190","Text":"I can replace upper sum by lower sum."},{"Start":"05:25.190 ","End":"05:28.580","Text":"This is equal to this and this is equal to this."},{"Start":"05:28.580 ","End":"05:33.030","Text":"This, again by what we showed,"},{"Start":"05:33.030 ","End":"05:34.730","Text":"here it is even,"},{"Start":"05:34.730 ","End":"05:40.340","Text":"this plus this is less than or equal to the lower integral of f plus g. Now,"},{"Start":"05:40.340 ","End":"05:45.050","Text":"the lower integral is always less than or equal to the upper integral."},{"Start":"05:45.050 ","End":"05:48.620","Text":"So this must equal this."},{"Start":"05:48.620 ","End":"05:51.770","Text":"Because, well, if you look at this in inequality,"},{"Start":"05:51.770 ","End":"05:53.300","Text":"we\u0027ve completed the circle."},{"Start":"05:53.300 ","End":"05:55.070","Text":"This is less than or equal to this,"},{"Start":"05:55.070 ","End":"05:56.660","Text":"which is less than or equal to this,"},{"Start":"05:56.660 ","End":"05:58.505","Text":"which is less than or equal to this."},{"Start":"05:58.505 ","End":"06:04.425","Text":"So all the less than or equals have to become equalities."},{"Start":"06:04.425 ","End":"06:08.160","Text":"In fact, this equals this."},{"Start":"06:08.160 ","End":"06:10.965","Text":"F plus g is integrable."},{"Start":"06:10.965 ","End":"06:13.760","Text":"That\u0027s 1 part because we have to actually show"},{"Start":"06:13.760 ","End":"06:17.810","Text":"the formula that the integral of the sum is the sum of the integrals."},{"Start":"06:17.810 ","End":"06:19.880","Text":"Now like I said earlier,"},{"Start":"06:19.880 ","End":"06:22.100","Text":"all these less than or equals to,"},{"Start":"06:22.100 ","End":"06:25.160","Text":"actually both of them become equalities."},{"Start":"06:25.160 ","End":"06:33.770","Text":"I\u0027ve mark that in red because of the fact that this is less than or equal to this,"},{"Start":"06:33.770 ","End":"06:35.870","Text":"yeah, we said that they\u0027re all equals."},{"Start":"06:35.870 ","End":"06:38.660","Text":"Now if you look at this again,"},{"Start":"06:38.660 ","End":"06:42.395","Text":"because f and g are integrable,"},{"Start":"06:42.395 ","End":"06:45.845","Text":"this is equal to this is equal to the"},{"Start":"06:45.845 ","End":"06:50.180","Text":"integral without upper or lower for f and this equals this"},{"Start":"06:50.180 ","End":"06:54.620","Text":"equals the integral for g. The integral of f plus g"},{"Start":"06:54.620 ","End":"06:59.480","Text":"which is the common value of the upper and lower integrals when they\u0027re equal,"},{"Start":"06:59.480 ","End":"07:07.110","Text":"is equal to integral of f plus the integral of g. That concludes the proof."}],"ID":24696},{"Watched":false,"Name":"Exercise 8 - Properties of Definite Integrals","Duration":"2m 11s","ChapterTopicVideoID":23770,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23770.jpeg","UploadDate":"2021-01-08T11:12:09.4100000","DurationForVideoObject":"PT2M11S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.890","Text":"In this exercise, we\u0027re continuing with properties of integrability."},{"Start":"00:04.890 ","End":"00:08.789","Text":"In this case, we have 2 integrable functions,"},{"Start":"00:08.789 ","End":"00:14.310","Text":"f and g, and we have to prove that f times g is integrable on a, b."},{"Start":"00:14.310 ","End":"00:17.850","Text":"Also, f divided by g is integrable on a,"},{"Start":"00:17.850 ","End":"00:22.515","Text":"b provided we have some restriction on g,"},{"Start":"00:22.515 ","End":"00:27.150","Text":"the absolute value has to be kept away from 0."},{"Start":"00:27.150 ","End":"00:29.190","Text":"Not just bigger than 0,"},{"Start":"00:29.190 ","End":"00:34.325","Text":"but always bigger or equal to some constant which is bigger than 0."},{"Start":"00:34.325 ","End":"00:36.350","Text":"Start with part a."},{"Start":"00:36.350 ","End":"00:38.720","Text":"Now we already showed in another exercise that"},{"Start":"00:38.720 ","End":"00:41.630","Text":"the square of an integrable function is integrable,"},{"Start":"00:41.630 ","End":"00:43.190","Text":"and we\u0027ll use that."},{"Start":"00:43.190 ","End":"00:46.490","Text":"We also showed that integration is linear."},{"Start":"00:46.490 ","End":"00:50.545","Text":"In other words, you can add or multiply by a constant."},{"Start":"00:50.545 ","End":"00:58.715","Text":"By simple algebra, x times y is a 1/4 of x plus y squared minus x minus y squared."},{"Start":"00:58.715 ","End":"01:02.725","Text":"Then using linearity and squaring f, g is integrable."},{"Start":"01:02.725 ","End":"01:04.680","Text":"F and g are integrable,"},{"Start":"01:04.680 ","End":"01:07.880","Text":"hence so is f plus g. If this is,"},{"Start":"01:07.880 ","End":"01:09.275","Text":"then its squared is."},{"Start":"01:09.275 ","End":"01:11.645","Text":"Similarly, f minus g is integrable,"},{"Start":"01:11.645 ","End":"01:12.965","Text":"and its square is."},{"Start":"01:12.965 ","End":"01:15.920","Text":"Then the difference of 2 integral functions."},{"Start":"01:15.920 ","End":"01:18.140","Text":"Then multiplying by a constant."},{"Start":"01:18.140 ","End":"01:19.880","Text":"A lot of steps here,"},{"Start":"01:19.880 ","End":"01:23.365","Text":"but in each step we maintain integrability."},{"Start":"01:23.365 ","End":"01:26.090","Text":"The thing is that there is no simple formula."},{"Start":"01:26.090 ","End":"01:29.270","Text":"You can\u0027t say that the integral of f times g is the integral of"},{"Start":"01:29.270 ","End":"01:32.930","Text":"f times the integral of g. We just know that it\u0027s integrable."},{"Start":"01:32.930 ","End":"01:38.675","Text":"Now for part b, we\u0027re going to use the results of another exercise."},{"Start":"01:38.675 ","End":"01:41.965","Text":"Here\u0027s what the exercise looks like."},{"Start":"01:41.965 ","End":"01:44.640","Text":"Now back to ours."},{"Start":"01:44.640 ","End":"01:49.020","Text":"Under these conditions, 1 over g is integrable."},{"Start":"01:49.020 ","End":"01:56.310","Text":"Then we can express f over g as f times 1 over g. From part a,"},{"Start":"01:56.310 ","End":"01:57.645","Text":"which is the product,"},{"Start":"01:57.645 ","End":"02:00.630","Text":"and from the other exercise for the reciprocal,"},{"Start":"02:00.630 ","End":"02:04.850","Text":"combined gives f times 1 over g is integrable,"},{"Start":"02:04.850 ","End":"02:11.370","Text":"meaning that f over g is integrable. That\u0027s it."}],"ID":24697},{"Watched":false,"Name":"Exercise 9","Duration":"5m 38s","ChapterTopicVideoID":23771,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23771.jpeg","UploadDate":"2021-01-08T11:14:00.4830000","DurationForVideoObject":"PT5M38S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.480","Text":"In this exercise, we\u0027re going to prove that the reciprocal of"},{"Start":"00:03.480 ","End":"00:08.940","Text":"an integrable function is integrable under certain mild conditions."},{"Start":"00:08.940 ","End":"00:14.880","Text":"More precisely, suppose that f is integrable on a b,"},{"Start":"00:14.880 ","End":"00:18.510","Text":"and there is some positive constant c,"},{"Start":"00:18.510 ","End":"00:25.730","Text":"such that the absolute value of f greater or equal to c on the whole interval a, b."},{"Start":"00:25.730 ","End":"00:29.825","Text":"It\u0027s not enough simply to say that this is positive."},{"Start":"00:29.825 ","End":"00:31.970","Text":"Alternative to this condition,"},{"Start":"00:31.970 ","End":"00:35.390","Text":"we can say that f is integrable and non 0,"},{"Start":"00:35.390 ","End":"00:38.150","Text":"and that 1 over f is bounded."},{"Start":"00:38.150 ","End":"00:42.020","Text":"In this case, we have to show that 1 over f,"},{"Start":"00:42.020 ","End":"00:46.435","Text":"call it g, is also integrable on a b."},{"Start":"00:46.435 ","End":"00:50.300","Text":"The main tool we\u0027ll be using here is Riemann\u0027s criterion,"},{"Start":"00:50.300 ","End":"00:55.100","Text":"which is an if and only if condition for bounded functions to be integrable."},{"Start":"00:55.100 ","End":"00:57.380","Text":"This is basically what it says."},{"Start":"00:57.380 ","End":"01:01.740","Text":"A bounded function f will be integrable on a,"},{"Start":"01:01.740 ","End":"01:05.690","Text":"b if and only if for every Epsilon there\u0027s"},{"Start":"01:05.690 ","End":"01:12.760","Text":"a partition such that the upper sum minus the lower sum is less than Epsilon."},{"Start":"01:12.760 ","End":"01:16.970","Text":"Now, f of x is bigger or equal to c,"},{"Start":"01:16.970 ","End":"01:19.340","Text":"an absolute value, so g of x,"},{"Start":"01:19.340 ","End":"01:20.765","Text":"which is the reciprocal,"},{"Start":"01:20.765 ","End":"01:26.040","Text":"is less than or equal to 1 over c. That shows that g is"},{"Start":"01:26.040 ","End":"01:27.380","Text":"bounded on a, b."},{"Start":"01:27.380 ","End":"01:32.665","Text":"Boundedness is what we need before we can even use the criterion."},{"Start":"01:32.665 ","End":"01:36.125","Text":"Let Epsilon be given positive,"},{"Start":"01:36.125 ","End":"01:40.820","Text":"we have to find the partition such that the upper sum minus"},{"Start":"01:40.820 ","End":"01:46.600","Text":"the lower sum for g on this partition is less than Epsilon."},{"Start":"01:46.600 ","End":"01:49.920","Text":"We know that f is integrable."},{"Start":"01:49.920 ","End":"01:53.360","Text":"We can choose p such that the upper sum minus"},{"Start":"01:53.360 ","End":"01:56.720","Text":"the lower sum is less than a different Epsilon."},{"Start":"01:56.720 ","End":"02:00.440","Text":"The Epsilon for this, we\u0027ll choose to be c squared Epsilon."},{"Start":"02:00.440 ","End":"02:02.825","Text":"At the end we\u0027ll see why we chose this."},{"Start":"02:02.825 ","End":"02:10.380","Text":"Denote P equals x_0 to x_n and i_i is the interval x_i minus 1, x_i."},{"Start":"02:10.380 ","End":"02:13.700","Text":"We have big M_i and little m_i as"},{"Start":"02:13.700 ","End":"02:17.630","Text":"the supremum and infimum of f and the different letter L,"},{"Start":"02:17.630 ","End":"02:19.490","Text":"big L_i, little l_i,"},{"Start":"02:19.490 ","End":"02:24.085","Text":"the supremum and infimum for g on this interval."},{"Start":"02:24.085 ","End":"02:30.295","Text":"Now let\u0027s estimate big L_i minus little l_i."},{"Start":"02:30.295 ","End":"02:39.680","Text":"This is equal to supremum of g minus the infimum of g on the interval I_i."},{"Start":"02:39.680 ","End":"02:42.020","Text":"Now here what we need to be careful,"},{"Start":"02:42.020 ","End":"02:43.505","Text":"It\u0027s a little tricky."},{"Start":"02:43.505 ","End":"02:48.350","Text":"The supremum of g is the supremum of g of x for all x in this interval."},{"Start":"02:48.350 ","End":"02:52.355","Text":"You could say the infimum of g is the infimum of g of x."},{"Start":"02:52.355 ","End":"02:55.850","Text":"But the supremum minus the infimum,"},{"Start":"02:55.850 ","End":"02:58.895","Text":"it\u0027s going to be different x here and here,"},{"Start":"02:58.895 ","End":"03:03.185","Text":"we have to take this as g of x and this g of y."},{"Start":"03:03.185 ","End":"03:06.200","Text":"Basically the maximum minus the minimum is"},{"Start":"03:06.200 ","End":"03:11.930","Text":"the greatest g of x minus g of y for 2 different x and y."},{"Start":"03:11.930 ","End":"03:14.810","Text":"This is equal to the supremum where x and y belong"},{"Start":"03:14.810 ","End":"03:17.695","Text":"to the integral of g of x minus g of y."},{"Start":"03:17.695 ","End":"03:20.450","Text":"We can put an absolute value here,"},{"Start":"03:20.450 ","End":"03:25.730","Text":"because the supremum is going to occur when g of x is bigger than g of y."},{"Start":"03:25.730 ","End":"03:31.315","Text":"Otherwise we could invert x and y so we can take absolute value here."},{"Start":"03:31.315 ","End":"03:34.835","Text":"This is equal to because g is 1 over f,"},{"Start":"03:34.835 ","End":"03:38.720","Text":"we have 1 over f of x minus 1 over f of y."},{"Start":"03:38.720 ","End":"03:46.775","Text":"Doing some algebra, cross-multiplying we have it would be f of y minus f of x."},{"Start":"03:46.775 ","End":"03:50.030","Text":"But it\u0027s an absolute value so we can invert."},{"Start":"03:50.030 ","End":"03:57.335","Text":"Now recall that f is bigger or equal to c. This is bigger or equal to c and so is this."},{"Start":"03:57.335 ","End":"04:03.440","Text":"The denominator is bigger or equal to c squared so if we replace it by c squared,"},{"Start":"04:03.440 ","End":"04:07.130","Text":"we have to take less than or equal to because it\u0027s in the denominator."},{"Start":"04:07.130 ","End":"04:10.990","Text":"Just like before, we can drop the absolute value."},{"Start":"04:10.990 ","End":"04:20.525","Text":"Again, as before, this is the supremum of f minus the infimum of f over c squared."},{"Start":"04:20.525 ","End":"04:25.090","Text":"This is big M_i and this is little m _i."},{"Start":"04:25.090 ","End":"04:29.700","Text":"What we get is that L_i minus l_i is less"},{"Start":"04:29.700 ","End":"04:34.015","Text":"than or equal to big M_i minus little m_i over c squared."},{"Start":"04:34.015 ","End":"04:40.595","Text":"Now we\u0027re going to use this inequality and sum it from 1 to n. Let\u0027s start from here."},{"Start":"04:40.595 ","End":"04:49.365","Text":"The upper sum minus the lower sum for g. This partition is the sum of big L_i,"},{"Start":"04:49.365 ","End":"04:51.700","Text":"little l_i times Delta x _i."},{"Start":"04:51.700 ","End":"04:56.480","Text":"Now we apply this inequality and we can write 1 over c"},{"Start":"04:56.480 ","End":"05:01.890","Text":"squared times big M_i minus little m_i Delta x _i."},{"Start":"05:01.890 ","End":"05:07.565","Text":"This sum is exactly the upper sum minus the lower sum."},{"Start":"05:07.565 ","End":"05:10.790","Text":"But for the function f and above,"},{"Start":"05:10.790 ","End":"05:15.110","Text":"we had that this is less than c squared Epsilon."},{"Start":"05:15.110 ","End":"05:18.650","Text":"If we go back c. Yeah,"},{"Start":"05:18.650 ","End":"05:24.450","Text":"c squared Epsilon and 1 over c squared times c squared Epsilon is Epsilon."},{"Start":"05:24.450 ","End":"05:26.400","Text":"This is less than Epsilon."},{"Start":"05:26.400 ","End":"05:29.260","Text":"We have that this is less than Epsilon."},{"Start":"05:29.260 ","End":"05:32.855","Text":"Then by Riemann\u0027s criterion in the other direction,"},{"Start":"05:32.855 ","End":"05:39.280","Text":"it shows that g is integrable on a, b. We\u0027re done."}],"ID":24698},{"Watched":false,"Name":"Exercise 10","Duration":"1m 50s","ChapterTopicVideoID":23772,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23772.jpeg","UploadDate":"2021-01-08T11:14:35.5770000","DurationForVideoObject":"PT1M50S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.680","Text":"In this exercise, we show that if we have 2 integrable functions,"},{"Start":"00:04.680 ","End":"00:06.840","Text":"f and g on an interval,"},{"Start":"00:06.840 ","End":"00:11.310","Text":"then their maximum and their minimum as functions are also"},{"Start":"00:11.310 ","End":"00:16.845","Text":"integrable called the maximum Phi and the minimum Psi."},{"Start":"00:16.845 ","End":"00:23.040","Text":"We\u0027re given a hint that we can express the maximum of 2 numbers a and"},{"Start":"00:23.040 ","End":"00:30.240","Text":"b as the expression a plus b plus the absolute value of a minus b,"},{"Start":"00:30.240 ","End":"00:32.440","Text":"all this over 2."},{"Start":"00:32.440 ","End":"00:35.450","Text":"A diagram could help here."},{"Start":"00:35.450 ","End":"00:37.070","Text":"If we have 2 numbers,"},{"Start":"00:37.070 ","End":"00:41.200","Text":"a and b and I\u0027m not saying here which is a and which is b,"},{"Start":"00:41.200 ","End":"00:46.385","Text":"in either case, the center is a plus b over 2."},{"Start":"00:46.385 ","End":"00:52.820","Text":"The distance from the middle to either of these other points,"},{"Start":"00:52.820 ","End":"00:57.365","Text":"a and b is a half of the absolute value of a minus b."},{"Start":"00:57.365 ","End":"01:02.030","Text":"The absolute value of a minus b is this distance and half of it is this."},{"Start":"01:02.030 ","End":"01:05.660","Text":"Basically, we\u0027re taking the middle plus or minus"},{"Start":"01:05.660 ","End":"01:09.735","Text":"half the distance gives us the maximum and the minimum."},{"Start":"01:09.735 ","End":"01:12.980","Text":"The maximum is as written here,"},{"Start":"01:12.980 ","End":"01:18.470","Text":"and the minimum is just same thing but with a minus here,"},{"Start":"01:18.470 ","End":"01:21.530","Text":"plus here, there\u0027s a minus here."},{"Start":"01:21.530 ","End":"01:27.860","Text":"Now, we already know certain properties of integrability that if you have 2 functions,"},{"Start":"01:27.860 ","End":"01:32.280","Text":"then their sum is, the difference is,"},{"Start":"01:32.280 ","End":"01:35.020","Text":"the absolute value is,"},{"Start":"01:35.020 ","End":"01:38.015","Text":"adding them, multiplying by a constant,"},{"Start":"01:38.015 ","End":"01:43.955","Text":"just applying the linearity properties and the absolute value property,"},{"Start":"01:43.955 ","End":"01:50.530","Text":"we get that Phi and Psi are both integrable too and that\u0027s it."}],"ID":24699},{"Watched":false,"Name":"Exercise 11","Duration":"8m 59s","ChapterTopicVideoID":23773,"CourseChapterTopicPlaylistID":257168,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23773.jpeg","UploadDate":"2021-01-08T11:17:02.1370000","DurationForVideoObject":"PT8M59S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.180","Text":"This exercise is a bit theoretical,"},{"Start":"00:03.180 ","End":"00:05.655","Text":"kind of lengthy, but bear with me."},{"Start":"00:05.655 ","End":"00:09.330","Text":"Let f be bounded on the interval a,"},{"Start":"00:09.330 ","End":"00:15.525","Text":"b, and usual notation for a partition P on a, b."},{"Start":"00:15.525 ","End":"00:18.720","Text":"If we give an Epsilon bigger than 0,"},{"Start":"00:18.720 ","End":"00:20.655","Text":"and this partition P,"},{"Start":"00:20.655 ","End":"00:26.480","Text":"we define 2 subsets of the indices,"},{"Start":"00:26.480 ","End":"00:35.085","Text":"1 through n. We call them A_Epsilon of P and B_Epsilon of P. We define them as follows,"},{"Start":"00:35.085 ","End":"00:41.870","Text":"i will belong to A if the oscillation,"},{"Start":"00:41.870 ","End":"00:48.550","Text":"that\u0027s the supremum minus the infimum on that interval is less than Epsilon."},{"Start":"00:48.550 ","End":"00:53.510","Text":"Conversely, if that oscillation is bigger or equal to Epsilon,"},{"Start":"00:53.510 ","End":"00:57.245","Text":"then the index goes in the B set."},{"Start":"00:57.245 ","End":"01:03.600","Text":"Another definition is s_Epsilon of P will be the sum of"},{"Start":"01:03.600 ","End":"01:13.985","Text":"the widths of the intervals with big oscillation where the i is in the B set."},{"Start":"01:13.985 ","End":"01:16.220","Text":"Here\u0027s what we have to show,"},{"Start":"01:16.220 ","End":"01:20.850","Text":"a bounded function f is integrable on a,"},{"Start":"01:20.850 ","End":"01:27.530","Text":"b if and only if for every positive Epsilon and Tau,"},{"Start":"01:27.530 ","End":"01:31.250","Text":"we can find a positive Delta such"},{"Start":"01:31.250 ","End":"01:37.415","Text":"that whenever the mesh of a partition is less than Delta,"},{"Start":"01:37.415 ","End":"01:42.515","Text":"s_Epsilon for this partition is less than Tau."},{"Start":"01:42.515 ","End":"01:48.065","Text":"We have to show 2 things because this is an if and only if so we\u0027ll take each direction."},{"Start":"01:48.065 ","End":"01:54.290","Text":"Now, the fact that f is bounded is given either way."},{"Start":"01:54.290 ","End":"01:59.450","Text":"One condition is that f is integrable on a,"},{"Start":"01:59.450 ","End":"02:06.310","Text":"b, and the equivalent that we have to prove is all of this."},{"Start":"02:06.310 ","End":"02:09.735","Text":"For short, I\u0027ll call this the condition."},{"Start":"02:09.735 ","End":"02:12.770","Text":"We\u0027ve got to show that f is integrable on a,"},{"Start":"02:12.770 ","End":"02:15.954","Text":"b if and only if the condition holds."},{"Start":"02:15.954 ","End":"02:17.910","Text":"We start with the if,"},{"Start":"02:17.910 ","End":"02:20.940","Text":"that\u0027s the arrow in this direction."},{"Start":"02:20.940 ","End":"02:23.360","Text":"We assume that this condition holds,"},{"Start":"02:23.360 ","End":"02:26.540","Text":"we have to prove that f is integrable on a,"},{"Start":"02:26.540 ","End":"02:30.200","Text":"b and we\u0027re going to use Riemann\u0027s criterion for that."},{"Start":"02:30.200 ","End":"02:35.705","Text":"We have to show that given any Epsilon nought bigger than 0,"},{"Start":"02:35.705 ","End":"02:39.620","Text":"we can find a partition such that the upper minus"},{"Start":"02:39.620 ","End":"02:43.970","Text":"the lower sum is less than this Epsilon nought."},{"Start":"02:43.970 ","End":"02:49.520","Text":"I use this notation because I\u0027m saving Epsilon for the Epsilon here."},{"Start":"02:49.520 ","End":"02:53.510","Text":"We\u0027ll get to the Epsilon t Delta later."},{"Start":"02:53.510 ","End":"02:59.660","Text":"Meanwhile, let\u0027s interpret what this upper minus lower sum is."},{"Start":"02:59.660 ","End":"03:05.930","Text":"Assuming all the standard expressions like M_i is the supremum of f on"},{"Start":"03:05.930 ","End":"03:13.310","Text":"the interval x_i minus 1 to x_i and P is the partition x nought through x_n and so on,"},{"Start":"03:13.310 ","End":"03:15.775","Text":"just the usual notation."},{"Start":"03:15.775 ","End":"03:19.080","Text":"This minus this is this as usual."},{"Start":"03:19.080 ","End":"03:23.690","Text":"Now we can break this up into 2 sums because the indices from 1 to n"},{"Start":"03:23.690 ","End":"03:28.625","Text":"break up into those that are in the A set and those that are in the B set."},{"Start":"03:28.625 ","End":"03:35.760","Text":"In this sum, the oscillation M_i minus m_i is less than Epsilon."},{"Start":"03:35.760 ","End":"03:42.605","Text":"That\u0027s by the definition of A_Epsilon of P. In the other case,"},{"Start":"03:42.605 ","End":"03:46.050","Text":"f is bounded by M,"},{"Start":"03:46.050 ","End":"03:50.990","Text":"so the supremum minus the infimum is"},{"Start":"03:50.990 ","End":"03:57.295","Text":"bounded by twice M Delta x_i here and here."},{"Start":"03:57.295 ","End":"04:01.995","Text":"What we get is that this is less than"},{"Start":"04:01.995 ","End":"04:08.870","Text":"Epsilon times the sum of all the Delta x_is, b minus a."},{"Start":"04:08.870 ","End":"04:10.310","Text":"We\u0027re taking less of them,"},{"Start":"04:10.310 ","End":"04:16.525","Text":"so it\u0027s still less than b minus a and the Epsilon is the Epsilon here."},{"Start":"04:16.525 ","End":"04:20.265","Text":"Here we have the 2M from here,"},{"Start":"04:20.265 ","End":"04:23.880","Text":"and the sum of the Delta x_i,"},{"Start":"04:23.880 ","End":"04:28.560","Text":"where i is in B_Epsilon of P. This is"},{"Start":"04:28.560 ","End":"04:33.830","Text":"the definition of s_Epsilon of P. I\u0027ll go back and check that."},{"Start":"04:33.830 ","End":"04:35.030","Text":"Yeah, there it is."},{"Start":"04:35.030 ","End":"04:37.985","Text":"That\u0027s the definition I\u0027m using over here."},{"Start":"04:37.985 ","End":"04:42.595","Text":"Now it\u0027s time to choose the Epsilon and Tau."},{"Start":"04:42.595 ","End":"04:46.160","Text":"We\u0027ll take both of them to be this expression,"},{"Start":"04:46.160 ","End":"04:50.515","Text":"Epsilon nought over b minus a plus 2M."},{"Start":"04:50.515 ","End":"04:54.230","Text":"We\u0027ll let Delta be as in the condition."},{"Start":"04:54.230 ","End":"04:57.225","Text":"It says that given an Epsilon and Tau,"},{"Start":"04:57.225 ","End":"04:58.830","Text":"there exists a Delta."},{"Start":"04:58.830 ","End":"05:01.354","Text":"That\u0027s the Delta that\u0027s promised."},{"Start":"05:01.354 ","End":"05:09.230","Text":"Now, choose any partition with the mesh or norm of P less than Delta."},{"Start":"05:09.230 ","End":"05:12.470","Text":"We are going to show now that this minus this,"},{"Start":"05:12.470 ","End":"05:16.180","Text":"the upper minus lower sum is less than Epsilon nought."},{"Start":"05:16.180 ","End":"05:21.730","Text":"S Epsilon of P is less than Tau because"},{"Start":"05:21.730 ","End":"05:28.100","Text":"that\u0027s what the condition promises that if mesh of p is less than Delta,"},{"Start":"05:28.100 ","End":"05:33.810","Text":"then f Epsilon of P is less than Tau and Tau is equal to Epsilon."},{"Start":"05:33.810 ","End":"05:38.210","Text":"The upper minus lower sum is less than,"},{"Start":"05:38.210 ","End":"05:42.190","Text":"now I\u0027m just copying from here."},{"Start":"05:42.190 ","End":"05:46.770","Text":"S Epsilon of P from here,"},{"Start":"05:46.770 ","End":"05:49.260","Text":"it\u0027s less than Epsilon."},{"Start":"05:49.260 ","End":"05:52.380","Text":"If I replace this by Epsilon,"},{"Start":"05:52.380 ","End":"05:56.040","Text":"then I can take the Epsilon out of the brackets. I think you follow."},{"Start":"05:56.040 ","End":"06:01.820","Text":"From here we see that this expression is equal to Epsilon nought,"},{"Start":"06:01.820 ","End":"06:04.575","Text":"just a simple algebra."},{"Start":"06:04.575 ","End":"06:10.790","Text":"What we have now is that the upper sum minus the lower sum is less than Epsilon nought,"},{"Start":"06:10.790 ","End":"06:16.105","Text":"which shows by the Riemann criterion that f is integrable."},{"Start":"06:16.105 ","End":"06:18.840","Text":"That\u0027s the 1st half done."},{"Start":"06:18.840 ","End":"06:24.215","Text":"Now the other direction where we\u0027re given that f is integrable,"},{"Start":"06:24.215 ","End":"06:27.080","Text":"and we have to prove the condition,"},{"Start":"06:27.080 ","End":"06:30.220","Text":"go back and look at what the condition is."},{"Start":"06:30.220 ","End":"06:35.375","Text":"Let Epsilon and Tau both positive be given,"},{"Start":"06:35.375 ","End":"06:38.269","Text":"we have to find the Delta."},{"Start":"06:38.269 ","End":"06:41.720","Text":"Recall that we had a proposition that I\u0027ve highlighted"},{"Start":"06:41.720 ","End":"06:46.505","Text":"here that if f is integrable on a, b,"},{"Start":"06:46.505 ","End":"06:51.050","Text":"then for any Epsilon there is a Delta such"},{"Start":"06:51.050 ","End":"06:55.880","Text":"that whenever a partition has mesh less than Delta,"},{"Start":"06:55.880 ","End":"07:01.055","Text":"the upper sum minus the lower sum for that partition is less than Epsilon."},{"Start":"07:01.055 ","End":"07:05.085","Text":"Now, let Epsilon for this proposition,"},{"Start":"07:05.085 ","End":"07:11.310","Text":"be Epsilon times Tau for what was given here."},{"Start":"07:11.310 ","End":"07:16.430","Text":"Nought Epsilon will take Epsilon Tau and this Delta that\u0027s guaranteed by"},{"Start":"07:16.430 ","End":"07:21.290","Text":"this proposition will be the Delta that we need for the condition."},{"Start":"07:21.290 ","End":"07:26.185","Text":"You have to show that if the partition has norm less than Delta,"},{"Start":"07:26.185 ","End":"07:30.895","Text":"then this expression s_Epsilon of P is less than Tau."},{"Start":"07:30.895 ","End":"07:35.230","Text":"Let\u0027s start, assume norm of P less than Delta."},{"Start":"07:35.230 ","End":"07:40.845","Text":"That means by the proposition that U minus L,"},{"Start":"07:40.845 ","End":"07:44.780","Text":"upper minus lower sum for this P is less than the Epsilon here,"},{"Start":"07:44.780 ","End":"07:47.575","Text":"but the Epsilon here is Epsilon Tau."},{"Start":"07:47.575 ","End":"07:53.660","Text":"The sum of supremum minus infimum times the width of the interval,"},{"Start":"07:53.660 ","End":"07:58.190","Text":"the sum of all these is going to be less than this because I mean this equals this."},{"Start":"07:58.190 ","End":"08:02.630","Text":"Then since these are all positive terms to the sum,"},{"Start":"08:02.630 ","End":"08:04.460","Text":"we can take less terms,"},{"Start":"08:04.460 ","End":"08:06.770","Text":"not all the i from 1 to n,"},{"Start":"08:06.770 ","End":"08:12.950","Text":"but just the i in B Epsilon of P that will be less than or equal to this,"},{"Start":"08:12.950 ","End":"08:15.470","Text":"which will be less than Epsilon Tau."},{"Start":"08:15.470 ","End":"08:24.460","Text":"Now each of these is bigger or equal to Epsilon by the definition of the B Epsilon."},{"Start":"08:24.460 ","End":"08:28.760","Text":"If we shrink this by replacing it by"},{"Start":"08:28.760 ","End":"08:33.050","Text":"Epsilon we\u0027ll get something that\u0027s possibly smaller anyway,"},{"Start":"08:33.050 ","End":"08:35.815","Text":"still less than Epsilon Tau."},{"Start":"08:35.815 ","End":"08:40.845","Text":"Now divide both sides by Epsilon so we have the sum of"},{"Start":"08:40.845 ","End":"08:46.730","Text":"the widths of the intervals in B Epsilon that\u0027s less than Tau."},{"Start":"08:46.730 ","End":"08:50.570","Text":"But this is what we called s Epsilon of P"},{"Start":"08:50.570 ","End":"08:54.920","Text":"and this is exactly what we had to show in the condition."},{"Start":"08:54.920 ","End":"08:57.220","Text":"We\u0027ve seen that we have finished the 2nd direction,"},{"Start":"08:57.220 ","End":"09:00.630","Text":"and so we finished the exercise."}],"ID":24700}],"Thumbnail":null,"ID":257168},{"Name":"Further Exercises - Criterion for Integrability","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"3m 38s","ChapterTopicVideoID":23798,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23798.jpeg","UploadDate":"2021-01-11T08:37:13.4370000","DurationForVideoObject":"PT3M38S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.775","Text":"In this exercise, f is a bounded function on the interval a, b."},{"Start":"00:05.775 ","End":"00:10.910","Text":"Let\u0027s suppose that there is a partition P of the interval a, b,"},{"Start":"00:10.910 ","End":"00:15.210","Text":"such that L of P, f equals U of P, f."},{"Start":"00:15.210 ","End":"00:21.270","Text":"We have to show that f is a constant function on the interval a, b."},{"Start":"00:21.270 ","End":"00:27.449","Text":"Let\u0027s say the partition is x-naught, x_1 up to x_n,"},{"Start":"00:27.449 ","End":"00:32.160","Text":"where the x-naught is a and the x_n is b."},{"Start":"00:32.160 ","End":"00:35.760","Text":"They\u0027re ordered in increasing order, of course."},{"Start":"00:35.760 ","End":"00:43.650","Text":"Now the U minus the L for this partition in this function is"},{"Start":"00:43.650 ","End":"00:51.780","Text":"the sum from 1 to n of M_i minus m_i times Delta x_i."},{"Start":"00:51.780 ","End":"00:57.810","Text":"To remind you that M_i is the supremum,"},{"Start":"00:57.810 ","End":"01:01.154","Text":"m_i is the infimum in each interval,"},{"Start":"01:01.154 ","End":"01:05.325","Text":"and Delta x_i is the width of the interval."},{"Start":"01:05.325 ","End":"01:08.465","Text":"Of course, Delta x_i is nonnegative"},{"Start":"01:08.465 ","End":"01:12.740","Text":"because the elements of the partition are in increasing order,"},{"Start":"01:12.740 ","End":"01:15.785","Text":"so each one minus the previous is bigger or equal to 0."},{"Start":"01:15.785 ","End":"01:19.760","Text":"Also, supremum is certainly bigger or equal to infimum,"},{"Start":"01:19.760 ","End":"01:22.195","Text":"so this is bigger or equal to 0."},{"Start":"01:22.195 ","End":"01:28.400","Text":"If we have the sum of nonnegative quantities being 0,"},{"Start":"01:28.400 ","End":"01:31.645","Text":"then each of these has to equal 0."},{"Start":"01:31.645 ","End":"01:33.890","Text":"If the product is 0,"},{"Start":"01:33.890 ","End":"01:36.995","Text":"then either this is 0 or this is 0."},{"Start":"01:36.995 ","End":"01:42.295","Text":"Anyway, we\u0027re going to show that f is constant on each of these intervals."},{"Start":"01:42.295 ","End":"01:48.060","Text":"Like I said, either M_i equals m_i or Delta x_i is 0,"},{"Start":"01:48.060 ","End":"01:50.750","Text":"in either case, we\u0027ll get that f is constant."},{"Start":"01:50.750 ","End":"01:52.490","Text":"In the one case,"},{"Start":"01:52.490 ","End":"01:57.470","Text":"if the supremum equals the infimum of a function on an interval,"},{"Start":"01:57.470 ","End":"02:00.590","Text":"certainly, it\u0027s constant because the function is"},{"Start":"02:00.590 ","End":"02:03.920","Text":"sandwiched between the infimum and the supremum,"},{"Start":"02:03.920 ","End":"02:05.665","Text":"and these 2 are the same."},{"Start":"02:05.665 ","End":"02:08.180","Text":"Also, if Delta x_i is 0,"},{"Start":"02:08.180 ","End":"02:09.800","Text":"then the width of the interval is 0,"},{"Start":"02:09.800 ","End":"02:11.810","Text":"so the intervals just a point."},{"Start":"02:11.810 ","End":"02:14.477","Text":"Certainly, a function on a point is a constant."},{"Start":"02:14.477 ","End":"02:16.040","Text":"It only has 1 value."},{"Start":"02:16.040 ","End":"02:17.390","Text":"Now, in either case,"},{"Start":"02:17.390 ","End":"02:20.075","Text":"whether it\u0027s because M_i equals m_i"},{"Start":"02:20.075 ","End":"02:22.040","Text":"or because Delta is 0,"},{"Start":"02:22.040 ","End":"02:26.035","Text":"we get that f is a constant and we\u0027ll call that constant c_i."},{"Start":"02:26.035 ","End":"02:30.020","Text":"At the moment, f is constant on each of the intervals,"},{"Start":"02:30.020 ","End":"02:33.230","Text":"but the constant c_i might not all be the same."},{"Start":"02:33.230 ","End":"02:36.680","Text":"We want to show that it\u0027s the same c_i on each i."},{"Start":"02:36.680 ","End":"02:41.890","Text":"The thing is that each pair of intervals overlaps at the endpoint."},{"Start":"02:41.890 ","End":"02:44.340","Text":"x_i minus 1, x_i"},{"Start":"02:44.340 ","End":"02:46.480","Text":"and x_i, x_i plus 1,"},{"Start":"02:46.480 ","End":"02:48.930","Text":"they overlap at the point x_i."},{"Start":"02:48.930 ","End":"02:57.615","Text":"Overlap means that it\u0027s the same constant because c_i is the constant for this interval,"},{"Start":"02:57.615 ","End":"03:00.585","Text":"and it\u0027s also f at any point."},{"Start":"03:00.585 ","End":"03:02.475","Text":"Let\u0027s say the right endpoint."},{"Start":"03:02.475 ","End":"03:06.750","Text":"c_i plus 1 is the constant on this interval."},{"Start":"03:06.750 ","End":"03:12.960","Text":"It\u0027s going to be the same as any of the x\u0027s here and choose the left endpoint."},{"Start":"03:14.020 ","End":"03:17.580","Text":"Once it\u0027s c_i and once it\u0027s c_i plus 1,"},{"Start":"03:17.580 ","End":"03:20.010","Text":"so these 2 are equal."},{"Start":"03:20.010 ","End":"03:23.595","Text":"Each 2 adjacent c_i\u0027s are the same,"},{"Start":"03:23.595 ","End":"03:26.265","Text":"so all the c_i\u0027s are equal."},{"Start":"03:26.265 ","End":"03:29.790","Text":"Since each point is in one of these intervals,"},{"Start":"03:29.790 ","End":"03:31.190","Text":"it\u0027s equal to one of the c_i,"},{"Start":"03:31.190 ","End":"03:32.510","Text":"but they\u0027re all the same,"},{"Start":"03:32.510 ","End":"03:36.190","Text":"so f is constant on a, b."},{"Start":"03:36.190 ","End":"03:38.440","Text":"We are done."}],"ID":24740},{"Watched":false,"Name":"Exercise 2","Duration":"10m 39s","ChapterTopicVideoID":23799,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23799.jpeg","UploadDate":"2021-01-11T08:42:12.2930000","DurationForVideoObject":"PT10M39S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.330","Text":"This exercise is a 3 in 1."},{"Start":"00:03.330 ","End":"00:05.190","Text":"In each of the 3 cases,"},{"Start":"00:05.190 ","End":"00:08.550","Text":"we have to evaluate the upper and lower integrals of"},{"Start":"00:08.550 ","End":"00:13.335","Text":"f. F is defined differently in each of these actions."},{"Start":"00:13.335 ","End":"00:18.030","Text":"We have to show that f is integrable and to find"},{"Start":"00:18.030 ","End":"00:23.294","Text":"the actual value of the integral of f on that interval."},{"Start":"00:23.294 ","End":"00:31.085","Text":"In Part a, f is a constant function equals always to Alpha on an interval a, b."},{"Start":"00:31.085 ","End":"00:36.780","Text":"In Part b, f is defined piece-wise at 1/2,"},{"Start":"00:36.780 ","End":"00:38.460","Text":"it\u0027s equal to 10,"},{"Start":"00:38.460 ","End":"00:40.095","Text":"but less than 1/2,"},{"Start":"00:40.095 ","End":"00:46.540","Text":"it\u0027s equal to 0 and greater than 1/2 it\u0027s equal to 1."},{"Start":"00:46.540 ","End":"00:51.490","Text":"This is on the interval from 0-1."},{"Start":"00:51.490 ","End":"00:58.565","Text":"The third part is a linear function f of x equals x on the interval 0, 1."},{"Start":"00:58.565 ","End":"01:06.320","Text":"We\u0027ll start with a and let\u0027s take any partition P of the interval a, b."},{"Start":"01:06.320 ","End":"01:09.470","Text":"Note that because we have a constant function,"},{"Start":"01:09.470 ","End":"01:19.885","Text":"the supremum and the infimum of the values in the interval x_i minus 1 x_i is Alpha."},{"Start":"01:19.885 ","End":"01:21.420","Text":"I mean the function\u0027s constant,"},{"Start":"01:21.420 ","End":"01:24.895","Text":"it\u0027s maximum and minimum is Alpha everywhere."},{"Start":"01:24.895 ","End":"01:32.695","Text":"The lower sum for this partition is the sum of M_i Delta x_i, which is,"},{"Start":"01:32.695 ","End":"01:36.540","Text":"you can take Alpha outside the brackets to solve the M_i or Alpha,"},{"Start":"01:36.540 ","End":"01:38.455","Text":"just the sum of the Deltas,"},{"Start":"01:38.455 ","End":"01:43.280","Text":"and this sums up with the interval which is b minus a."},{"Start":"01:43.280 ","End":"01:47.450","Text":"That\u0027s the lower and the same thing for the upper."},{"Start":"01:47.450 ","End":"01:51.275","Text":"It\u0027s exactly the same computation just with m here."},{"Start":"01:51.275 ","End":"01:57.450","Text":"The lower and the upper for a particular partition agree."},{"Start":"01:57.450 ","End":"02:02.990","Text":"The supremum of these and the infimum of these is also Alpha b minus a."},{"Start":"02:02.990 ","End":"02:07.315","Text":"The upper integral is the lower integral and that\u0027s Alpha b minus a,"},{"Start":"02:07.315 ","End":"02:11.640","Text":"especially the area of a rectangle and we\u0027ll just write a summary line."},{"Start":"02:11.640 ","End":"02:17.900","Text":"F is integrable and the integral from a to b of f of x dx is Alpha times b minus a."},{"Start":"02:17.900 ","End":"02:21.795","Text":"Now we\u0027ll move on to b."},{"Start":"02:21.795 ","End":"02:25.370","Text":"Let\u0027s just look at it again before it scrolls off."},{"Start":"02:25.370 ","End":"02:28.385","Text":"The function is 0 up to 1/2,"},{"Start":"02:28.385 ","End":"02:32.030","Text":"not including at 1/2 exactly it\u0027s 10 and then it goes"},{"Start":"02:32.030 ","End":"02:36.720","Text":"down to 1 from the right of 1/2 up to 1."},{"Start":"02:36.720 ","End":"02:43.415","Text":"We\u0027ll take the partition to be equal depths and we\u0027ll take n even,"},{"Start":"02:43.415 ","End":"02:46.250","Text":"say n equals 2k, in that way,"},{"Start":"02:46.250 ","End":"02:50.340","Text":"the middle value will be exactly 1/2."},{"Start":"02:50.340 ","End":"02:55.470","Text":"When n is k, k over n is k over 2k is 1/2."},{"Start":"02:55.470 ","End":"02:58.995","Text":"In general, x_i is i over n,"},{"Start":"02:58.995 ","End":"03:02.595","Text":"Delta x_i is the width of each is 1 over n,"},{"Start":"03:02.595 ","End":"03:06.265","Text":"and 1/2 falls at x_k."},{"Start":"03:06.265 ","End":"03:12.095","Text":"We\u0027ll make a table for the values of little m_i and big M_i,"},{"Start":"03:12.095 ","End":"03:14.345","Text":"the infimum and the supremum."},{"Start":"03:14.345 ","End":"03:19.440","Text":"If we\u0027re totally to the left of 1/2,"},{"Start":"03:19.440 ","End":"03:21.900","Text":"when i is less than k,"},{"Start":"03:21.900 ","End":"03:25.575","Text":"x_i and x_i minus 1 missed the 1/2."},{"Start":"03:25.575 ","End":"03:28.680","Text":"Both of these are 0."},{"Start":"03:28.680 ","End":"03:31.850","Text":"When i is exactly k,"},{"Start":"03:31.850 ","End":"03:34.685","Text":"we have from k minus 1 to k,"},{"Start":"03:34.685 ","End":"03:37.460","Text":"the interval and everything kicks up"},{"Start":"03:37.460 ","End":"03:43.310","Text":"the right border is where f is 0 except at the right border where it\u0027s 10."},{"Start":"03:43.310 ","End":"03:45.560","Text":"In contrast to the following interval,"},{"Start":"03:45.560 ","End":"03:48.950","Text":"which is from x_k to x_k plus 1,"},{"Start":"03:48.950 ","End":"03:52.700","Text":"they\u0027re just the left endpoint is 1/2,"},{"Start":"03:52.700 ","End":"03:55.205","Text":"and so f of it is 10."},{"Start":"03:55.205 ","End":"03:56.540","Text":"The rest of the interval,"},{"Start":"03:56.540 ","End":"03:58.190","Text":"the function is 1."},{"Start":"03:58.190 ","End":"04:01.535","Text":"Here we have the minimum 1 and the maximum 10."},{"Start":"04:01.535 ","End":"04:03.770","Text":"Finally, further to the right,"},{"Start":"04:03.770 ","End":"04:08.620","Text":"everything\u0027s 1 so the minimum and maximum are both 1."},{"Start":"04:08.620 ","End":"04:17.129","Text":"Let\u0027s start by computing U of P_n and f later we\u0027ll do the L. Now for the upper,"},{"Start":"04:17.129 ","End":"04:20.650","Text":"we\u0027re interested in the M_i."},{"Start":"04:20.650 ","End":"04:26.269","Text":"It starts off being 0 on the left values, left intervals,"},{"Start":"04:26.269 ","End":"04:28.695","Text":"then the 1 just before 1/2,"},{"Start":"04:28.695 ","End":"04:31.750","Text":"it\u0027s equal to 10 and the 1 just to the right of 1/2,"},{"Start":"04:31.750 ","End":"04:36.305","Text":"It\u0027s also 10, and then it drops off to 1."},{"Start":"04:36.305 ","End":"04:42.390","Text":"If we take the sum of m_i Delta x_i,"},{"Start":"04:42.390 ","End":"04:44.795","Text":"the very first 1 is 0, I haven\u0027t written it."},{"Start":"04:44.795 ","End":"04:51.310","Text":"Then we have 1 interval where it\u0027s Delta x_k times 10,"},{"Start":"04:51.310 ","End":"04:55.535","Text":"then another 1, Delta x_k plus 1 times 10,"},{"Start":"04:55.535 ","End":"05:00.755","Text":"and then the rest of the intervals from k plus 2 to n are just 1."},{"Start":"05:00.755 ","End":"05:03.045","Text":"It\u0027s 1 times Delta x_i."},{"Start":"05:03.045 ","End":"05:07.430","Text":"This is the sum we get and if we want to take the L,"},{"Start":"05:07.430 ","End":"05:10.250","Text":"the lowest sum for this partition,"},{"Start":"05:10.250 ","End":"05:13.750","Text":"then what we\u0027re interested is in for L,"},{"Start":"05:13.750 ","End":"05:18.880","Text":"we\u0027re interested in m. It starts off by being 0."},{"Start":"05:18.880 ","End":"05:23.895","Text":"Then the 1 just before 1/2 is still 0."},{"Start":"05:23.895 ","End":"05:26.530","Text":"The 1 just to the right of 1/2 is 1,"},{"Start":"05:26.530 ","End":"05:29.205","Text":"and so on for the rest of them, it\u0027s 1."},{"Start":"05:29.205 ","End":"05:32.880","Text":"I just need to put a color on the U."},{"Start":"05:33.260 ","End":"05:35.880","Text":"Let\u0027s compute this 1."},{"Start":"05:35.880 ","End":"05:40.640","Text":"Now, all the intervals all have width of 1 over"},{"Start":"05:40.640 ","End":"05:46.910","Text":"n and so what we get here is 10 of them plus 10 of them."},{"Start":"05:46.910 ","End":"05:48.200","Text":"Then from here to here,"},{"Start":"05:48.200 ","End":"05:50.390","Text":"we have k minus 1 of them,"},{"Start":"05:50.390 ","End":"05:59.660","Text":"and the width of each is 1 over n. What we get is 10 plus 10 minus 1 is 19 plus"},{"Start":"05:59.660 ","End":"06:03.830","Text":"k over n. K over n is 1/2 cause n is"},{"Start":"06:03.830 ","End":"06:09.675","Text":"2k and plus 19 over n. That\u0027s what this comes out to."},{"Start":"06:09.675 ","End":"06:14.255","Text":"For the L, what we get is this 1 gives us 0,"},{"Start":"06:14.255 ","End":"06:20.970","Text":"this 1 0, we have 1 interval just to the right of 1/2 and there the minimum is 1."},{"Start":"06:20.970 ","End":"06:26.190","Text":"That\u0027s 1 of those and the rest of them, it\u0027s also 1."},{"Start":"06:26.190 ","End":"06:34.710","Text":"What we have is 1 plus k minus 1 and each of these is 1 over n. This just comes out to be"},{"Start":"06:34.710 ","End":"06:43.895","Text":"1/2 independent of n. The upper integral is the infimum of all of these."},{"Start":"06:43.895 ","End":"06:46.655","Text":"It\u0027s less than or equal to each of these."},{"Start":"06:46.655 ","End":"06:49.400","Text":"It\u0027s less than or equal to 1/2 plus 19 over"},{"Start":"06:49.400 ","End":"06:53.930","Text":"n. This is true for all n and if it\u0027s true for all n,"},{"Start":"06:53.930 ","End":"06:58.150","Text":"then it\u0027s less than or equal to 1/2 cause that\u0027s the limit."},{"Start":"06:58.150 ","End":"07:02.660","Text":"Similarly, the lower integral is bigger or"},{"Start":"07:02.660 ","End":"07:06.920","Text":"equal to each of these L. It\u0027s bigger or equal to 1/2."},{"Start":"07:06.920 ","End":"07:10.640","Text":"From these 2, we get an inequality"},{"Start":"07:10.640 ","End":"07:14.720","Text":"that starts with 1/2 and ends with 1/2 and it\u0027s always less than or equal to,"},{"Start":"07:14.720 ","End":"07:17.270","Text":"which means that it has to be equality throughout."},{"Start":"07:17.270 ","End":"07:19.325","Text":"If it\u0027s equality throughout,"},{"Start":"07:19.325 ","End":"07:22.225","Text":"then this equals 1/2,"},{"Start":"07:22.225 ","End":"07:24.950","Text":"and this condition that the lower integral and the"},{"Start":"07:24.950 ","End":"07:27.590","Text":"upper integral are the same means that the function"},{"Start":"07:27.590 ","End":"07:33.220","Text":"is integrable and the common value is the value of the integral."},{"Start":"07:33.220 ","End":"07:37.545","Text":"The integral of this f of x from 0-1 is 1/2."},{"Start":"07:37.545 ","End":"07:41.365","Text":"We have still Part c. In Part c,"},{"Start":"07:41.365 ","End":"07:45.410","Text":"we had f of x equals x on the interval 0,"},{"Start":"07:45.410 ","End":"07:48.215","Text":"1, f of x equals x."},{"Start":"07:48.215 ","End":"07:56.449","Text":"Now we\u0027ll take the partition to be equally divided from 0-1 in n steps,"},{"Start":"07:56.449 ","End":"08:00.860","Text":"each of them 1 over n. The lower sum for"},{"Start":"08:00.860 ","End":"08:05.990","Text":"this partition is 1 over n. In each of the intervals,"},{"Start":"08:05.990 ","End":"08:08.540","Text":"the least is the left endpoint,"},{"Start":"08:08.540 ","End":"08:10.220","Text":"f of x equals x."},{"Start":"08:10.220 ","End":"08:14.465","Text":"The infimum is the left-hand point and the supremum is a right endpoint."},{"Start":"08:14.465 ","End":"08:16.464","Text":"For the lower sum,"},{"Start":"08:16.464 ","End":"08:18.540","Text":"we take 0, 1 over n,"},{"Start":"08:18.540 ","End":"08:24.320","Text":"2 over n up to n minus 1 over n and the upper is taking"},{"Start":"08:24.320 ","End":"08:32.405","Text":"the right endpoint in each pair and the width in each case is 1 over n. This is equal to,"},{"Start":"08:32.405 ","End":"08:38.320","Text":"now we know the sum of the numbers from 1 to anything,"},{"Start":"08:38.320 ","End":"08:44.315","Text":"say n minus 1 you take this multiplied by the following integer and divide by 2."},{"Start":"08:44.315 ","End":"08:47.120","Text":"On the denominator, we have n and we have n from here,"},{"Start":"08:47.120 ","End":"08:53.345","Text":"so that\u0027s n squared and we have n minus 1 n over 2 as the sum from 1 to n minus 1."},{"Start":"08:53.345 ","End":"08:55.085","Text":"Here we have similar,"},{"Start":"08:55.085 ","End":"08:58.550","Text":"just that the sum is from 1 to n. It\u0027s n,"},{"Start":"08:58.550 ","End":"09:02.995","Text":"n plus 1, and the same denominator 2n squared."},{"Start":"09:02.995 ","End":"09:09.350","Text":"Now, this sum, if we take the 1/2 out and we divide top and bottom by n,"},{"Start":"09:09.350 ","End":"09:11.120","Text":"we get n minus 1 over n,"},{"Start":"09:11.120 ","End":"09:12.980","Text":"which is 1 minus 1 over n,"},{"Start":"09:12.980 ","End":"09:15.710","Text":"and that tends to 1/2."},{"Start":"09:15.710 ","End":"09:21.860","Text":"It actually ascends, it\u0027s an increasing sequence and its limit is 1/2."},{"Start":"09:21.860 ","End":"09:25.070","Text":"On this side, if we simplify it,"},{"Start":"09:25.070 ","End":"09:29.060","Text":"we get 1 plus 1 over n and also goes to 1/2 from above."},{"Start":"09:29.060 ","End":"09:31.570","Text":"It\u0027s a decreasing sequence."},{"Start":"09:31.570 ","End":"09:35.190","Text":"The supremum of all these,"},{"Start":"09:35.190 ","End":"09:40.730","Text":"is 1/2 and the supremum of these is still less than or equal to"},{"Start":"09:40.730 ","End":"09:48.620","Text":"the lower integral because each of these is less than the lower integral."},{"Start":"09:48.620 ","End":"09:52.880","Text":"The lower integral is less than or equal to the upper integral and"},{"Start":"09:52.880 ","End":"09:57.410","Text":"the upper integral is less than or equal to each of these,"},{"Start":"09:57.410 ","End":"10:00.170","Text":"and hence to its infimum."},{"Start":"10:00.170 ","End":"10:03.575","Text":"This is also equal to 1/2."},{"Start":"10:03.575 ","End":"10:06.380","Text":"Just like in part b, we have 1/2,"},{"Start":"10:06.380 ","End":"10:08.090","Text":"less or equal to this, less or equal to this,"},{"Start":"10:08.090 ","End":"10:10.285","Text":"less or equal to 1/2."},{"Start":"10:10.285 ","End":"10:14.900","Text":"In other words, each of these has got to equal 1/2 and so the"},{"Start":"10:14.900 ","End":"10:20.605","Text":"integral exists and is equal to 1/2."},{"Start":"10:20.605 ","End":"10:24.710","Text":"Just by the way, for those who know how to do definite integrals,"},{"Start":"10:24.710 ","End":"10:28.545","Text":"the integral of x from 0-1,"},{"Start":"10:28.545 ","End":"10:33.875","Text":"here\u0027s the calculation for those who know integration and it comes out to be 1/2."},{"Start":"10:33.875 ","End":"10:37.755","Text":"That\u0027s just to check to make sure."},{"Start":"10:37.755 ","End":"10:40.510","Text":"We are done."}],"ID":24741},{"Watched":false,"Name":"Exercise 3","Duration":"2m 42s","ChapterTopicVideoID":23800,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23800.jpeg","UploadDate":"2021-01-11T08:43:24.3470000","DurationForVideoObject":"PT2M42S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.340","Text":"In this exercise, f is a bounded function on the interval a, b."},{"Start":"00:05.340 ","End":"00:10.095","Text":"We have a sequence of partitions P_n on this interval,"},{"Start":"00:10.095 ","End":"00:17.766","Text":"such that the upper sum minus the lower sum tend to 0 as n goes to infinity."},{"Start":"00:17.766 ","End":"00:18.575","Text":"2 parts."},{"Start":"00:18.575 ","End":"00:24.060","Text":"In a, we have to show that f is integrable on a, b, you should say."},{"Start":"00:24.060 ","End":"00:30.720","Text":"Part b is to show that the limit of the upper sum for P_n"},{"Start":"00:30.720 ","End":"00:34.380","Text":"as well as the limit for the lower sum of P_n,"},{"Start":"00:34.380 ","End":"00:37.320","Text":"both are equal to the integral,"},{"Start":"00:37.320 ","End":"00:39.835","Text":"which we know exists from a."},{"Start":"00:39.835 ","End":"00:41.960","Text":"Now, for each n,"},{"Start":"00:41.960 ","End":"00:45.720","Text":"the lower sum for the partition P_n"},{"Start":"00:45.720 ","End":"00:50.140","Text":"is less than or equal to the lower integral."},{"Start":"00:50.140 ","End":"00:53.840","Text":"The lower integral is always less than or equal to the upper integral"},{"Start":"00:53.840 ","End":"00:58.205","Text":"and it\u0027s less than or equal to the upper sum for a partition."},{"Start":"00:58.205 ","End":"01:01.472","Text":"Now, clearly, if you have these inequalities"},{"Start":"01:01.472 ","End":"01:03.665","Text":"and this minus this"},{"Start":"01:03.665 ","End":"01:08.395","Text":"is less than or equal to this, minus this,"},{"Start":"01:08.395 ","End":"01:14.460","Text":"this difference tends to 0 as n goes to infinity."},{"Start":"01:15.470 ","End":"01:18.324","Text":"By the sandwich theorem,"},{"Start":"01:18.324 ","End":"01:20.255","Text":"this tends to 0."},{"Start":"01:20.255 ","End":"01:22.025","Text":"Well, it doesn\u0027t depend on n"},{"Start":"01:22.025 ","End":"01:24.140","Text":"and therefore, it is 0."},{"Start":"01:24.140 ","End":"01:29.745","Text":"This minus this, which means that this equals this."},{"Start":"01:29.745 ","End":"01:35.980","Text":"f is integrable, and there exists the integral from a to b of f of x, dx."},{"Start":"01:35.980 ","End":"01:39.165","Text":"Now on to part b."},{"Start":"01:39.165 ","End":"01:42.060","Text":"The integral, which exists,"},{"Start":"01:42.060 ","End":"01:48.065","Text":"is less than or equal to any upper sum for partition P_n"},{"Start":"01:48.065 ","End":"01:52.065","Text":"and bigger or equal to the lower sum for the partition P_n,"},{"Start":"01:52.065 ","End":"01:54.490","Text":"or for any partition."},{"Start":"01:55.100 ","End":"01:59.615","Text":"If we subtract this L from everything,"},{"Start":"01:59.615 ","End":"02:04.500","Text":"we get that 0 is less than or equal to this minus L,"},{"Start":"02:04.500 ","End":"02:07.440","Text":"which is less than or equal to U minus L."},{"Start":"02:07.440 ","End":"02:13.035","Text":"This were told goes to 0 as n goes to infinity."},{"Start":"02:13.035 ","End":"02:16.755","Text":"That means, again, by the sandwich theorem,"},{"Start":"02:16.755 ","End":"02:24.065","Text":"that the integral minus the lower sum for P_n goes to 0."},{"Start":"02:24.065 ","End":"02:32.295","Text":"That\u0027s equivalent to saying that this sequence L of P_n and f goes to this integral."},{"Start":"02:32.295 ","End":"02:34.650","Text":"Similarly, it\u0027s almost identical,"},{"Start":"02:34.650 ","End":"02:35.999","Text":"I won\u0027t go into the details,"},{"Start":"02:35.999 ","End":"02:38.070","Text":"the U of P_n and f"},{"Start":"02:38.070 ","End":"02:39.960","Text":"goes to the integral."},{"Start":"02:39.960 ","End":"02:43.840","Text":"We\u0027re done."}],"ID":24742},{"Watched":false,"Name":"Exercise 4","Duration":"5m 44s","ChapterTopicVideoID":23801,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23801.jpeg","UploadDate":"2021-01-11T08:46:41.1400000","DurationForVideoObject":"PT5M44S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.820","Text":"This exercise is a 3 in 1."},{"Start":"00:02.820 ","End":"00:05.610","Text":"In each case, we have to show that the function"},{"Start":"00:05.610 ","End":"00:10.035","Text":"f is integrable using the Riemann criterion."},{"Start":"00:10.035 ","End":"00:14.730","Text":"In A, f of x is x on the interval 0, 1."},{"Start":"00:14.730 ","End":"00:21.345","Text":"On B, it\u0027s x squared on 0, 1 and then C, it\u0027s 1 over x on the interval 1, 2."},{"Start":"00:21.345 ","End":"00:23.475","Text":"In each of the 3 cases,"},{"Start":"00:23.475 ","End":"00:28.950","Text":"it\u0027s sufficient to find partitions P_n such that U of P_n,"},{"Start":"00:28.950 ","End":"00:31.050","Text":"f minus L of P_n,"},{"Start":"00:31.050 ","End":"00:33.720","Text":"and f turns to 0."},{"Start":"00:33.720 ","End":"00:39.280","Text":"This is because of a previous exercise and I\u0027ll remind you,"},{"Start":"00:39.280 ","End":"00:43.430","Text":"this was the exercise in case you missed it."},{"Start":"00:43.430 ","End":"00:47.150","Text":"Okay, back to ours. Here we are."},{"Start":"00:47.150 ","End":"00:51.665","Text":"So let\u0027s start with a and let\u0027s take the partition P_n,"},{"Start":"00:51.665 ","End":"00:54.800","Text":"just divide the interval 0,"},{"Start":"00:54.800 ","End":"00:57.080","Text":"1 into n equal pieces."},{"Start":"00:57.080 ","End":"01:01.745","Text":"Now we have n intervals here and in each of them,"},{"Start":"01:01.745 ","End":"01:03.575","Text":"because f of x equals x,"},{"Start":"01:03.575 ","End":"01:06.500","Text":"the minimum is the left endpoint of"},{"Start":"01:06.500 ","End":"01:11.049","Text":"the interval and the maximum is the right endpoint of the interval."},{"Start":"01:11.049 ","End":"01:14.975","Text":"What we get is that the lower sum,"},{"Start":"01:14.975 ","End":"01:17.610","Text":"we take the left 0, 1 over n,"},{"Start":"01:17.610 ","End":"01:23.910","Text":"2 over n up to n minus 1 over n. The width is the same for each."},{"Start":"01:23.910 ","End":"01:26.915","Text":"It\u0027s 1 over n and the sum of these,"},{"Start":"01:26.915 ","End":"01:30.515","Text":"using the formula for the sum from 1"},{"Start":"01:30.515 ","End":"01:33.470","Text":"to n. In this case n minus 1."},{"Start":"01:33.470 ","End":"01:37.910","Text":"We take the n minus 1 multiplied by the next integer and divide by 2."},{"Start":"01:37.910 ","End":"01:43.040","Text":"The n squared comes from n here and ends in the denominator here."},{"Start":"01:43.040 ","End":"01:47.015","Text":"Similarly, the upper sum,"},{"Start":"01:47.015 ","End":"01:49.460","Text":"we just take the right endpoints,"},{"Start":"01:49.460 ","End":"01:52.370","Text":"so we have from 1 to n, and that\u0027s n,"},{"Start":"01:52.370 ","End":"01:56.390","Text":"n plus 1 over 2 and n squared as before."},{"Start":"01:56.390 ","End":"02:01.265","Text":"The difference is this minus this,"},{"Start":"02:01.265 ","End":"02:03.300","Text":"and to simplify it,"},{"Start":"02:03.300 ","End":"02:06.335","Text":"do the subtraction, comes out to be 1 over n,"},{"Start":"02:06.335 ","End":"02:08.360","Text":"and this turns to 0,"},{"Start":"02:08.360 ","End":"02:10.565","Text":"and that\u0027s what we wanted."},{"Start":"02:10.565 ","End":"02:13.975","Text":"So that concludes part A,"},{"Start":"02:13.975 ","End":"02:17.900","Text":"and part B is somewhat similar."},{"Start":"02:17.900 ","End":"02:20.240","Text":"We use the same partition,"},{"Start":"02:20.240 ","End":"02:25.505","Text":"just equal steps, 1 over n each time."},{"Start":"02:25.505 ","End":"02:30.305","Text":"As before, the minimum is on the left"},{"Start":"02:30.305 ","End":"02:34.400","Text":"of the interval and the maximum occurs on the right of the interval."},{"Start":"02:34.400 ","End":"02:38.330","Text":"I also want to remind you of a formula we\u0027re going to use that the sum of"},{"Start":"02:38.330 ","End":"02:43.680","Text":"the squares from 1 squared up to n squared is given here."},{"Start":"02:43.680 ","End":"02:46.660","Text":"Now, the lowest sum, like I said,"},{"Start":"02:46.660 ","End":"02:50.050","Text":"we take the squares of the left endpoint,"},{"Start":"02:50.050 ","End":"02:53.185","Text":"which is from 0 up to n minus 1 over n,"},{"Start":"02:53.185 ","End":"02:56.905","Text":"each of them squared and this is what we get from this formula."},{"Start":"02:56.905 ","End":"03:00.435","Text":"The upper sum, we get from the right endpoints,"},{"Start":"03:00.435 ","End":"03:03.780","Text":"1 over n, 2 over n up to n over n squared."},{"Start":"03:03.780 ","End":"03:08.095","Text":"We get this and if we subtract,"},{"Start":"03:08.095 ","End":"03:13.040","Text":"we get this minus this and that comes out."},{"Start":"03:13.040 ","End":"03:15.250","Text":"I just put a common denominator,"},{"Start":"03:15.250 ","End":"03:24.970","Text":"divide top and bottom by n here and 1 of these n. So we\u0027re left with n squared."},{"Start":"03:25.400 ","End":"03:30.345","Text":"Now, this product is n squared plus 3n plus 2."},{"Start":"03:30.345 ","End":"03:32.030","Text":"This product is very similar,"},{"Start":"03:32.030 ","End":"03:34.895","Text":"10 squared minus 3n plus 2."},{"Start":"03:34.895 ","End":"03:41.100","Text":"When we subtract, we just get 3n minus minus 3n, which is 6n."},{"Start":"03:41.100 ","End":"03:43.560","Text":"So we have 6n over 6n squared,"},{"Start":"03:43.560 ","End":"03:46.694","Text":"which is 1 over n, which turns to 0."},{"Start":"03:46.694 ","End":"03:50.780","Text":"That\u0027s what we wanted to show in each of the 3 cases."},{"Start":"03:50.780 ","End":"03:52.885","Text":"So we\u0027ve done it for part B."},{"Start":"03:52.885 ","End":"03:57.140","Text":"Let\u0027s move on to part C. In this case,"},{"Start":"03:57.140 ","End":"03:58.640","Text":"we have a different interval."},{"Start":"03:58.640 ","End":"04:01.415","Text":"It\u0027s not from 0-1, it\u0027s from 1-2."},{"Start":"04:01.415 ","End":"04:04.100","Text":"But again, we\u0027ll take equal spaces,"},{"Start":"04:04.100 ","End":"04:08.220","Text":"1 plus something over n in each case."},{"Start":"04:08.220 ","End":"04:13.290","Text":"We have n equal steps to get from 1-2."},{"Start":"04:13.290 ","End":"04:16.320","Text":"The function is 1 over x."},{"Start":"04:16.320 ","End":"04:18.225","Text":"This time it\u0027s decreasing."},{"Start":"04:18.225 ","End":"04:21.695","Text":"The maximum is on the left and the minimum is on the right."},{"Start":"04:21.695 ","End":"04:26.860","Text":"The lowest sum is 1 over this and then 1 over this."},{"Start":"04:26.860 ","End":"04:29.505","Text":"Now if we do a bit of algebra, we can do it in our heads."},{"Start":"04:29.505 ","End":"04:35.880","Text":"This is n plus 1 over n. So we invert it for the 1 over x."},{"Start":"04:35.880 ","End":"04:37.620","Text":"Here n plus 2 over n,"},{"Start":"04:37.620 ","End":"04:39.240","Text":"so it\u0027s n over n plus 2."},{"Start":"04:39.240 ","End":"04:43.640","Text":"In general, we\u0027re going to get always n over n plus 1, n plus 2,"},{"Start":"04:43.640 ","End":"04:48.675","Text":"and so on, up to 2n, n terms altogether."},{"Start":"04:48.675 ","End":"04:50.670","Text":"For the upper sum,"},{"Start":"04:50.670 ","End":"04:52.860","Text":"we\u0027ll just get something very similar,"},{"Start":"04:52.860 ","End":"04:55.875","Text":"just we start with the left endpoint."},{"Start":"04:55.875 ","End":"05:00.540","Text":"We start from 1 over this,"},{"Start":"05:00.540 ","End":"05:03.150","Text":"which is actually n over n,"},{"Start":"05:03.150 ","End":"05:06.315","Text":"then 1 over this which is n over n plus 1."},{"Start":"05:06.315 ","End":"05:07.835","Text":"Basically the same as here,"},{"Start":"05:07.835 ","End":"05:09.335","Text":"just that we stop here."},{"Start":"05:09.335 ","End":"05:10.774","Text":"Now we want the difference,"},{"Start":"05:10.774 ","End":"05:16.045","Text":"the U minus L. So it\u0027s 1 over n. Basically,"},{"Start":"05:16.045 ","End":"05:21.470","Text":"everything\u0027s the same except that here we have this,"},{"Start":"05:21.470 ","End":"05:23.240","Text":"and here we have this,"},{"Start":"05:23.240 ","End":"05:25.190","Text":"and everything else is the same."},{"Start":"05:25.190 ","End":"05:32.070","Text":"We just have n over n minus n over 2n times the 1 over n for the difference."},{"Start":"05:32.070 ","End":"05:34.035","Text":"This is 1 minus 1/2,"},{"Start":"05:34.035 ","End":"05:36.990","Text":"which is 1/2, 1/2 times 1 over n, 1 over 2n."},{"Start":"05:36.990 ","End":"05:40.590","Text":"Anyway, turns to 0 and that\u0027s what we had to show,"},{"Start":"05:40.590 ","End":"05:44.620","Text":"and so we are done."}],"ID":24743},{"Watched":false,"Name":"Exercise 5","Duration":"2m 9s","ChapterTopicVideoID":23802,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23802.jpeg","UploadDate":"2021-01-11T08:47:42.0400000","DurationForVideoObject":"PT2M9S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.030","Text":"In this exercise, we have 3 functions, f_1, f_2,"},{"Start":"00:03.030 ","End":"00:06.690","Text":"and f. All 3 of them are bounded on the interval 0,"},{"Start":"00:06.690 ","End":"00:09.975","Text":"1, and f_1 is less than or equal to f,"},{"Start":"00:09.975 ","End":"00:12.000","Text":"which is less than or equal to f_2,"},{"Start":"00:12.000 ","End":"00:13.680","Text":"on all the interval."},{"Start":"00:13.680 ","End":"00:16.650","Text":"Now suppose f_1 and f_2 are integrable,"},{"Start":"00:16.650 ","End":"00:18.750","Text":"and that the integral of f_1 equals"},{"Start":"00:18.750 ","End":"00:21.515","Text":"the integral of f_2 and is equal to I."},{"Start":"00:21.515 ","End":"00:26.530","Text":"We have to show that f is also integrable,"},{"Start":"00:26.530 ","End":"00:30.410","Text":"and to find its integral from 0 to 1."},{"Start":"00:30.410 ","End":"00:35.015","Text":"Now take a partition P of 0, 1,"},{"Start":"00:35.015 ","End":"00:40.895","Text":"then L of P and f_1 is less than or equal to L of P and f,"},{"Start":"00:40.895 ","End":"00:44.900","Text":"because f_1 is less than or equal to f. Now this is"},{"Start":"00:44.900 ","End":"00:50.030","Text":"less than or equal to the lower integral of f,"},{"Start":"00:50.030 ","End":"00:53.060","Text":"and that\u0027s less than or equal to the upper integral of f."},{"Start":"00:53.060 ","End":"00:55.880","Text":"This is always true for any f."},{"Start":"00:55.880 ","End":"00:58.670","Text":"This is less than or equal to any particular"},{"Start":"00:58.670 ","End":"01:01.325","Text":"upper sum U of P and F,"},{"Start":"01:01.325 ","End":"01:04.760","Text":"which is less than or equal to U of P and f_2,"},{"Start":"01:04.760 ","End":"01:07.760","Text":"because f is less than or equal to f_2."},{"Start":"01:07.760 ","End":"01:13.880","Text":"What we get, is that the integral of f_1,"},{"Start":"01:13.880 ","End":"01:16.805","Text":"which is the supremum of all of these,"},{"Start":"01:16.805 ","End":"01:18.970","Text":"but each of these is less than or equal to this,"},{"Start":"01:18.970 ","End":"01:22.265","Text":"so the supremum is also less than or equal to this,"},{"Start":"01:22.265 ","End":"01:26.090","Text":"and is less than or equal to this as before."},{"Start":"01:26.090 ","End":"01:30.270","Text":"This is less than or equal to each of these use,"},{"Start":"01:30.270 ","End":"01:33.235","Text":"so it\u0027s less than or equal to the infimum."},{"Start":"01:33.235 ","End":"01:37.840","Text":"The infimum is the integral of f_2."},{"Start":"01:37.840 ","End":"01:40.835","Text":"This and this are both equal to I."},{"Start":"01:40.835 ","End":"01:42.830","Text":"We have I is less than or equal to this,"},{"Start":"01:42.830 ","End":"01:45.110","Text":"less than or equal to this, less than or equal to I,"},{"Start":"01:45.110 ","End":"01:47.855","Text":"which means that we have equality throughout."},{"Start":"01:47.855 ","End":"01:50.435","Text":"If we have equality throughout,"},{"Start":"01:50.435 ","End":"01:52.540","Text":"then it means that this equals this,"},{"Start":"01:52.540 ","End":"01:54.830","Text":"and so the integral of f exists,"},{"Start":"01:54.830 ","End":"01:57.355","Text":"and furthermore it\u0027s equal to I."},{"Start":"01:57.355 ","End":"01:59.600","Text":"This is intuitive. I mean,"},{"Start":"01:59.600 ","End":"02:01.730","Text":"if the integral of f_1 is I and the integral of"},{"Start":"02:01.730 ","End":"02:04.430","Text":"f_2 is I and f is sandwiched between them,"},{"Start":"02:04.430 ","End":"02:06.785","Text":"you\u0027d expect it also to be I."},{"Start":"02:06.785 ","End":"02:09.390","Text":"Okay. We\u0027re done."}],"ID":24744},{"Watched":false,"Name":"Exercise 6","Duration":"7m 52s","ChapterTopicVideoID":23803,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23803.jpeg","UploadDate":"2021-01-11T08:51:36.0400000","DurationForVideoObject":"PT7M52S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.780","Text":"In this exercise, f is defined on the interval"},{"Start":"00:03.780 ","End":"00:10.020","Text":"0,1 in such a way that f of x is equal to x whenever x is rational,"},{"Start":"00:10.020 ","End":"00:14.640","Text":"but f of x equals 0 for all the irrational x."},{"Start":"00:14.640 ","End":"00:21.945","Text":"We have to evaluate the upper and lower integrals of f and show that f is not integrable."},{"Start":"00:21.945 ","End":"00:28.830","Text":"Actually, the upper and lower are going to be unequal and therefore not integrable,"},{"Start":"00:28.830 ","End":"00:33.240","Text":"and all the work will be done for the upper 1,"},{"Start":"00:33.240 ","End":"00:34.690","Text":"which is harder to calculate,"},{"Start":"00:34.690 ","End":"00:36.060","Text":"the lower 1 is trivial."},{"Start":"00:36.060 ","End":"00:38.265","Text":"Anyway, we\u0027ll start with a difficult 1 first,"},{"Start":"00:38.265 ","End":"00:43.739","Text":"and we\u0027ll take partition P_n to be as follows,"},{"Start":"00:43.739 ","End":"00:47.670","Text":"what we often do, just break it up into n equal pieces."},{"Start":"00:47.670 ","End":"00:53.965","Text":"The width of each partition is 1 over n. M_i,"},{"Start":"00:53.965 ","End":"01:00.775","Text":"the supremum of the interval from x_i minus 1 to x_i is equal to x_i."},{"Start":"01:00.775 ","End":"01:07.545","Text":"This is because x_i it\u0027s actually i over n, which is rational."},{"Start":"01:07.545 ","End":"01:09.390","Text":"That\u0027s the biggest we can get."},{"Start":"01:09.390 ","End":"01:12.920","Text":"It\u0027s increasing on the rational and for the irrational to 0,"},{"Start":"01:12.920 ","End":"01:23.000","Text":"so the supremum is x_i or i over n. The upper sum for this partition is 1 over n,"},{"Start":"01:23.000 ","End":"01:24.140","Text":"which is the width,"},{"Start":"01:24.140 ","End":"01:30.870","Text":"times the sum of the right side of the partition, the x_i,"},{"Start":"01:30.870 ","End":"01:36.055","Text":"which is the i over n. We sum them from 1 to n. What we get,"},{"Start":"01:36.055 ","End":"01:38.510","Text":"the numerators add up to n,"},{"Start":"01:38.510 ","End":"01:39.770","Text":"n plus 1 over 2,"},{"Start":"01:39.770 ","End":"01:41.270","Text":"and here we have an n and an n,"},{"Start":"01:41.270 ","End":"01:43.310","Text":"makes it n squared."},{"Start":"01:43.310 ","End":"01:47.660","Text":"This comes out to be a 1/2 plus 1/2n."},{"Start":"01:47.660 ","End":"01:53.270","Text":"Now the upper integral of f is less than or equal to each 1 of these,"},{"Start":"01:53.270 ","End":"01:56.810","Text":"and therefore to the infimum of these on all integers"},{"Start":"01:56.810 ","End":"02:02.000","Text":"n. This is equal to just what we calculated,"},{"Start":"02:02.000 ","End":"02:04.310","Text":"the infimum of a 1/2 plus 1/2n,"},{"Start":"02:04.310 ","End":"02:05.540","Text":"which is a 1/2,"},{"Start":"02:05.540 ","End":"02:08.149","Text":"just like the limit as n goes to infinity."},{"Start":"02:08.149 ","End":"02:14.030","Text":"Now actually, this upper integral will turn out to be exactly equal to a 1/2."},{"Start":"02:14.030 ","End":"02:17.285","Text":"What we\u0027ll do is we\u0027ll show the reverse inequality."},{"Start":"02:17.285 ","End":"02:20.600","Text":"We show that this is bigger or equal to 1/2,"},{"Start":"02:20.600 ","End":"02:22.205","Text":"then we\u0027ll get equality."},{"Start":"02:22.205 ","End":"02:24.540","Text":"Let P, x_0, x_1,"},{"Start":"02:24.540 ","End":"02:28.110","Text":"x_2 up to x_n be any partition of 0,1."},{"Start":"02:28.110 ","End":"02:36.230","Text":"Now the upper sum U of p and f is by definition the sum of M_i Delta x_i."},{"Start":"02:36.230 ","End":"02:39.960","Text":"I claim that each of"},{"Start":"02:39.960 ","End":"02:49.390","Text":"these M_i Delta x_i is equal to x_i times x_i minus x_i minus 1."},{"Start":"02:49.760 ","End":"02:52.489","Text":"Let\u0027s prove this claim."},{"Start":"02:52.489 ","End":"02:56.420","Text":"If the interval is a trivial interval,"},{"Start":"02:56.420 ","End":"02:57.920","Text":"meaning it\u0027s just a point,"},{"Start":"02:57.920 ","End":"03:00.065","Text":"the 2 end points are equal,"},{"Start":"03:00.065 ","End":"03:07.655","Text":"then it\u0027s clear, because this comes out to be 0 and this comes out to be 0."},{"Start":"03:07.655 ","End":"03:10.040","Text":"If these 2 are not equal,"},{"Start":"03:10.040 ","End":"03:13.115","Text":"then this is strictly less than this."},{"Start":"03:13.115 ","End":"03:22.635","Text":"We can choose a sequence of rationals in the interval such that r_n tends to x_i."},{"Start":"03:22.635 ","End":"03:29.710","Text":"Any real number can be approached on the left or on the right by a sequence of rationals."},{"Start":"03:29.710 ","End":"03:31.980","Text":"F of r_n equals r_n."},{"Start":"03:31.980 ","End":"03:36.600","Text":"By the definition of f for rationals f of x equals x."},{"Start":"03:36.600 ","End":"03:43.565","Text":"R_n is less than or equal to the supremum of f on the interval,"},{"Start":"03:43.565 ","End":"03:52.005","Text":"so what we have is that M_i is bigger or equal to the limit of the r_n\u0027s,"},{"Start":"03:52.005 ","End":"03:57.030","Text":"meaning if M_i is bigger or equal to each r_n and it\u0027s bigger or equal to the limit,"},{"Start":"03:57.030 ","End":"03:59.645","Text":"and the limit, as it says here,"},{"Start":"03:59.645 ","End":"04:04.070","Text":"is x_i, so this is an inequality in 1 direction."},{"Start":"04:04.070 ","End":"04:07.895","Text":"Now we show the inequality in the other direction."},{"Start":"04:07.895 ","End":"04:11.210","Text":"If we take any x in this interval,"},{"Start":"04:11.210 ","End":"04:14.960","Text":"then f of x is either equal to 0 or"},{"Start":"04:14.960 ","End":"04:18.740","Text":"to x depending on whether x is rational or irrational."},{"Start":"04:18.740 ","End":"04:20.750","Text":"But whether it\u0027s 0, whether it\u0027s x,"},{"Start":"04:20.750 ","End":"04:22.910","Text":"it\u0027s still less than or equal to x,"},{"Start":"04:22.910 ","End":"04:27.145","Text":"and x is less than or equal to the right end point,"},{"Start":"04:27.145 ","End":"04:32.360","Text":"so f of x is less than or equal to x_i for each x in the interval."},{"Start":"04:32.360 ","End":"04:36.390","Text":"M_i is the supremum of such f of x,"},{"Start":"04:36.390 ","End":"04:40.160","Text":"so it\u0027s also less than or equal to x_i."},{"Start":"04:40.160 ","End":"04:47.300","Text":"M_i equals x_i, and so M_i equals x_i."},{"Start":"04:47.300 ","End":"04:49.160","Text":"That\u0027s really all we had to show."},{"Start":"04:49.160 ","End":"04:51.310","Text":"The remaining bit is obvious,"},{"Start":"04:51.310 ","End":"04:55.165","Text":"Delta x is by definition x_i minus x_i minus 1."},{"Start":"04:55.165 ","End":"05:00.469","Text":"We\u0027ve proved that claim that this equals this."},{"Start":"05:00.469 ","End":"05:03.720","Text":"That\u0027s the QED."},{"Start":"05:03.970 ","End":"05:08.030","Text":"This u of p and f, we can,"},{"Start":"05:08.030 ","End":"05:12.095","Text":"instead of this, replace M_i by x_i."},{"Start":"05:12.095 ","End":"05:17.030","Text":"This is equal to the sum of this,"},{"Start":"05:17.030 ","End":"05:18.620","Text":"that\u0027s what we just did here."},{"Start":"05:18.620 ","End":"05:27.790","Text":"Now let\u0027s multiply out x_i times x_i is x_i squared minus the sum of x_i, x_i minus 1."},{"Start":"05:27.790 ","End":"05:31.280","Text":"We\u0027re going to apply an inequality to this bit."},{"Start":"05:31.280 ","End":"05:37.385","Text":"In general, ab is less than or equal to 1/2 of a squared plus b squared."},{"Start":"05:37.385 ","End":"05:41.930","Text":"If you multiply both sides by 2 and bring the 2ab over here,"},{"Start":"05:41.930 ","End":"05:44.660","Text":"it just says that a squared minus 2ab plus b"},{"Start":"05:44.660 ","End":"05:48.965","Text":"squared is non-negative and it\u0027s a perfect square."},{"Start":"05:48.965 ","End":"05:51.290","Text":"Think about that."},{"Start":"05:51.290 ","End":"05:54.200","Text":"If we apply that here,"},{"Start":"05:54.200 ","End":"05:56.240","Text":"notice that there\u0027s a minus here,"},{"Start":"05:56.240 ","End":"05:57.620","Text":"so if it was minus ab,"},{"Start":"05:57.620 ","End":"05:59.590","Text":"it would be bigger or equal to."},{"Start":"05:59.590 ","End":"06:04.305","Text":"This is bigger or equal to this bit as is,"},{"Start":"06:04.305 ","End":"06:09.570","Text":"minus a 1/2, x_i squared plus x_i minus 1 squared."},{"Start":"06:09.570 ","End":"06:14.700","Text":"X_i squared minus a 1/2 of x squared is just a 1/2 of x_i squared."},{"Start":"06:14.700 ","End":"06:19.710","Text":"Here we have minus a 1/2 of x_i minus 1 squared."},{"Start":"06:19.710 ","End":"06:23.190","Text":"This comes out to be this."},{"Start":"06:23.190 ","End":"06:26.225","Text":"This is a telescoping series."},{"Start":"06:26.225 ","End":"06:29.930","Text":"We get x_n squared minus x_n minus 1 squared plus x_n"},{"Start":"06:29.930 ","End":"06:33.710","Text":"minus 1 squared minus the lower 1 squared."},{"Start":"06:33.710 ","End":"06:35.600","Text":"Each time we\u0027re subtracting the previous,"},{"Start":"06:35.600 ","End":"06:38.370","Text":"so we have the first squared minus the last squared,"},{"Start":"06:38.370 ","End":"06:39.890","Text":"is all we\u0027re left with,"},{"Start":"06:39.890 ","End":"06:48.840","Text":"which is this, but the x_n squared is just 1 squared and x_0 squared is 0 squared."},{"Start":"06:48.840 ","End":"06:52.005","Text":"We just get this as equal to 1/2."},{"Start":"06:52.005 ","End":"06:54.675","Text":"U of P, f is bigger or equal to 1/2,"},{"Start":"06:54.675 ","End":"06:59.090","Text":"so the infimum of this is bigger or equal to a 1/2,"},{"Start":"06:59.090 ","End":"07:01.975","Text":"and the infimum of this is the upper integral,"},{"Start":"07:01.975 ","End":"07:05.520","Text":"and earlier we had the opposite inequality,"},{"Start":"07:05.520 ","End":"07:09.830","Text":"and so the upper integral is exactly equal to a 1/2."},{"Start":"07:09.830 ","End":"07:11.270","Text":"That was a difficult 1."},{"Start":"07:11.270 ","End":"07:14.670","Text":"The other 1, the lower integral,"},{"Start":"07:14.670 ","End":"07:17.840","Text":"it\u0027s pretty clear that in any interval,"},{"Start":"07:17.840 ","End":"07:20.134","Text":"if its width is not 0,"},{"Start":"07:20.134 ","End":"07:24.335","Text":"then the lower sum is just 0,"},{"Start":"07:24.335 ","End":"07:25.730","Text":"so L of P,"},{"Start":"07:25.730 ","End":"07:28.340","Text":"f is 0 for all partitions,"},{"Start":"07:28.340 ","End":"07:34.190","Text":"and that means that the lower integral is 0."},{"Start":"07:34.190 ","End":"07:36.335","Text":"We take the infimum of all of these,"},{"Start":"07:36.335 ","End":"07:40.470","Text":"and it\u0027s also equal to 0."},{"Start":"07:40.470 ","End":"07:42.585","Text":"The upper integral is a 1/2,"},{"Start":"07:42.585 ","End":"07:45.315","Text":"the lower integral is 0,"},{"Start":"07:45.315 ","End":"07:48.530","Text":"and this proves that f is not integrable because"},{"Start":"07:48.530 ","End":"07:52.890","Text":"the upper and lower integrals differ. We\u0027re done."}],"ID":24745},{"Watched":false,"Name":"Exercise 7","Duration":"14m 35s","ChapterTopicVideoID":23804,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23804.jpeg","UploadDate":"2021-01-11T08:58:25.3030000","DurationForVideoObject":"PT14M35S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.940","Text":"This exercise has 4 parts."},{"Start":"00:02.940 ","End":"00:10.680","Text":"They all relate to a function f from the unit interval to the unit interval,"},{"Start":"00:10.680 ","End":"00:12.419","Text":"and it\u0027s defined as follows."},{"Start":"00:12.419 ","End":"00:16.860","Text":"If x is irrational or 0 or 1,"},{"Start":"00:16.860 ","End":"00:20.610","Text":"for convenience, then f of x is 0."},{"Start":"00:20.610 ","End":"00:22.815","Text":"Otherwise if x is rational,"},{"Start":"00:22.815 ","End":"00:24.540","Text":"and not 0 or 1,"},{"Start":"00:24.540 ","End":"00:27.705","Text":"then we can write x as p over q,"},{"Start":"00:27.705 ","End":"00:30.630","Text":"where p and q have no common factors,"},{"Start":"00:30.630 ","End":"00:32.190","Text":"and in this case,"},{"Start":"00:32.190 ","End":"00:34.785","Text":"we define f of x to be 1 over q,"},{"Start":"00:34.785 ","End":"00:37.380","Text":"so just throw out the p basically."},{"Start":"00:37.380 ","End":"00:40.320","Text":"Now, part a is as follows."},{"Start":"00:40.320 ","End":"00:44.540","Text":"For any integer capital N,"},{"Start":"00:44.540 ","End":"00:46.850","Text":"we consider the set A_N,"},{"Start":"00:46.850 ","End":"00:51.170","Text":"which is all the x in the open interval 0, 1,"},{"Start":"00:51.170 ","End":"00:52.730","Text":"x is p over q,"},{"Start":"00:52.730 ","End":"00:54.460","Text":"as in the definition,"},{"Start":"00:54.460 ","End":"00:58.685","Text":"where q is less than or equal to capital N,"},{"Start":"00:58.685 ","End":"01:03.140","Text":"we have to show that the set A_N is finite."},{"Start":"01:03.140 ","End":"01:12.350","Text":"In part b, we\u0027re given some capital N natural number and Epsilon bigger than 0,"},{"Start":"01:12.350 ","End":"01:16.000","Text":"we have to show that there exist intervals,"},{"Start":"01:16.000 ","End":"01:19.050","Text":"this m of them, x_1, x_2,"},{"Start":"01:19.050 ","End":"01:22.080","Text":"then x_3, x_4 up to x_2m minus 1,"},{"Start":"01:22.080 ","End":"01:28.470","Text":"x_2m, such that all these x_i are in increasing order,"},{"Start":"01:28.470 ","End":"01:32.900","Text":"and the set A_N from part a is"},{"Start":"01:32.900 ","End":"01:38.850","Text":"contained in the union of the corresponding open intervals,"},{"Start":"01:38.850 ","End":"01:42.620","Text":"and that furthermore, the sum of the length of"},{"Start":"01:42.620 ","End":"01:47.180","Text":"these intervals is less than or equal to Epsilon over 2."},{"Start":"01:47.180 ","End":"01:53.265","Text":"In part c, we have to show that the function f that we defined is integrable,"},{"Start":"01:53.265 ","End":"01:57.830","Text":"and then part d isn\u0027t clear immediately"},{"Start":"01:57.830 ","End":"02:02.600","Text":"how it related to the function f. But we have to find 2 integrable functions,"},{"Start":"02:02.600 ","End":"02:05.180","Text":"g and h on the interval 0,"},{"Start":"02:05.180 ","End":"02:09.380","Text":"1 such that their composition is not integrable."},{"Start":"02:09.380 ","End":"02:19.369","Text":"It turns out we can use this f to help us here in part d. Let\u0027s take a look at part a."},{"Start":"02:19.369 ","End":"02:22.375","Text":"Now, x is p over q,"},{"Start":"02:22.375 ","End":"02:24.640","Text":"q is less than or equal to N,"},{"Start":"02:24.640 ","End":"02:26.854","Text":"and therefore so is p,"},{"Start":"02:26.854 ","End":"02:30.055","Text":"because p is less than or equal to q."},{"Start":"02:30.055 ","End":"02:33.904","Text":"Altogether there are N combinations for q,"},{"Start":"02:33.904 ","End":"02:42.290","Text":"and N combinations at most for p. Altogether there\u0027s at most N squared members,"},{"Start":"02:42.290 ","End":"02:45.580","Text":"and certainly it\u0027s finite."},{"Start":"02:45.580 ","End":"02:48.915","Text":"Let\u0027s move on to part b."},{"Start":"02:48.915 ","End":"02:53.055","Text":"Now we\u0027ve shown that A_N is finite."},{"Start":"02:53.055 ","End":"02:56.310","Text":"Let\u0027s say that it has m members,"},{"Start":"02:56.310 ","End":"03:00.030","Text":"and we\u0027ll order them a_1 less than a_2,"},{"Start":"03:00.030 ","End":"03:01.980","Text":"and so on up to a_m,"},{"Start":"03:01.980 ","End":"03:06.509","Text":"and just throw in 0 and 1 into this chain,"},{"Start":"03:06.509 ","End":"03:08.990","Text":"and we want to construct Delta."},{"Start":"03:08.990 ","End":"03:15.170","Text":"We\u0027ll start off with d. D will be the least distance between 2 distinct points,"},{"Start":"03:15.170 ","End":"03:16.940","Text":"so the finite number of points,"},{"Start":"03:16.940 ","End":"03:20.845","Text":"there\u0027s only a finite number of pairs of points which give a distance,"},{"Start":"03:20.845 ","End":"03:23.210","Text":"and d is the smallest to them,"},{"Start":"03:23.210 ","End":"03:24.754","Text":"and we choose Delta."},{"Start":"03:24.754 ","End":"03:28.714","Text":"We want Delta to be less than 1/2 of d,"},{"Start":"03:28.714 ","End":"03:34.495","Text":"and we also want Delta to be less than Epsilon over 4m,"},{"Start":"03:34.495 ","End":"03:36.080","Text":"and you\u0027ll see why."},{"Start":"03:36.080 ","End":"03:43.160","Text":"Next we want to build the intervals or the x_1 to x_2m,"},{"Start":"03:43.160 ","End":"03:45.095","Text":"and we\u0027ll do that as follows."},{"Start":"03:45.095 ","End":"03:47.260","Text":"For i equals 1 to m,"},{"Start":"03:47.260 ","End":"03:49.035","Text":"we\u0027ll build a pair of points,"},{"Start":"03:49.035 ","End":"03:52.440","Text":"x_2i minus 1 and x_2i as follows."},{"Start":"03:52.440 ","End":"03:57.225","Text":"One of them is a_i minus Delta and one of them is a_i plus Delta,"},{"Start":"03:57.225 ","End":"04:02.370","Text":"just an interval of radius Delta around a_i."},{"Start":"04:02.370 ","End":"04:06.780","Text":"This gives us x_1 to x_2m,"},{"Start":"04:06.780 ","End":"04:09.300","Text":"and we throw in 0 and 1,"},{"Start":"04:09.300 ","End":"04:13.460","Text":"and the claim is that these inequalities hold,"},{"Start":"04:13.460 ","End":"04:16.550","Text":"just distinguish between the odds and evens."},{"Start":"04:16.550 ","End":"04:19.630","Text":"For the even intervals,"},{"Start":"04:19.630 ","End":"04:22.040","Text":"x_2i minus 1 to x_2i,"},{"Start":"04:22.040 ","End":"04:26.460","Text":"that\u0027s clear because it\u0027s a_i minus Delta and a_i plus Delta,"},{"Start":"04:26.460 ","End":"04:29.055","Text":"so certainly this is less than this."},{"Start":"04:29.055 ","End":"04:34.650","Text":"Also, let\u0027s say from x_2 to x_3,"},{"Start":"04:34.650 ","End":"04:38.175","Text":"like from x_2i to x_2i plus 1,"},{"Start":"04:38.175 ","End":"04:42.960","Text":"we have 2 consecutive a\u0027s and from this one we add Delta,"},{"Start":"04:42.960 ","End":"04:45.230","Text":"from this one we subtract Delta."},{"Start":"04:45.230 ","End":"04:50.680","Text":"But a_i and a_i plus 1, are apart,"},{"Start":"04:50.680 ","End":"04:54.275","Text":"at least d, which is bigger than 2 Delta,"},{"Start":"04:54.275 ","End":"04:56.450","Text":"so there\u0027s no overlap."},{"Start":"04:56.450 ","End":"04:58.675","Text":"This really is less than this,"},{"Start":"04:58.675 ","End":"05:01.490","Text":"and we really have to check at the end points also."},{"Start":"05:01.490 ","End":"05:03.770","Text":"I won\u0027t go into the details,"},{"Start":"05:03.770 ","End":"05:06.980","Text":"basically because d is bigger than Delta."},{"Start":"05:06.980 ","End":"05:10.655","Text":"Then at the endpoints we also have the inequality."},{"Start":"05:10.655 ","End":"05:12.890","Text":"That was one of the 3 things we had to show."},{"Start":"05:12.890 ","End":"05:17.625","Text":"If I go back up, we had to show this."},{"Start":"05:17.625 ","End":"05:23.120","Text":"The next thing we have to show is that A_N is contained in these open intervals."},{"Start":"05:23.120 ","End":"05:27.200","Text":"But that\u0027s fairly clear because each a_i,"},{"Start":"05:27.200 ","End":"05:30.140","Text":"i goes from 1 to m,"},{"Start":"05:30.140 ","End":"05:32.955","Text":"is in the interval that we built,"},{"Start":"05:32.955 ","End":"05:36.110","Text":"a_i minus Delta, a_i plus Delta,"},{"Start":"05:36.110 ","End":"05:39.425","Text":"which is the interval x_2i minus 1, x_2i."},{"Start":"05:39.425 ","End":"05:44.730","Text":"Each of these contains one of the a_i\u0027s and so A_N,"},{"Start":"05:44.730 ","End":"05:48.500","Text":"which is all of the a_i\u0027s together is in this union,"},{"Start":"05:48.500 ","End":"05:52.625","Text":"and the last of the 3 things we had to show that\u0027s this one, the 1/3 one,"},{"Start":"05:52.625 ","End":"05:54.665","Text":"is that the sum of these,"},{"Start":"05:54.665 ","End":"05:59.540","Text":"it\u0027s actually x_2 minus x_1 plus x_4 minus x_3 without the absolute values."},{"Start":"05:59.540 ","End":"06:01.310","Text":"Anyway, these are the distances,"},{"Start":"06:01.310 ","End":"06:05.925","Text":"and each of these has the same distance to Delta."},{"Start":"06:05.925 ","End":"06:08.340","Text":"That\u0027s how we built these intervals,"},{"Start":"06:08.340 ","End":"06:10.140","Text":"and we have m of them,"},{"Start":"06:10.140 ","End":"06:19.365","Text":"so m times 2 Delta is 2m Delta but Delta\u0027s less than Epsilon over 4m."},{"Start":"06:19.365 ","End":"06:21.510","Text":"So if we multiply this by 2m,"},{"Start":"06:21.510 ","End":"06:30.735","Text":"we\u0027re still less than Epsilon over 2 now as required,"},{"Start":"06:30.735 ","End":"06:32.655","Text":"so that\u0027s part b."},{"Start":"06:32.655 ","End":"06:34.855","Text":"Now we come to part c,"},{"Start":"06:34.855 ","End":"06:37.535","Text":"which as you may recall,"},{"Start":"06:37.535 ","End":"06:40.705","Text":"is where we have to show that f is integrable."},{"Start":"06:40.705 ","End":"06:48.380","Text":"We\u0027ll use partitions, we\u0027ll show that for each Epsilon we can find a partition of 0,"},{"Start":"06:48.380 ","End":"06:53.080","Text":"1 such that the upper sum minus the lower sum is less than Epsilon."},{"Start":"06:53.080 ","End":"06:56.080","Text":"This is sufficient for integrability."},{"Start":"06:56.080 ","End":"07:03.590","Text":"Given Epsilon, we\u0027ll choose N such that 1 over N is less than Epsilon over 2."},{"Start":"07:03.590 ","End":"07:06.220","Text":"We can always do this if you choose N bigger"},{"Start":"07:06.220 ","End":"07:12.295","Text":"than 2 over Epsilon by the Archimedean property."},{"Start":"07:12.295 ","End":"07:15.265","Text":"Anyway, corresponding to this N,"},{"Start":"07:15.265 ","End":"07:17.545","Text":"we choose the partition,"},{"Start":"07:17.545 ","End":"07:20.070","Text":"just like in part b,"},{"Start":"07:20.070 ","End":"07:22.740","Text":"but including the 0 and the 1,"},{"Start":"07:22.740 ","End":"07:26.445","Text":"which we label as x naught x_2m plus 1,"},{"Start":"07:26.445 ","End":"07:28.965","Text":"with the properties as in b."},{"Start":"07:28.965 ","End":"07:34.815","Text":"Now, let\u0027s suppose that x is in the interval,"},{"Start":"07:34.815 ","End":"07:38.400","Text":"but is not in this union."},{"Start":"07:38.400 ","End":"07:43.770","Text":"Earlier we showed that this union contains the set A_N."},{"Start":"07:43.770 ","End":"07:46.935","Text":"If x is not in this union,"},{"Start":"07:46.935 ","End":"07:51.000","Text":"then it can\u0027t belong to A_N."},{"Start":"07:51.000 ","End":"07:52.755","Text":"We\u0027ll distinguish 2 cases,"},{"Start":"07:52.755 ","End":"07:56.070","Text":"we\u0027re still in x naught here,"},{"Start":"07:56.070 ","End":"08:00.630","Text":"would say either x is rational or x is irrational."},{"Start":"08:00.630 ","End":"08:02.625","Text":"Now if x is rational,"},{"Start":"08:02.625 ","End":"08:06.390","Text":"then f of x is sum 1 over q,"},{"Start":"08:06.390 ","End":"08:09.960","Text":"but q has to be bigger or equal to N,"},{"Start":"08:09.960 ","End":"08:15.660","Text":"otherwise, it would be in A_N because that\u0027s the set of"},{"Start":"08:15.660 ","End":"08:22.680","Text":"all those rationals where q is less than N. F of x could be 1 over q,"},{"Start":"08:22.680 ","End":"08:26.310","Text":"and if x is irrational then f of x is 0."},{"Start":"08:26.310 ","End":"08:32.940","Text":"In either case, f of x is bigger or equal to 0 and less than or equal to 1"},{"Start":"08:32.940 ","End":"08:38.790","Text":"over N. The 1 over N comes from the fact that if q is bigger or equal to N,"},{"Start":"08:38.790 ","End":"08:42.405","Text":"then 1 over q is less than or equal to 1 over N. This holds"},{"Start":"08:42.405 ","End":"08:46.800","Text":"outside the union of these m intervals."},{"Start":"08:46.800 ","End":"08:50.235","Text":"We\u0027ll call these the odd intervals,"},{"Start":"08:50.235 ","End":"08:52.860","Text":"the ones that start with an odd index."},{"Start":"08:52.860 ","End":"08:55.320","Text":"Outside of these, if you think about it,"},{"Start":"08:55.320 ","End":"08:58.650","Text":"and if you take the interval 0 and 1 and remove these open intervals,"},{"Start":"08:58.650 ","End":"09:06.060","Text":"we\u0027re left with a union of the closed even intervals from 0-1,"},{"Start":"09:06.060 ","End":"09:10.020","Text":"from 2-3, from 4-5, and so on."},{"Start":"09:10.020 ","End":"09:13.860","Text":"These are of the form x_2i, x_2i plus 1."},{"Start":"09:13.860 ","End":"09:19.425","Text":"On these intervals, because f is between 0 and 1 over N,"},{"Start":"09:19.425 ","End":"09:23.910","Text":"then the maximum minus the minimum or the supremum minus"},{"Start":"09:23.910 ","End":"09:30.420","Text":"the infimum is less than or equal to 1 over N. Now on the odd intervals,"},{"Start":"09:30.420 ","End":"09:37.545","Text":"which are these, but the closed intervals throwing the end points,"},{"Start":"09:37.545 ","End":"09:42.690","Text":"we have that the supremum minus the infimum is less than or equal to 1,"},{"Start":"09:42.690 ","End":"09:45.840","Text":"because the function is always between 0-1."},{"Start":"09:45.840 ","End":"09:49.845","Text":"Now we\u0027re going to evaluate this upper sum minus the lower sum."},{"Start":"09:49.845 ","End":"09:54.705","Text":"We break it up into the 2m sub-intervals."},{"Start":"09:54.705 ","End":"10:01.650","Text":"We have the sum of M_k minus little m_k times the width Delta x_k,"},{"Start":"10:01.650 ","End":"10:05.250","Text":"and I will break this up into odds and evens,"},{"Start":"10:05.250 ","End":"10:14.130","Text":"so we have the sum from 1-m of these terms where k is 2i even."},{"Start":"10:14.130 ","End":"10:18.525","Text":"Then here is where k is 2i plus 1, the odd."},{"Start":"10:18.525 ","End":"10:24.600","Text":"As we saw, this difference on the even intervals is less than or equal to 1,"},{"Start":"10:24.600 ","End":"10:34.215","Text":"and on the odd intervals it\u0027s less than or equal to 1 over N. What we can say is that,"},{"Start":"10:34.215 ","End":"10:41.295","Text":"this part is less than or equal to 1 times the sum of the x_2i,"},{"Start":"10:41.295 ","End":"10:52.365","Text":"and this part is less than or equal to 1 over N times the sum of the Delta x_2i plus 1."},{"Start":"10:52.365 ","End":"10:56.130","Text":"Now we\u0027ve already shown that the sum of these,"},{"Start":"10:56.130 ","End":"11:01.920","Text":"that\u0027s exactly this x_2 minus x_1 plus x_4 minus x_3."},{"Start":"11:01.920 ","End":"11:06.420","Text":"We showed that this sum is less than or equal to Epsilon over 2 in part"},{"Start":"11:06.420 ","End":"11:13.140","Text":"c. We know that they\u0027re part of 0 1,"},{"Start":"11:13.140 ","End":"11:15.960","Text":"the total length can\u0027t be more than 1,"},{"Start":"11:15.960 ","End":"11:17.760","Text":"so yeah, it\u0027s less than 1."},{"Start":"11:17.760 ","End":"11:19.755","Text":"From these 2 inequalities,"},{"Start":"11:19.755 ","End":"11:24.345","Text":"what we get is that the first part is less than or equal to Epsilon over 2,"},{"Start":"11:24.345 ","End":"11:26.910","Text":"the second part is less than 1 over N,"},{"Start":"11:26.910 ","End":"11:30.780","Text":"so we get Epsilon over 2 plus 1 over N,"},{"Start":"11:30.780 ","End":"11:35.535","Text":"and this is less than Epsilon because 1 over N"},{"Start":"11:35.535 ","End":"11:40.905","Text":"is less than Epsilon over 2 by the way, we chose it."},{"Start":"11:40.905 ","End":"11:51.090","Text":"That concludes part c. Now we move on to part d. We need to find an example of g and h,"},{"Start":"11:51.090 ","End":"11:54.465","Text":"which are integrable, but the composition isn\u0027t."},{"Start":"11:54.465 ","End":"11:58.380","Text":"Let\u0027s take h to be the f from above,"},{"Start":"11:58.380 ","End":"12:08.955","Text":"and we\u0027ll define g to be equal to 1 everywhere except where x is 0 and then g is 0."},{"Start":"12:08.955 ","End":"12:11.790","Text":"We showed that f is integrable,"},{"Start":"12:11.790 ","End":"12:13.950","Text":"which means that h is integrable,"},{"Start":"12:13.950 ","End":"12:22.530","Text":"and g is also integrable because it\u0027s practically the same as the constant function 1,"},{"Start":"12:22.530 ","End":"12:26.745","Text":"except to the single point which doesn\u0027t affect the integrability."},{"Start":"12:26.745 ","End":"12:30.180","Text":"By the way, the value of the integral is the"},{"Start":"12:30.180 ","End":"12:34.230","Text":"same as the value of the constant function 1, and that\u0027s 1."},{"Start":"12:34.230 ","End":"12:38.655","Text":"Now, what is g composed with h?"},{"Start":"12:38.655 ","End":"12:42.210","Text":"Well, h is f, and if you go back to the definition of f,"},{"Start":"12:42.210 ","End":"12:51.180","Text":"you\u0027ll see that f of x is 0 if and only if x is irrational or 0 or 1,"},{"Start":"12:51.180 ","End":"12:53.085","Text":"we threw in also."},{"Start":"12:53.085 ","End":"12:54.880","Text":"All the rest of the cases,"},{"Start":"12:54.880 ","End":"12:59.925","Text":"if h or f is non-zero,"},{"Start":"12:59.925 ","End":"13:02.700","Text":"then g of it is equal to 1."},{"Start":"13:02.700 ","End":"13:05.745","Text":"Think about this, so this is g composed with"},{"Start":"13:05.745 ","End":"13:10.920","Text":"h. Basically it\u0027s the function 1 if x is rational on 0,"},{"Start":"13:10.920 ","End":"13:15.555","Text":"if x is irrational except for 2 points at the end,"},{"Start":"13:15.555 ","End":"13:17.985","Text":"which works the other way."},{"Start":"13:17.985 ","End":"13:22.710","Text":"For any partition P of 0, 1,"},{"Start":"13:22.710 ","End":"13:25.650","Text":"the lowest sum is 0,"},{"Start":"13:25.650 ","End":"13:28.035","Text":"I claim and the upper sum is 1."},{"Start":"13:28.035 ","End":"13:31.410","Text":"This is because we can assume the partition doesn\u0027t have"},{"Start":"13:31.410 ","End":"13:36.975","Text":"any trivial intervals that strictly increasing on the x_i,"},{"Start":"13:36.975 ","End":"13:38.850","Text":"that the widths are positive,"},{"Start":"13:38.850 ","End":"13:44.145","Text":"just throw away any degenerate intervals with 0."},{"Start":"13:44.145 ","End":"13:46.620","Text":"In any such interval,"},{"Start":"13:46.620 ","End":"13:50.685","Text":"there are rationals and irrationals,"},{"Start":"13:50.685 ","End":"13:53.490","Text":"not including 1 and 0."},{"Start":"13:53.490 ","End":"14:01.575","Text":"The supremum or maximum is 1 and the infimum is 0 in each such non-trivial interval."},{"Start":"14:01.575 ","End":"14:08.760","Text":"That means that, for the upper sum we\u0027re taking 1 and for the lowest sum we\u0027re taking 0."},{"Start":"14:08.760 ","End":"14:12.300","Text":"Each time we get 0 and 1."},{"Start":"14:12.300 ","End":"14:16.020","Text":"The supremum of these and the infimum of these"},{"Start":"14:16.020 ","End":"14:20.445","Text":"are 0 and 1 respectively and 0 is not equal to 1,"},{"Start":"14:20.445 ","End":"14:23.970","Text":"so the lower integral is not equal to the upper integral."},{"Start":"14:23.970 ","End":"14:28.575","Text":"G composed with h is not integrable,"},{"Start":"14:28.575 ","End":"14:31.200","Text":"and that\u0027s what we had to show."},{"Start":"14:31.200 ","End":"14:36.670","Text":"That completes part d and the whole question finally."}],"ID":24746},{"Watched":false,"Name":"Exercise 8","Duration":"5m 1s","ChapterTopicVideoID":23805,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23805.jpeg","UploadDate":"2021-01-11T09:00:20.8870000","DurationForVideoObject":"PT5M1S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.161","Text":"In this exercise, f is a real valued function and it\u0027s integrable on a, b,"},{"Start":"00:07.161 ","End":"00:11.685","Text":"and we have a subinterval c, d of a, b."},{"Start":"00:11.685 ","End":"00:16.330","Text":"We have to show that f is integrable also on c, d."},{"Start":"00:16.330 ","End":"00:20.054","Text":"We\u0027ll use the Riemann Criterion for integrability"},{"Start":"00:20.054 ","End":"00:23.760","Text":"and show that for all epsilon bigger than 0,"},{"Start":"00:23.760 ","End":"00:28.992","Text":"there exists a partition P of the interval c, d"},{"Start":"00:28.992 ","End":"00:35.480","Text":"such that the upper sum minus the lower sum is less than epsilon."},{"Start":"00:35.480 ","End":"00:38.125","Text":"Let epsilon be given."},{"Start":"00:38.125 ","End":"00:41.220","Text":"Since f is integrable on a, b,"},{"Start":"00:41.220 ","End":"00:43.340","Text":"then by the Riemann Criterion,"},{"Start":"00:43.340 ","End":"00:46.464","Text":"there\u0027s a partition, call it, P_1 of a, b"},{"Start":"00:46.464 ","End":"00:54.375","Text":"such that the upper sum minus the lower sum of the partition P_1 is less than epsilon."},{"Start":"00:54.375 ","End":"00:57.260","Text":"Now, a partition, P_1 in this case,"},{"Start":"00:57.260 ","End":"01:03.830","Text":"is a set of points that includes a and b and some other points in between."},{"Start":"01:03.830 ","End":"01:05.630","Text":"To this set of points,"},{"Start":"01:05.630 ","End":"01:08.735","Text":"we can add 2 points, c and d,"},{"Start":"01:08.735 ","End":"01:11.085","Text":"if they\u0027re not already in there."},{"Start":"01:11.085 ","End":"01:16.780","Text":"We can take the union of P_1 with c and d, call that P_2."},{"Start":"01:16.780 ","End":"01:19.120","Text":"That\u0027s a refinement of P_1."},{"Start":"01:19.120 ","End":"01:21.710","Text":"Refinement just means adding more points."},{"Start":"01:21.710 ","End":"01:25.820","Text":"P_2 is still a partition of the same a, b."},{"Start":"01:25.820 ","End":"01:29.570","Text":"One of the properties of refinement is that"},{"Start":"01:29.570 ","End":"01:37.520","Text":"the upper sum can possibly shrink and the lowest sum can possibly increase."},{"Start":"01:37.520 ","End":"01:38.890","Text":"In case you haven\u0027t seen this,"},{"Start":"01:38.890 ","End":"01:41.375","Text":"let me give a little explanation."},{"Start":"01:41.375 ","End":"01:45.440","Text":"Refinement can always be done one point at a time."},{"Start":"01:45.440 ","End":"01:52.520","Text":"This part of the diagram shows 1 interval in the graph above it,"},{"Start":"01:52.520 ","End":"01:57.620","Text":"and the lower sum is the green one"},{"Start":"01:57.620 ","End":"02:00.605","Text":"and the upper sum is the gray,"},{"Start":"02:00.605 ","End":"02:02.735","Text":"which includes this green."},{"Start":"02:02.735 ","End":"02:05.210","Text":"Then we add a partition point,"},{"Start":"02:05.210 ","End":"02:08.770","Text":"that would be the point here,"},{"Start":"02:08.770 ","End":"02:16.940","Text":"and then the lower sum made up of 2 bits is all the green part,"},{"Start":"02:16.940 ","End":"02:20.870","Text":"and the upper sum is all the shaded part,"},{"Start":"02:20.870 ","End":"02:22.955","Text":"the gray including the green."},{"Start":"02:22.955 ","End":"02:28.440","Text":"We can see that the green has increased and that the gray has decreased."},{"Start":"02:28.440 ","End":"02:32.525","Text":"I mean the green has gained this bit and the gray has lost this bit."},{"Start":"02:32.525 ","End":"02:35.885","Text":"That\u0027s a pictorial explanation for this."},{"Start":"02:35.885 ","End":"02:40.550","Text":"Now, let\u0027s write the partition P_2 as sequence,"},{"Start":"02:40.550 ","End":"02:45.425","Text":"x-naught, x_1, up to x_n in increasing order,"},{"Start":"02:45.425 ","End":"02:49.825","Text":"and x-naught is a and x_n is b."},{"Start":"02:49.825 ","End":"02:53.490","Text":"Since c and d are in P_2,"},{"Start":"02:53.490 ","End":"02:58.079","Text":"then c is sum x_i and d is sum x_j,"},{"Start":"02:58.079 ","End":"03:01.602","Text":"where i and j are between 0 and n."},{"Start":"03:01.602 ","End":"03:04.940","Text":"Now we define the P that we\u0027re looking for,"},{"Start":"03:04.940 ","End":"03:13.700","Text":"which is just to take the portion here from c to d and then from x_i up to x_j."},{"Start":"03:13.700 ","End":"03:22.743","Text":"What we get is that the upper sum minus the lower sum for this partition of c, d"},{"Start":"03:22.743 ","End":"03:32.530","Text":"is the sum where k goes, actually, from i plus 1 to j of M_k minus m_k Delta x_k."},{"Start":"03:32.530 ","End":"03:34.319","Text":"M is the supremum,"},{"Start":"03:34.319 ","End":"03:36.920","Text":"m is the infimum in the interval,"},{"Start":"03:36.920 ","End":"03:40.415","Text":"Delta is the width of the interval."},{"Start":"03:40.415 ","End":"03:45.499","Text":"This is less than or equal to the sum,"},{"Start":"03:45.499 ","End":"03:50.270","Text":"a bigger sum on x-naught all the way to x_n."},{"Start":"03:50.270 ","End":"03:54.707","Text":"The P_2 partition here, k, goes from 1 to n."},{"Start":"03:54.707 ","End":"03:58.280","Text":"Now, there are more terms here than here."},{"Start":"03:58.280 ","End":"04:03.289","Text":"Each of the terms here is positive."},{"Start":"04:03.289 ","End":"04:04.951","Text":"The Delta is positive"},{"Start":"04:04.951 ","End":"04:09.080","Text":"and the M minus m, well, is nonnegative anyway."},{"Start":"04:09.080 ","End":"04:12.185","Text":"We get something less than or equal to."},{"Start":"04:12.185 ","End":"04:16.320","Text":"We\u0027ve only increased it if we add more terms."},{"Start":"04:16.940 ","End":"04:25.925","Text":"This is equal to the upper sum minus the lower sum for the partition P_2,"},{"Start":"04:25.925 ","End":"04:28.780","Text":"which is a partition of a, b."},{"Start":"04:28.780 ","End":"04:31.905","Text":"This is less than or equal to."},{"Start":"04:31.905 ","End":"04:34.885","Text":"Replace P_2 by P_1."},{"Start":"04:34.885 ","End":"04:39.890","Text":"This is so from here to here because of these 2 inequalities."},{"Start":"04:39.890 ","End":"04:44.715","Text":"We\u0027ve increased U and we\u0027ve decreased L,"},{"Start":"04:44.715 ","End":"04:48.240","Text":"so this inequality holds."},{"Start":"04:48.240 ","End":"04:53.315","Text":"This already we\u0027ve seen above is less than epsilon."},{"Start":"04:53.315 ","End":"04:55.655","Text":"Here it is, less than epsilon."},{"Start":"04:55.655 ","End":"04:57.845","Text":"This is what we had to show."},{"Start":"04:57.845 ","End":"05:02.310","Text":"That concludes this exercise."}],"ID":24747},{"Watched":false,"Name":"Exercise 9","Duration":"10m 38s","ChapterTopicVideoID":23806,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23806.jpeg","UploadDate":"2021-01-11T09:06:35.8130000","DurationForVideoObject":"PT10M38S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.885","Text":"The main goal of this exercise is actually part b,"},{"Start":"00:03.885 ","End":"00:05.895","Text":"to show that if f is integrable,"},{"Start":"00:05.895 ","End":"00:09.360","Text":"then so are the absolute value of f and f squared."},{"Start":"00:09.360 ","End":"00:12.465","Text":"Part a will help us to solve part b."},{"Start":"00:12.465 ","End":"00:15.960","Text":"In part a, we\u0027re given that f is bounded on an interval c,"},{"Start":"00:15.960 ","End":"00:20.940","Text":"d. Big M is the supremum of f and"},{"Start":"00:20.940 ","End":"00:26.655","Text":"m prime is the supremum of absolute value of f on the interval."},{"Start":"00:26.655 ","End":"00:34.665","Text":"Similarly, little m, little m prime are the infimum of f and f prime respectively."},{"Start":"00:34.665 ","End":"00:40.320","Text":"We have to show that this inequality holds."},{"Start":"00:40.320 ","End":"00:43.550","Text":"M prime minus little m prime is less than or equal to m"},{"Start":"00:43.550 ","End":"00:48.590","Text":"minus m. Choose any x and y in the interval c,"},{"Start":"00:48.590 ","End":"00:52.880","Text":"d, then f of x is less than or equal to m because that\u0027s"},{"Start":"00:52.880 ","End":"00:57.500","Text":"the supremum and f of y is bigger or equal to little m because that\u0027s the infimum."},{"Start":"00:57.500 ","End":"01:00.275","Text":"We can subtract and get this inequality,"},{"Start":"01:00.275 ","End":"01:07.730","Text":"f is less than or equal to big M and minus f is less than or equal to minus m. Similarly,"},{"Start":"01:07.730 ","End":"01:10.040","Text":"f of y minus f of x is"},{"Start":"01:10.040 ","End":"01:14.690","Text":"less than or equal to big M minus little m actually follows from symmetry."},{"Start":"01:14.690 ","End":"01:17.120","Text":"I mean, there\u0027s no reason to call this 1x and this 1y."},{"Start":"01:17.120 ","End":"01:19.220","Text":"You could call this 1x and this 1y,"},{"Start":"01:19.220 ","End":"01:22.695","Text":"and you get the same inequality."},{"Start":"01:22.695 ","End":"01:26.330","Text":"If these 2 are true, this and this,"},{"Start":"01:26.330 ","End":"01:31.100","Text":"then the absolute value of f of x minus f of y is less than or equal"},{"Start":"01:31.100 ","End":"01:36.170","Text":"to M minus m. If a is smaller than something and minus a is smaller than something,"},{"Start":"01:36.170 ","End":"01:40.100","Text":"then the absolute value of a is smaller or equal to that same thing."},{"Start":"01:40.100 ","End":"01:45.625","Text":"One of the variants of the triangle inequality gives us this inequality,"},{"Start":"01:45.625 ","End":"01:50.600","Text":"and so what we get is that this is less than or equal to this,"},{"Start":"01:50.600 ","End":"01:52.880","Text":"but this is less than or equal to M minus m,"},{"Start":"01:52.880 ","End":"01:54.890","Text":"so we get this inequality."},{"Start":"01:54.890 ","End":"02:00.100","Text":"What we want to do now is get from this line to this line here."},{"Start":"02:00.100 ","End":"02:04.675","Text":"Let\u0027s choose some sequences, x_n and y_n,"},{"Start":"02:04.675 ","End":"02:09.305","Text":"such that the absolute value of f of x_n goes to M prime,"},{"Start":"02:09.305 ","End":"02:14.855","Text":"and the absolute value of f of y_n goes to little m prime."},{"Start":"02:14.855 ","End":"02:16.205","Text":"You can always do that."},{"Start":"02:16.205 ","End":"02:18.495","Text":"This is the supremum,"},{"Start":"02:18.495 ","End":"02:21.015","Text":"this is the infimum."},{"Start":"02:21.015 ","End":"02:25.290","Text":"What we get is M prime minus m prime."},{"Start":"02:25.290 ","End":"02:27.650","Text":"This one is equal to this limit,"},{"Start":"02:27.650 ","End":"02:30.125","Text":"and this one is equal to this limit."},{"Start":"02:30.125 ","End":"02:35.875","Text":"Now, the difference of the limits is equal to the limit of the difference."},{"Start":"02:35.875 ","End":"02:41.960","Text":"For each n, this is less than or equal to big M minus little m."},{"Start":"02:41.960 ","End":"02:45.380","Text":"Because this is less than or equal to m and"},{"Start":"02:45.380 ","End":"02:48.770","Text":"this is bigger or equal to little m which is being subtracted."},{"Start":"02:48.770 ","End":"02:55.550","Text":"Think about it. This is less than or equal to big M minus little m,"},{"Start":"02:55.550 ","End":"02:57.170","Text":"because for each n,"},{"Start":"02:57.170 ","End":"02:59.690","Text":"this is less than or equal to then when we take the limit,"},{"Start":"02:59.690 ","End":"03:01.865","Text":"it\u0027s still less than or equal to."},{"Start":"03:01.865 ","End":"03:05.615","Text":"That\u0027s part a. Now let\u0027s get on to part b."},{"Start":"03:05.615 ","End":"03:08.770","Text":"Part b is subdivided into 2 parts."},{"Start":"03:08.770 ","End":"03:11.450","Text":"One is about the absolute value of f,"},{"Start":"03:11.450 ","End":"03:14.560","Text":"and then the second half of it is f squared."},{"Start":"03:14.560 ","End":"03:18.745","Text":"Let\u0027s start with absolute value of f is it integrable? We have to show."},{"Start":"03:18.745 ","End":"03:21.500","Text":"We\u0027re going to use the Riemann criterion."},{"Start":"03:21.500 ","End":"03:24.619","Text":"Let Epsilon be any positive number."},{"Start":"03:24.619 ","End":"03:28.700","Text":"Then since f is integrable, that was given,"},{"Start":"03:28.700 ","End":"03:32.525","Text":"there\u0027s a partition P such that"},{"Start":"03:32.525 ","End":"03:38.450","Text":"the upper sum for P and F minus the lower sum for P and F is less than Epsilon."},{"Start":"03:38.450 ","End":"03:43.115","Text":"Now we can write U and L in terms of the definition with the sigma,"},{"Start":"03:43.115 ","End":"03:45.155","Text":"it\u0027s equal to this sum."},{"Start":"03:45.155 ","End":"03:46.835","Text":"Where just to remind you,"},{"Start":"03:46.835 ","End":"03:49.789","Text":"big M_i is the supremum on the interval"},{"Start":"03:49.789 ","End":"03:53.075","Text":"and little m_i is the infimum of f on the interval."},{"Start":"03:53.075 ","End":"03:58.595","Text":"If we apply part a with the interval x_i minus 1 x_i,"},{"Start":"03:58.595 ","End":"04:00.875","Text":"taking the place of c, d,"},{"Start":"04:00.875 ","End":"04:09.135","Text":"then we can replace M_i minus little m_i with M_i prime minus little m_i prime,"},{"Start":"04:09.135 ","End":"04:12.979","Text":"where the M_is are defined as follows,"},{"Start":"04:12.979 ","End":"04:14.945","Text":"just as in part a."},{"Start":"04:14.945 ","End":"04:19.100","Text":"Now what\u0027s written here is exactly the upper sum"},{"Start":"04:19.100 ","End":"04:24.125","Text":"minus the lowest sum for the partition and for the absolute value of f,"},{"Start":"04:24.125 ","End":"04:27.710","Text":"because we just have absolute value of f instead of f here."},{"Start":"04:27.710 ","End":"04:31.040","Text":"By the Riemann criterion with the Epsilon,"},{"Start":"04:31.040 ","End":"04:37.340","Text":"we found for each Epsilon a partition P such that U minus L for P,"},{"Start":"04:37.340 ","End":"04:40.100","Text":"an absolute value of f is less than Epsilon,"},{"Start":"04:40.100 ","End":"04:44.075","Text":"which means the absolute value of f is integrable."},{"Start":"04:44.075 ","End":"04:47.600","Text":"Now on to the second half of part b,"},{"Start":"04:47.600 ","End":"04:50.735","Text":"we want to show that f squared is integrable."},{"Start":"04:50.735 ","End":"04:57.170","Text":"We\u0027ll use part a as well as the first half of part b to do this."},{"Start":"04:57.170 ","End":"05:01.715","Text":"As before, we\u0027ll use the Riemann criterion for integrability."},{"Start":"05:01.715 ","End":"05:04.580","Text":"We\u0027ll define a constant K. We\u0027ll need this"},{"Start":"05:04.580 ","End":"05:09.855","Text":"later as the supremum of absolute value of f on the whole AB."},{"Start":"05:09.855 ","End":"05:12.800","Text":"Epsilon be any positive number."},{"Start":"05:12.800 ","End":"05:16.370","Text":"We want to find a partition such that, well,"},{"Start":"05:16.370 ","End":"05:19.355","Text":"by the first half part b,"},{"Start":"05:19.355 ","End":"05:22.640","Text":"f absolute value is integrable."},{"Start":"05:22.640 ","End":"05:28.660","Text":"We have a partition P of the interval a, b,"},{"Start":"05:28.660 ","End":"05:30.965","Text":"such that U minus L,"},{"Start":"05:30.965 ","End":"05:34.490","Text":"this partition and absolute value of f is less"},{"Start":"05:34.490 ","End":"05:38.590","Text":"than and we can choose instead of Epsilon, Epsilon over 2K."},{"Start":"05:38.590 ","End":"05:45.215","Text":"Now let\u0027s consider U minus L for the partition and for the function f squared."},{"Start":"05:45.215 ","End":"05:48.365","Text":"This is equal to this sum."},{"Start":"05:48.365 ","End":"05:52.025","Text":"Where I have to tell you what the m double prime are."},{"Start":"05:52.025 ","End":"05:56.645","Text":"M double prime is the supremum of the function in question,"},{"Start":"05:56.645 ","End":"06:00.980","Text":"namely f squared and little m double prime is the infimum of"},{"Start":"06:00.980 ","End":"06:05.610","Text":"f squared on the interval corresponding to i."},{"Start":"06:05.610 ","End":"06:10.955","Text":"This is for each i from 1 to n. I claim that"},{"Start":"06:10.955 ","End":"06:17.900","Text":"this M double prime is equal to M single prime squared,"},{"Start":"06:17.900 ","End":"06:21.080","Text":"both for uppercase M and for lower case m,"},{"Start":"06:21.080 ","End":"06:23.600","Text":"where the single prime is,"},{"Start":"06:23.600 ","End":"06:26.330","Text":"as in the first half of this question,"},{"Start":"06:26.330 ","End":"06:30.835","Text":"it\u0027s the supremum and infimum of absolute value."},{"Start":"06:30.835 ","End":"06:33.620","Text":"Double prime refers to the square function,"},{"Start":"06:33.620 ","End":"06:36.320","Text":"the single prime for the absolute value function."},{"Start":"06:36.320 ","End":"06:37.790","Text":"We\u0027ll prove this later."},{"Start":"06:37.790 ","End":"06:41.435","Text":"I\u0027ll just highlight it for now, this and this."},{"Start":"06:41.435 ","End":"06:43.730","Text":"I just don\u0027t want to stop the flow."},{"Start":"06:43.730 ","End":"06:45.610","Text":"We prove it at the end,"},{"Start":"06:45.610 ","End":"06:49.530","Text":"this will help us to expand this sum."},{"Start":"06:49.530 ","End":"06:55.479","Text":"I\u0027ll just highlight this bit that\u0027s same as here."},{"Start":"06:55.479 ","End":"07:04.130","Text":"This is equal to replace M_i double-prime by M_i prime squared and similarly for the"},{"Start":"07:04.130 ","End":"07:08.360","Text":"lowercase m. Now we can use the difference of squares formula from"},{"Start":"07:08.360 ","End":"07:12.979","Text":"algebra to get this expanded like so."},{"Start":"07:12.979 ","End":"07:18.365","Text":"I claim that this here is less than or equal to 2K."},{"Start":"07:18.365 ","End":"07:26.900","Text":"First of all, M_i prime is the supremum of absolute value of f on this interval."},{"Start":"07:26.900 ","End":"07:30.635","Text":"But this interval is part of the interval a, b."},{"Start":"07:30.635 ","End":"07:33.860","Text":"So it\u0027s less than or equal to the K,"},{"Start":"07:33.860 ","End":"07:36.775","Text":"which is the supremum on the whole interval."},{"Start":"07:36.775 ","End":"07:41.130","Text":"Little m prime i is less than or equal to"},{"Start":"07:41.130 ","End":"07:46.280","Text":"big M_i prime because the infimum is less than the supremum."},{"Start":"07:46.280 ","End":"07:53.565","Text":"Each of these is less than or equal to K. We can replace this by 2K."},{"Start":"07:53.565 ","End":"08:01.130","Text":"Now what we can do is substitute what\u0027s highlighted in green here into this sum here."},{"Start":"08:01.130 ","End":"08:07.100","Text":"We get that the U minus L for P and f squared"},{"Start":"08:07.100 ","End":"08:14.495","Text":"is less than or equal to this sum with the green replaced by this expression."},{"Start":"08:14.495 ","End":"08:19.750","Text":"We have 2K and then we have this bit here."},{"Start":"08:19.750 ","End":"08:22.385","Text":"We have a less than or equal to now."},{"Start":"08:22.385 ","End":"08:25.925","Text":"This sigma is exactly"},{"Start":"08:25.925 ","End":"08:32.960","Text":"the U minus L for the partition and for absolute value of f. To go back up,"},{"Start":"08:32.960 ","End":"08:35.990","Text":"we\u0027ll see that Epsilon over 2K is"},{"Start":"08:35.990 ","End":"08:41.495","Text":"this U minus L and this expression is exactly what is here."},{"Start":"08:41.495 ","End":"08:43.445","Text":"This is equal to Epsilon."},{"Start":"08:43.445 ","End":"08:49.280","Text":"U minus L for p and f squared is less than or equal to Epsilon."},{"Start":"08:49.280 ","End":"08:51.949","Text":"This is for all Epsilon bigger than 0."},{"Start":"08:51.949 ","End":"08:58.640","Text":"We\u0027re basically done except that I still owe you the proof that this holds."},{"Start":"08:58.640 ","End":"09:00.395","Text":"I\u0027ll show you this now."},{"Start":"09:00.395 ","End":"09:04.565","Text":"If you take any x in the interval x_i minus 1x_i,"},{"Start":"09:04.565 ","End":"09:09.425","Text":"then the absolute value of f of x is less than or equal to the supremum,"},{"Start":"09:09.425 ","End":"09:11.885","Text":"which is M_i prime."},{"Start":"09:11.885 ","End":"09:16.845","Text":"Then squaring both sides, we have non-negatives,"},{"Start":"09:16.845 ","End":"09:18.735","Text":"we can squared the inequality,"},{"Start":"09:18.735 ","End":"09:22.670","Text":"and so f squared equals M_i prime squared."},{"Start":"09:22.670 ","End":"09:26.060","Text":"This is true for all x in the interval."},{"Start":"09:26.060 ","End":"09:31.220","Text":"The supremum of this is still less than or equal to this."},{"Start":"09:31.220 ","End":"09:35.735","Text":"But the supremum of this is what we defined as M_i double prime."},{"Start":"09:35.735 ","End":"09:40.100","Text":"We have an inequality 1 way now will show the reverse inequality."},{"Start":"09:40.100 ","End":"09:43.970","Text":"For each x in the interval f squared of"},{"Start":"09:43.970 ","End":"09:48.860","Text":"x is less than or equal to the supremum of f squared,"},{"Start":"09:48.860 ","End":"09:50.775","Text":"which is M_i double prime."},{"Start":"09:50.775 ","End":"09:53.959","Text":"Then we can take the square root of both sides."},{"Start":"09:53.959 ","End":"09:58.955","Text":"We have non-negative quantities that we can take the square root."},{"Start":"09:58.955 ","End":"10:01.625","Text":"Square root of this is absolute value of f,"},{"Start":"10:01.625 ","End":"10:03.935","Text":"square root of this just the square root."},{"Start":"10:03.935 ","End":"10:06.320","Text":"This is true for all x."},{"Start":"10:06.320 ","End":"10:13.039","Text":"The supremum of this is less than or equal to this."},{"Start":"10:13.039 ","End":"10:14.390","Text":"It\u0027s true for each x,"},{"Start":"10:14.390 ","End":"10:15.950","Text":"so is true for the supremum,"},{"Start":"10:15.950 ","End":"10:19.190","Text":"which means that M_i prime,"},{"Start":"10:19.190 ","End":"10:20.960","Text":"which is equal to this,"},{"Start":"10:20.960 ","End":"10:22.370","Text":"is less than or equal to this."},{"Start":"10:22.370 ","End":"10:24.890","Text":"Then we have non-negative quantities,"},{"Start":"10:24.890 ","End":"10:26.090","Text":"we can square them."},{"Start":"10:26.090 ","End":"10:30.425","Text":"This squared is less than or equal to this."},{"Start":"10:30.425 ","End":"10:33.140","Text":"Now we have the 2 directions of the inequality."},{"Start":"10:33.140 ","End":"10:38.010","Text":"We have equality. We\u0027re really done."}],"ID":24748},{"Watched":false,"Name":"Exercise 10","Duration":"2m 10s","ChapterTopicVideoID":23807,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23807.jpeg","UploadDate":"2021-01-11T09:07:12.8830000","DurationForVideoObject":"PT2M10S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.655","Text":"This exercise has 2 parts,"},{"Start":"00:02.655 ","End":"00:05.850","Text":"in each part we have to find an example of"},{"Start":"00:05.850 ","End":"00:09.344","Text":"a function from the unit interval to the reals."},{"Start":"00:09.344 ","End":"00:16.995","Text":"In part a, we want f not to be integrable and yet absolute value of f is integrable,"},{"Start":"00:16.995 ","End":"00:18.630","Text":"and in part b again,"},{"Start":"00:18.630 ","End":"00:20.065","Text":"f is not integrable,"},{"Start":"00:20.065 ","End":"00:21.950","Text":"but f squared is integrable."},{"Start":"00:21.950 ","End":"00:25.145","Text":"We have to give an example for each."},{"Start":"00:25.145 ","End":"00:32.600","Text":"For part a, we could define f to be so that f of x is"},{"Start":"00:32.600 ","End":"00:39.845","Text":"1 for x being rational and minus 1 for x irrational."},{"Start":"00:39.845 ","End":"00:42.995","Text":"It\u0027s similar to a function we\u0027ve encountered before,"},{"Start":"00:42.995 ","End":"00:46.570","Text":"where here it\u0027s 1 and here it\u0027s 0,"},{"Start":"00:46.570 ","End":"00:49.730","Text":"and just like the case where it\u0027s 1 and 0,"},{"Start":"00:49.730 ","End":"00:51.935","Text":"f is not integrable, It\u0027s very similar."},{"Start":"00:51.935 ","End":"00:59.360","Text":"You could also see that the lower integral is minus 1 because for any partition,"},{"Start":"00:59.360 ","End":"01:03.740","Text":"whenever 1 of the intervals has positive width,"},{"Start":"01:03.740 ","End":"01:08.375","Text":"we can choose x to be rational or irrational."},{"Start":"01:08.375 ","End":"01:10.895","Text":"In any positive interval,"},{"Start":"01:10.895 ","End":"01:13.355","Text":"there are rationals and irrationals."},{"Start":"01:13.355 ","End":"01:17.610","Text":"So if we just choose all the time, the rationals,"},{"Start":"01:17.610 ","End":"01:21.530","Text":"we\u0027ll get 1 and if we choose the irrationals each time,"},{"Start":"01:21.530 ","End":"01:23.515","Text":"then we\u0027ll get minus 1."},{"Start":"01:23.515 ","End":"01:24.970","Text":"This is the lowest sum,"},{"Start":"01:24.970 ","End":"01:28.295","Text":"this is the upper sum and the not equal f is not integrable."},{"Start":"01:28.295 ","End":"01:29.735","Text":"On the other hand,"},{"Start":"01:29.735 ","End":"01:33.770","Text":"the absolute value of f is always equal to 1,"},{"Start":"01:33.770 ","End":"01:35.900","Text":"because absolute value of 1 is 1,"},{"Start":"01:35.900 ","End":"01:37.940","Text":"absolute value of minus 1 is 1,"},{"Start":"01:37.940 ","End":"01:40.100","Text":"f is the constant function 1 on"},{"Start":"01:40.100 ","End":"01:44.545","Text":"the interval and the constant function is certainly integrable."},{"Start":"01:44.545 ","End":"01:47.655","Text":"That\u0027s part a. Part b,"},{"Start":"01:47.655 ","End":"01:50.570","Text":"we can use the same f because"},{"Start":"01:50.570 ","End":"01:55.070","Text":"the absolute value of f and f squared of the same thing in each case,"},{"Start":"01:55.070 ","End":"01:57.650","Text":"it\u0027s always equal to 1."},{"Start":"01:57.650 ","End":"02:03.980","Text":"So we already said that this is integrable of the constant function and so is this,"},{"Start":"02:03.980 ","End":"02:06.035","Text":"it\u0027s the same function."},{"Start":"02:06.035 ","End":"02:11.310","Text":"The same example is used for both and we are done."}],"ID":24749},{"Watched":false,"Name":"Exercise 11","Duration":"7m 22s","ChapterTopicVideoID":23808,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23808.jpeg","UploadDate":"2021-01-11T09:10:24.5900000","DurationForVideoObject":"PT7M22S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.880","Text":"In this exercise, we\u0027re given f and g,"},{"Start":"00:02.880 ","End":"00:05.490","Text":"2 integrable functions on a,b."},{"Start":"00:05.490 ","End":"00:08.400","Text":"In part a, we have to show that if f is less than or"},{"Start":"00:08.400 ","End":"00:11.340","Text":"equal to g for all x on the interval,"},{"Start":"00:11.340 ","End":"00:14.490","Text":"then the integral of f is less than or equal to the integral of"},{"Start":"00:14.490 ","End":"00:17.880","Text":"g. In part b we have to show that"},{"Start":"00:17.880 ","End":"00:21.270","Text":"the absolute value of the integral is"},{"Start":"00:21.270 ","End":"00:25.350","Text":"less than or equal to the integral of the absolute value."},{"Start":"00:25.350 ","End":"00:31.890","Text":"In part C, if f is sandwiched between little m and big M on the interval,"},{"Start":"00:31.890 ","End":"00:34.605","Text":"then this inequality holds."},{"Start":"00:34.605 ","End":"00:40.970","Text":"We\u0027ll use this to show that this inequality holds."},{"Start":"00:40.970 ","End":"00:44.230","Text":"This will be an example of this."},{"Start":"00:44.230 ","End":"00:48.920","Text":"One of the properties of the integral is that"},{"Start":"00:48.920 ","End":"00:53.300","Text":"the integral of the difference is the difference of the integrals."},{"Start":"00:53.300 ","End":"00:55.505","Text":"We\u0027ll use this in a moment."},{"Start":"00:55.505 ","End":"01:00.454","Text":"Also. g of x is bigger or equal to f of x everywhere,"},{"Start":"01:00.454 ","End":"01:06.274","Text":"which means that g of x minus f of x is non-negative on the interval."},{"Start":"01:06.274 ","End":"01:12.260","Text":"If that\u0027s true, then the integral of a non-negative,"},{"Start":"01:12.260 ","End":"01:15.710","Text":"in this case, g minus f is bigger or equal to 0."},{"Start":"01:15.710 ","End":"01:17.825","Text":"It\u0027s not a property of the Riemann integral."},{"Start":"01:17.825 ","End":"01:21.145","Text":"The integral of non-negative is non-negative."},{"Start":"01:21.145 ","End":"01:25.030","Text":"We can combine these and get that,"},{"Start":"01:25.030 ","End":"01:27.650","Text":"because this is bigger or equal to 0,"},{"Start":"01:27.650 ","End":"01:30.290","Text":"and this is equal to this,"},{"Start":"01:30.290 ","End":"01:33.140","Text":"then this is bigger or equal to 0."},{"Start":"01:33.140 ","End":"01:38.600","Text":"Which means just switching sides that the integral of f is less than or equal"},{"Start":"01:38.600 ","End":"01:44.900","Text":"to the integral of g. Now on to part B."},{"Start":"01:44.900 ","End":"01:48.050","Text":"For any number a is less than or equal to"},{"Start":"01:48.050 ","End":"01:52.505","Text":"the absolute value of a and bigger or equal to minus absolute value of a,"},{"Start":"01:52.505 ","End":"01:54.980","Text":"this is true for each x."},{"Start":"01:54.980 ","End":"01:59.015","Text":"By Part a, this integral"},{"Start":"01:59.015 ","End":"02:03.485","Text":"is less than or equal to this integral is less than or equal to this integral."},{"Start":"02:03.485 ","End":"02:06.880","Text":"Part A shows that the integrals preserve order."},{"Start":"02:06.880 ","End":"02:10.480","Text":"Now the integral of minus something,"},{"Start":"02:10.480 ","End":"02:12.579","Text":"we can take the minus outside,"},{"Start":"02:12.579 ","End":"02:14.484","Text":"it\u0027s minus the integral."},{"Start":"02:14.484 ","End":"02:17.920","Text":"This is true for generally any constant,"},{"Start":"02:17.920 ","End":"02:22.825","Text":"any constant c can be taken outside the integral in particular minus 1."},{"Start":"02:22.825 ","End":"02:26.020","Text":"From here, just taking the minus out here,"},{"Start":"02:26.020 ","End":"02:28.070","Text":"we get this inequality."},{"Start":"02:28.070 ","End":"02:36.445","Text":"That means that the absolute value of this is less than or equal to this."},{"Start":"02:36.445 ","End":"02:38.470","Text":"Because in general with algebra,"},{"Start":"02:38.470 ","End":"02:41.665","Text":"if you have b sandwiched between minus a and a,"},{"Start":"02:41.665 ","End":"02:45.925","Text":"then the absolute value of b is less than or equal to a."},{"Start":"02:45.925 ","End":"02:48.335","Text":"We get this just from algebra."},{"Start":"02:48.335 ","End":"02:50.805","Text":"That concludes Part B."},{"Start":"02:50.805 ","End":"02:56.180","Text":"In Part C, we\u0027re told that f of x is bounded below by little m and bounded above by"},{"Start":"02:56.180 ","End":"03:01.835","Text":"big M. Now we can look at these not just as constants but as constant functions."},{"Start":"03:01.835 ","End":"03:04.345","Text":"We can use part A."},{"Start":"03:04.345 ","End":"03:06.875","Text":"If we look at m and M as functions,"},{"Start":"03:06.875 ","End":"03:08.810","Text":"then we get this inequality."},{"Start":"03:08.810 ","End":"03:11.240","Text":"Take the integral of each of these 3 functions."},{"Start":"03:11.240 ","End":"03:14.300","Text":"The integral of a constant on an interval from a"},{"Start":"03:14.300 ","End":"03:17.765","Text":"to b is just b minus a, times that constant."},{"Start":"03:17.765 ","End":"03:19.880","Text":"Here we have b minus a times little m,"},{"Start":"03:19.880 ","End":"03:23.710","Text":"and here b minus a times big M. In the middle the integral."},{"Start":"03:23.710 ","End":"03:27.955","Text":"That\u0027s an example of this."},{"Start":"03:27.955 ","End":"03:32.720","Text":"In this example we have f of x is sine x over x of the interval a,"},{"Start":"03:32.720 ","End":"03:37.115","Text":"b is the interval from a quarter Pi to a third Pi."},{"Start":"03:37.115 ","End":"03:41.675","Text":"I claim that f is decreasing on this interval."},{"Start":"03:41.675 ","End":"03:44.330","Text":"This is just a shorthand way of saying decreasing."},{"Start":"03:44.330 ","End":"03:46.550","Text":"The way to show this,"},{"Start":"03:46.550 ","End":"03:52.319","Text":"the easiest way is to show that the derivative is negative on the interval."},{"Start":"03:52.900 ","End":"03:56.075","Text":"Do this by contradiction."},{"Start":"03:56.075 ","End":"04:01.840","Text":"Let\u0027s suppose that the derivative is 0 somewhere."},{"Start":"04:01.840 ","End":"04:05.340","Text":"Then we\u0027ll get a contradiction."},{"Start":"04:05.340 ","End":"04:09.260","Text":"Then we\u0027ll say that either f is always less than 0,"},{"Start":"04:09.260 ","End":"04:10.790","Text":"always bigger than 0,"},{"Start":"04:10.790 ","End":"04:12.890","Text":"and will eliminate 1 of the other 2."},{"Start":"04:12.890 ","End":"04:14.710","Text":"That\u0027s the strategy."},{"Start":"04:14.710 ","End":"04:18.814","Text":"Let\u0027s see if it\u0027s possible for f prime to be 0,"},{"Start":"04:18.814 ","End":"04:23.765","Text":"then differentiating using the quotient rule will get this as 0."},{"Start":"04:23.765 ","End":"04:26.569","Text":"The numerator is 0."},{"Start":"04:26.569 ","End":"04:30.125","Text":"If the numerator is 0,"},{"Start":"04:30.125 ","End":"04:32.870","Text":"then we can bring the sine x to the other side,"},{"Start":"04:32.870 ","End":"04:39.095","Text":"divide by cosine x and get the equation tangent x equals x."},{"Start":"04:39.095 ","End":"04:43.640","Text":"This is not possible in our interval because"},{"Start":"04:43.640 ","End":"04:48.050","Text":"in the main branch of tangent x from minus Pi over 2 and Pi over 2,"},{"Start":"04:48.050 ","End":"04:53.554","Text":"the line y equals x cuts this graph only once."},{"Start":"04:53.554 ","End":"04:57.110","Text":"Here\u0027s a picture of tangent x and x."},{"Start":"04:57.110 ","End":"05:01.160","Text":"Between pi over 4 and pi over 3 as here,"},{"Start":"05:01.160 ","End":"05:03.815","Text":"it\u0027s clear that the tangent is bigger."},{"Start":"05:03.815 ","End":"05:05.450","Text":"We could go into more detail."},{"Start":"05:05.450 ","End":"05:09.290","Text":"You could say the derivative of the tangent is secant squared,"},{"Start":"05:09.290 ","End":"05:13.870","Text":"which is always bigger than 1 except for 0."},{"Start":"05:13.870 ","End":"05:16.250","Text":"I think we\u0027ll just leave it at that,"},{"Start":"05:16.250 ","End":"05:20.460","Text":"that there is no solution to this in our interval."},{"Start":"05:20.460 ","End":"05:23.330","Text":"That leaves us with 2 possibilities."},{"Start":"05:23.330 ","End":"05:26.680","Text":"That either f prime is positive in the whole interval,"},{"Start":"05:26.680 ","End":"05:28.850","Text":"or f prime is negative in the whole interval."},{"Start":"05:28.850 ","End":"05:33.385","Text":"To check, we just have to substitute the value."},{"Start":"05:33.385 ","End":"05:38.470","Text":"Let\u0027s for example, substitute the left end point,"},{"Start":"05:38.470 ","End":"05:41.030","Text":"f prime at pi over 4."},{"Start":"05:41.030 ","End":"05:45.920","Text":"Well, leave you to follow the computation comes out to be negative,"},{"Start":"05:45.920 ","End":"05:49.670","Text":"which means that f prime is negative on all of this interval,"},{"Start":"05:49.670 ","End":"05:53.130","Text":"which means that f is decreasing there."},{"Start":"05:54.560 ","End":"06:00.320","Text":"You can take m to be f of pi over 3."},{"Start":"06:00.320 ","End":"06:03.695","Text":"The smallest value of f is at the right end point,"},{"Start":"06:03.695 ","End":"06:06.730","Text":"and the largest is at the left end point."},{"Start":"06:06.730 ","End":"06:09.735","Text":"Big M is f of Pi over 4."},{"Start":"06:09.735 ","End":"06:13.400","Text":"If we just substitute these values into f of x,"},{"Start":"06:13.400 ","End":"06:15.170","Text":"which is sine x over x,"},{"Start":"06:15.170 ","End":"06:17.560","Text":"we get little m is this."},{"Start":"06:17.560 ","End":"06:22.140","Text":"Sine of pi over 3 is root 3 over 2. We get this."},{"Start":"06:22.140 ","End":"06:23.960","Text":"Here we get 4 over Pi,"},{"Start":"06:23.960 ","End":"06:27.185","Text":"sine Pi over 4, sine Pi over 4 is root 2 over 2."},{"Start":"06:27.185 ","End":"06:29.215","Text":"We get this."},{"Start":"06:29.215 ","End":"06:38.090","Text":"Now, recalling the inequality which we just showed above and replacing it in our case,"},{"Start":"06:38.090 ","End":"06:42.295","Text":"we get that, this is little m here."},{"Start":"06:42.295 ","End":"06:43.900","Text":"This is the b minus a."},{"Start":"06:43.900 ","End":"06:50.120","Text":"This is the b minus a is the big M. Here\u0027s our integral."},{"Start":"06:50.570 ","End":"06:53.920","Text":"What that gives us, well,"},{"Start":"06:53.920 ","End":"06:55.900","Text":"this is common to both,"},{"Start":"06:55.900 ","End":"06:58.420","Text":"so it\u0027s 2 Pi over 3 minus Pi over 4,"},{"Start":"06:58.420 ","End":"06:59.710","Text":"that gives Pi over 12."},{"Start":"06:59.710 ","End":"07:01.825","Text":"Here and here we have Pi over 12."},{"Start":"07:01.825 ","End":"07:08.085","Text":"Little m is 3 root 3 over 2 Pi times pi over 12."},{"Start":"07:08.085 ","End":"07:13.090","Text":"That simplifies to this and big M simplifies to this."},{"Start":"07:13.090 ","End":"07:16.700","Text":"Just placing them here above,"},{"Start":"07:16.700 ","End":"07:18.830","Text":"we get what we wanted."},{"Start":"07:18.830 ","End":"07:20.815","Text":"This is what was required above,"},{"Start":"07:20.815 ","End":"07:23.150","Text":"and so we are done."}],"ID":24750},{"Watched":false,"Name":"Exercise 12","Duration":"5m 1s","ChapterTopicVideoID":23809,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23809.jpeg","UploadDate":"2021-01-11T09:13:07.2470000","DurationForVideoObject":"PT5M1S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.920","Text":"In this exercise, we have f a non-negative function on the interval a,"},{"Start":"00:04.920 ","End":"00:07.290","Text":"b and there are 2 parts."},{"Start":"00:07.290 ","End":"00:14.310","Text":"In a, we have to show that if f is continuous and the integral is 0,"},{"Start":"00:14.310 ","End":"00:20.030","Text":"then f must be 0 for all x in the interval."},{"Start":"00:20.030 ","End":"00:22.050","Text":"In part b, we dropped the assumption of"},{"Start":"00:22.050 ","End":"00:25.515","Text":"continuity and are asked to give a counterexample."},{"Start":"00:25.515 ","End":"00:32.220","Text":"In other words, we have to find f which is integrable on ab and the integral is 0."},{"Start":"00:32.220 ","End":"00:39.090","Text":"But at least for 1 x_naught in the interval f of x_naught is strictly positive."},{"Start":"00:39.090 ","End":"00:46.650","Text":"By contradiction, suppose that f of x_naught equals alpha, which is not 0,"},{"Start":"00:46.650 ","End":"00:50.840","Text":"and not 0, means that it\u0027s strictly positive"},{"Start":"00:50.840 ","End":"00:55.849","Text":"because the function is non-negative for some x_naught in the interval."},{"Start":"00:55.849 ","End":"00:59.825","Text":"At first, we will assume that x_naught is in the interior of the interval."},{"Start":"00:59.825 ","End":"01:01.610","Text":"In other words, not 1 of the endpoint and then we\u0027ll"},{"Start":"01:01.610 ","End":"01:04.780","Text":"say how to deal with the endpoint case."},{"Start":"01:04.780 ","End":"01:11.240","Text":"From the continuity, if it\u0027s equal to Alpha at x_naught,"},{"Start":"01:11.240 ","End":"01:12.739","Text":"we can find some neighborhood,"},{"Start":"01:12.739 ","End":"01:17.825","Text":"a Delta neighborhood of x_naught such that in this neighborhood,"},{"Start":"01:17.825 ","End":"01:20.195","Text":"f of x is bigger than,"},{"Start":"01:20.195 ","End":"01:22.340","Text":"let\u0027s say Alpha over 2."},{"Start":"01:22.340 ","End":"01:25.550","Text":"I mean, at this point it\u0027s bigger than Alpha over 2."},{"Start":"01:25.550 ","End":"01:29.180","Text":"We can find an interval using epsilon Delta."},{"Start":"01:29.180 ","End":"01:35.720","Text":"We can shrink Delta if necessary so that this interval fits into a,"},{"Start":"01:35.720 ","End":"01:38.285","Text":"b, the open interval."},{"Start":"01:38.285 ","End":"01:40.385","Text":"We can slightly modify this."},{"Start":"01:40.385 ","End":"01:43.280","Text":"Certainly, if this is contained in this,"},{"Start":"01:43.280 ","End":"01:48.835","Text":"then the endpoints will be contained in the closed interval."},{"Start":"01:48.835 ","End":"01:50.970","Text":"Also f of x well,"},{"Start":"01:50.970 ","End":"01:52.845","Text":"may not be strictly bigger than,"},{"Start":"01:52.845 ","End":"01:57.350","Text":"but at the endpoints it will certainly be greater or equal to from the continuity."},{"Start":"01:57.350 ","End":"02:02.225","Text":"Now, let\u0027s take a partition consisting of just 4 elements,"},{"Start":"02:02.225 ","End":"02:03.905","Text":"x_naught, x_1, x_2, x_3."},{"Start":"02:03.905 ","End":"02:07.400","Text":"We start from a, then we\u0027ll take the x_naught minus Delta,"},{"Start":"02:07.400 ","End":"02:10.285","Text":"then the x_naught plus Delta, and then b."},{"Start":"02:10.285 ","End":"02:11.869","Text":"Now, for this partition,"},{"Start":"02:11.869 ","End":"02:15.860","Text":"we can compute the lowest sum for the function f. It\u0027s"},{"Start":"02:15.860 ","End":"02:20.840","Text":"the sum from 1 to 3 of m_i Delta x_i."},{"Start":"02:20.840 ","End":"02:28.300","Text":"M_i is the infimum on the interval from x_i minus 1 to x_i."},{"Start":"02:28.300 ","End":"02:33.125","Text":"In the first case we\u0027re going from a to x_naught minus Delta."},{"Start":"02:33.125 ","End":"02:39.965","Text":"Certainly the infimum is bigger or equal to 0 because the function is non-negative."},{"Start":"02:39.965 ","End":"02:42.095","Text":"In the second part,"},{"Start":"02:42.095 ","End":"02:45.390","Text":"we know the Delta x_2,"},{"Start":"02:45.390 ","End":"02:47.240","Text":"which is going to be this minus this,"},{"Start":"02:47.240 ","End":"02:49.000","Text":"which is 2 Delta."},{"Start":"02:49.000 ","End":"02:51.770","Text":"Also from here to here,"},{"Start":"02:51.770 ","End":"02:54.710","Text":"f is bigger than Alpha over 2."},{"Start":"02:54.710 ","End":"03:01.550","Text":"In the last part also the infimum is at least 0 and each of these Deltas is positive."},{"Start":"03:01.550 ","End":"03:03.470","Text":"Altogether adding this up,"},{"Start":"03:03.470 ","End":"03:06.604","Text":"we get bigger or equal to this,"},{"Start":"03:06.604 ","End":"03:10.079","Text":"which is Alpha Delta, which is bigger than 0."},{"Start":"03:10.079 ","End":"03:13.325","Text":"Now, if a lower sum is bigger than 0,"},{"Start":"03:13.325 ","End":"03:17.360","Text":"then the integral is bigger than 0"},{"Start":"03:17.360 ","End":"03:21.320","Text":"also because the integral is bigger or equal to any lower sum."},{"Start":"03:21.320 ","End":"03:27.350","Text":"This gives us a contradiction because the integral was given to be equal to 0."},{"Start":"03:27.350 ","End":"03:32.570","Text":"Now, we\u0027re not quite done because we assumed that x_naught was in the open interval a,"},{"Start":"03:32.570 ","End":"03:37.270","Text":"b, we have to consider the case where x_naught is 1 of the endpoints."},{"Start":"03:37.270 ","End":"03:40.575","Text":"If x_naught is a, or x_naught is b,"},{"Start":"03:40.575 ","End":"03:42.455","Text":"the proof is very similar."},{"Start":"03:42.455 ","End":"03:46.535","Text":"Essentially, we use 1-sided intervals."},{"Start":"03:46.535 ","End":"03:49.490","Text":"You replace this interval by a,"},{"Start":"03:49.490 ","End":"03:53.590","Text":"a plus Delta or b minus Delta b."},{"Start":"03:53.590 ","End":"03:55.430","Text":"The rest of it is very similar."},{"Start":"03:55.430 ","End":"03:57.470","Text":"I\u0027ll leave you the details."},{"Start":"03:57.470 ","End":"04:02.375","Text":"That part a, now let\u0027s get on to part b."},{"Start":"04:02.375 ","End":"04:05.680","Text":"Here, we needed to give that counter-example."},{"Start":"04:05.680 ","End":"04:15.485","Text":"Let\u0027s let f of x be 1 when x is a and 0 on the rest of the interval a, b."},{"Start":"04:15.485 ","End":"04:22.840","Text":"Of course, it\u0027s a non-negative function because 1 and 0 are both non-negative."},{"Start":"04:22.840 ","End":"04:28.910","Text":"Now, our function f is very similar to the function that\u0027s constantly 0."},{"Start":"04:28.910 ","End":"04:31.205","Text":"If we took f equals 0 everywhere,"},{"Start":"04:31.205 ","End":"04:34.925","Text":"then it would be a constant function and it would be integrable."},{"Start":"04:34.925 ","End":"04:37.730","Text":"We just also did at 1 point."},{"Start":"04:37.730 ","End":"04:41.525","Text":"We know that when you do that it\u0027s still integrable."},{"Start":"04:41.525 ","End":"04:46.430","Text":"The integral is the same as the integral where the function is"},{"Start":"04:46.430 ","End":"04:51.350","Text":"constantly 0 and the function itself is as required."},{"Start":"04:51.350 ","End":"04:58.925","Text":"The integral is 0 and f of x_naught is bigger than 0 if we take x_naught to be a."},{"Start":"04:58.925 ","End":"05:02.520","Text":"This is an example and we\u0027re done."}],"ID":24751},{"Watched":false,"Name":"Exercise 13","Duration":"5m 14s","ChapterTopicVideoID":23810,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23810.jpeg","UploadDate":"2021-01-11T09:15:30.7230000","DurationForVideoObject":"PT5M14S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.950","Text":"In this exercise, f is a bounded function on the interval 0,1."},{"Start":"00:05.950 ","End":"00:10.185","Text":"Let\u0027s suppose that for any c between 0 and 1,"},{"Start":"00:10.185 ","End":"00:14.580","Text":"f is integrable on the closed interval from c to 1."},{"Start":"00:14.580 ","End":"00:19.075","Text":"In part a, we have to show that f is integrable on all of 0,1."},{"Start":"00:19.075 ","End":"00:21.795","Text":"It\u0027s like letting c tend to 0."},{"Start":"00:21.795 ","End":"00:25.050","Text":"In part b, we use part a,"},{"Start":"00:25.050 ","End":"00:29.865","Text":"to show that this function is integrable on 0,1,"},{"Start":"00:29.865 ","End":"00:36.665","Text":"where the function here is f_(x) is 0 if x is 0,"},{"Start":"00:36.665 ","End":"00:39.140","Text":"and sine of 1/x,"},{"Start":"00:39.140 ","End":"00:45.110","Text":"if x is in the interval from 0-1 excluding 0."},{"Start":"00:45.110 ","End":"00:50.270","Text":"If you\u0027re curious what it looks like, here is a sketch of this function."},{"Start":"00:50.270 ","End":"00:52.335","Text":"Onto part a."},{"Start":"00:52.335 ","End":"00:56.360","Text":"Let k be a bound for f,"},{"Start":"00:56.360 ","End":"01:00.020","Text":"meaning that the absolute value of f_(x) is less than or equal to"},{"Start":"01:00.020 ","End":"01:04.655","Text":"K on the interval, or you could say that f_(x) is"},{"Start":"01:04.655 ","End":"01:09.080","Text":"less than or equal to K and bigger equal to minus K. We\u0027ll use"},{"Start":"01:09.080 ","End":"01:14.390","Text":"the Riemann criterion to show that f is integrable on 0,1."},{"Start":"01:14.390 ","End":"01:18.354","Text":"Let Epsilon bigger than 0 be given."},{"Start":"01:18.354 ","End":"01:23.680","Text":"Our goal is to find the partition such that the difference between"},{"Start":"01:23.680 ","End":"01:29.390","Text":"the upper and the lower sums for this partition is less than Epsilon."},{"Start":"01:29.390 ","End":"01:34.450","Text":"Choose a natural number n such that n is bigger than this expression."},{"Start":"01:34.450 ","End":"01:36.910","Text":"Later, we\u0027ll see why this expression."},{"Start":"01:36.910 ","End":"01:41.365","Text":"You can use the Archimedean property to guarantee there is such an n."},{"Start":"01:41.365 ","End":"01:48.445","Text":"Rewriting this 2K is less than Epsilon over 2."},{"Start":"01:48.445 ","End":"01:53.830","Text":"F is integrable on the closed interval c,1"},{"Start":"01:53.830 ","End":"01:59.720","Text":"and c could be 1. Find the partition P_n,"},{"Start":"01:59.720 ","End":"02:05.250","Text":"starting at 1 and ending in 1 with the upper minus"},{"Start":"02:05.250 ","End":"02:07.910","Text":"the lowest sum for this partition less"},{"Start":"02:07.910 ","End":"02:11.375","Text":"than any epsilon or epsilon over 2, we could choose."},{"Start":"02:11.375 ","End":"02:17.600","Text":"We can use this partition P_n to complete it to a partition of all of"},{"Start":"02:17.600 ","End":"02:25.630","Text":"0,1 just by adding 1 more element that is 0 in front of the first element here."},{"Start":"02:25.630 ","End":"02:28.670","Text":"We\u0027ve got 0,1,"},{"Start":"02:28.670 ","End":"02:30.495","Text":"etc., up to 1."},{"Start":"02:30.495 ","End":"02:32.400","Text":"I guess the indexing will be off,"},{"Start":"02:32.400 ","End":"02:34.380","Text":"we\u0027ll start at x _minus 1,"},{"Start":"02:34.380 ","End":"02:37.090","Text":"this to be be x_0, and so on."},{"Start":"02:37.990 ","End":"02:43.760","Text":"We can compute U minus L for this new partition of"},{"Start":"02:43.760 ","End":"02:49.175","Text":"all of 0,1 in terms of the partition for 1 1."},{"Start":"02:49.175 ","End":"02:57.545","Text":"All we have when we take the sums of the M_i minus little m_i times Delta x_i,"},{"Start":"02:57.545 ","End":"03:00.830","Text":"we\u0027ll get an extra first term."},{"Start":"03:00.830 ","End":"03:04.009","Text":"There\u0027s an extra interval at the beginning of the partition,"},{"Start":"03:04.009 ","End":"03:06.635","Text":"the bit from 0 to 1,"},{"Start":"03:06.635 ","End":"03:09.425","Text":"all the other intervals will be the same."},{"Start":"03:09.425 ","End":"03:15.230","Text":"Now, the supremum, I called it M_0 because the index here is 0."},{"Start":"03:15.230 ","End":"03:20.359","Text":"The supremum on this interval is certainly less than or equal to K,"},{"Start":"03:20.359 ","End":"03:26.405","Text":"and the infimum is bigger or equal to minus K."},{"Start":"03:26.405 ","End":"03:33.620","Text":"Altogether, we have the tail bit and then this bit,"},{"Start":"03:33.620 ","End":"03:36.095","Text":"which has these inequalities."},{"Start":"03:36.095 ","End":"03:41.790","Text":"What we can say is that this is less than or equal to the largest it can be is here,"},{"Start":"03:41.790 ","End":"03:43.530","Text":"K, here take the minus K,"},{"Start":"03:43.530 ","End":"03:46.140","Text":"K minus minus K is 2K."},{"Start":"03:46.140 ","End":"03:54.075","Text":"Then 1 minus 0 is 1. We get this 2K and the rest of it is the same."},{"Start":"03:54.075 ","End":"04:01.190","Text":"We already said that this u minus l for P_n is less than Epsilon over 2,"},{"Start":"04:01.190 ","End":"04:07.535","Text":"so that\u0027s this 1, and also this 2K is less than Epsilon over 2 from here."},{"Start":"04:07.535 ","End":"04:13.500","Text":"We have less than Epsilon over 2 plus Epsilon over 2, which is Epsilon."},{"Start":"04:13.500 ","End":"04:18.090","Text":"We\u0027ve shown that this minus this is less than Epsilon."},{"Start":"04:18.090 ","End":"04:20.570","Text":"For all Epsilon bigger than 0,"},{"Start":"04:20.570 ","End":"04:24.785","Text":"there exists n such that this minus this plus an epsilon."},{"Start":"04:24.785 ","End":"04:27.350","Text":"By the Riemann criteria,"},{"Start":"04:27.350 ","End":"04:32.500","Text":"and it means that f is integrable on all of 0,1."},{"Start":"04:32.500 ","End":"04:35.484","Text":"Now, onto part b,"},{"Start":"04:35.484 ","End":"04:38.540","Text":"we have this function,"},{"Start":"04:38.540 ","End":"04:41.230","Text":"and here\u0027s the sketch."},{"Start":"04:41.230 ","End":"04:43.865","Text":"If we use part a,"},{"Start":"04:43.865 ","End":"04:48.060","Text":"we see that for any c that\u0027s strictly between 0 and 1,"},{"Start":"04:48.060 ","End":"04:56.360","Text":"f is continuous everywhere"},{"Start":"04:56.360 ","End":"04:59.735","Text":"as long as we exclude the 0."},{"Start":"04:59.735 ","End":"05:05.210","Text":"If it\u0027s continuous, then it\u0027s integrable, and in Part a,"},{"Start":"05:05.210 ","End":"05:07.500","Text":"if it\u0027s integrable and every c,1,"},{"Start":"05:07.500 ","End":"05:10.985","Text":"then it\u0027s integrable in all of 0,1."},{"Start":"05:10.985 ","End":"05:14.940","Text":"That completes b, and we\u0027re done."}],"ID":24752},{"Watched":false,"Name":"Exercise 14","Duration":"4m 6s","ChapterTopicVideoID":23811,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23811.jpeg","UploadDate":"2021-01-11T09:17:30.6500000","DurationForVideoObject":"PT4M6S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.995","Text":"In this exercise, f is a continuous function on the closed interval a,"},{"Start":"00:04.995 ","End":"00:09.225","Text":"b, and it has a certain property."},{"Start":"00:09.225 ","End":"00:17.385","Text":"That is that whenever the product fg is integrable for some integrable function g,"},{"Start":"00:17.385 ","End":"00:23.730","Text":"then the integral of fg is 0."},{"Start":"00:23.730 ","End":"00:26.490","Text":"If f has this property,"},{"Start":"00:26.490 ","End":"00:31.710","Text":"we have to show that f is identically 0."},{"Start":"00:31.710 ","End":"00:34.950","Text":"We\u0027ll do a proof by contradiction."},{"Start":"00:34.950 ","End":"00:40.610","Text":"Contradiction to f being identically 0 is that f"},{"Start":"00:40.610 ","End":"00:47.200","Text":"of x naught is non-zero for at least 1x naught in the interval a, b."},{"Start":"00:47.200 ","End":"00:51.170","Text":"Let\u0027s assume that f of x naught is bigger than"},{"Start":"00:51.170 ","End":"00:56.220","Text":"0 because the case less than 0 is so similar."},{"Start":"00:56.300 ","End":"01:01.685","Text":"X naught could be in the interior of the interval or it could be one of the endpoints."},{"Start":"01:01.685 ","End":"01:04.790","Text":"Let\u0027s see if we can reduce the number of cases."},{"Start":"01:04.790 ","End":"01:08.960","Text":"If x naught is one of the endpoints a or b,"},{"Start":"01:08.960 ","End":"01:11.345","Text":"then by the continuity,"},{"Start":"01:11.345 ","End":"01:14.110","Text":"if it\u0027s positive, one of the endpoints,"},{"Start":"01:14.110 ","End":"01:19.374","Text":"and we can find a point near x naught, call it x_1,"},{"Start":"01:19.374 ","End":"01:26.880","Text":"which is bigger than 0, and x_1 only being near a or b will be in the interior of a, b."},{"Start":"01:26.880 ","End":"01:30.150","Text":"We can consider this x_1 as our x naught."},{"Start":"01:30.150 ","End":"01:33.575","Text":"We can assume from the beginning that if this is the case,"},{"Start":"01:33.575 ","End":"01:35.750","Text":"the next naught is in the interior,"},{"Start":"01:35.750 ","End":"01:37.990","Text":"and then we just have one case."},{"Start":"01:37.990 ","End":"01:42.095","Text":"X naught is in the interior in the open interval,"},{"Start":"01:42.095 ","End":"01:44.765","Text":"and let the value be Alpha,"},{"Start":"01:44.765 ","End":"01:47.020","Text":"so Alpha is bigger than 0."},{"Start":"01:47.020 ","End":"01:50.705","Text":"Let\u0027s define Epsilon to be 1/2 of Alpha."},{"Start":"01:50.705 ","End":"01:56.000","Text":"Since f is continuous by the Epsilon Delta definition of continuity,"},{"Start":"01:56.000 ","End":"02:02.600","Text":"there exists some Delta bigger than 0 such that f of x is bigger than Alpha"},{"Start":"02:02.600 ","End":"02:09.570","Text":"over 2 for all x within Delta of x naught."},{"Start":"02:09.570 ","End":"02:11.660","Text":"Shrinking Delta if necessary,"},{"Start":"02:11.660 ","End":"02:19.170","Text":"we make sure that this interval is completely contained in the open interval a, b."},{"Start":"02:19.580 ","End":"02:23.510","Text":"If we just replace the open by the closed intervals,"},{"Start":"02:23.510 ","End":"02:28.534","Text":"f of x is bigger or equal to Alpha over 2 for all x in the closed interval,"},{"Start":"02:28.534 ","End":"02:32.080","Text":"which is contained in the closed interval a, b."},{"Start":"02:32.080 ","End":"02:36.560","Text":"Let\u0027s define our g of x. G of x,"},{"Start":"02:36.560 ","End":"02:44.515","Text":"we\u0027ll let it be 1 if x is in this interval that we just mentioned, and 0 otherwise."},{"Start":"02:44.515 ","End":"02:47.400","Text":"Of course, g is integrable."},{"Start":"02:47.400 ","End":"02:53.000","Text":"By the given the integral of fg is 0,"},{"Start":"02:53.000 ","End":"02:56.405","Text":"f times g, if you think about it,"},{"Start":"02:56.405 ","End":"03:01.110","Text":"is just 1 times f on this interval,"},{"Start":"03:01.110 ","End":"03:03.330","Text":"and 0 times f,"},{"Start":"03:03.330 ","End":"03:06.555","Text":"in other words, 0 outside that interval."},{"Start":"03:06.555 ","End":"03:13.240","Text":"This fg is just same as f on this interval and 0 otherwise."},{"Start":"03:13.310 ","End":"03:17.630","Text":"The integral of fg from a-b is just"},{"Start":"03:17.630 ","End":"03:21.200","Text":"the integral from x naught minus Delta to x naught plus"},{"Start":"03:21.200 ","End":"03:30.680","Text":"Delta of f. F of x is bigger or equal to Alpha over 2 on all this interval."},{"Start":"03:30.680 ","End":"03:37.225","Text":"The integral of f of xdx is at least the integral of Alpha over 2dx,"},{"Start":"03:37.225 ","End":"03:42.500","Text":"but the integral of a constant is just Alpha over 2 times the width of the interval,"},{"Start":"03:42.500 ","End":"03:44.000","Text":"which is 2 Delta."},{"Start":"03:44.000 ","End":"03:48.755","Text":"2 Delta times Alpha over 2 is Alpha Delta, which is positive."},{"Start":"03:48.755 ","End":"03:54.604","Text":"That\u0027s a contradiction because the integral is 0,"},{"Start":"03:54.604 ","End":"03:56.645","Text":"so it can\u0027t be positive."},{"Start":"03:56.645 ","End":"04:03.680","Text":"This contradiction came from the assumption that f is not identically 0,"},{"Start":"04:03.680 ","End":"04:07.110","Text":"and so it is, and we\u0027re done."}],"ID":24753},{"Watched":false,"Name":"Exercise 15","Duration":"4m 33s","ChapterTopicVideoID":23812,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23812.jpeg","UploadDate":"2021-01-11T09:18:59.3100000","DurationForVideoObject":"PT4M33S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.335","Text":"In this exercise, x and y are non-negative numbers."},{"Start":"00:04.335 ","End":"00:07.695","Text":"We have to show that this limit,"},{"Start":"00:07.695 ","End":"00:10.890","Text":"I won\u0027t read it out is equal to maximum of x and y."},{"Start":"00:10.890 ","End":"00:18.990","Text":"In part b, we have a continuous function f defined on ab and taking non-negative values."},{"Start":"00:18.990 ","End":"00:22.410","Text":"We have to show that this limit holds."},{"Start":"00:22.410 ","End":"00:24.060","Text":"Again, I won\u0027t read it out."},{"Start":"00:24.060 ","End":"00:26.700","Text":"Let\u0027s start with part a."},{"Start":"00:26.700 ","End":"00:34.620","Text":"Lets call the maximum of the 2 numbers M. What we can say is that M to the n well,"},{"Start":"00:34.620 ","End":"00:37.130","Text":"M is going to be either x or y."},{"Start":"00:37.130 ","End":"00:39.230","Text":"If it\u0027s equal to 1 of these 2,"},{"Start":"00:39.230 ","End":"00:41.210","Text":"then it\u0027s less than or equal to the sum."},{"Start":"00:41.210 ","End":"00:45.640","Text":"On the other hand, each of these is less than or equal to the maximum."},{"Start":"00:45.640 ","End":"00:48.490","Text":"It\u0027s less than or equal to 2M to the n."},{"Start":"00:48.820 ","End":"00:55.190","Text":"What we have if we take the nth root is that M is less than or equal to this,"},{"Start":"00:55.190 ","End":"00:57.640","Text":"less than or equal to this."},{"Start":"00:57.640 ","End":"01:00.635","Text":"Now let\u0027s let n go to infinity."},{"Start":"01:00.635 ","End":"01:03.845","Text":"Now, 2 to the 1 over n goes to 1,"},{"Start":"01:03.845 ","End":"01:07.120","Text":"as n goes to infinity it\u0027s like 2 to the 0."},{"Start":"01:07.120 ","End":"01:09.029","Text":"By the Sandwich Theorem,"},{"Start":"01:09.029 ","End":"01:13.150","Text":"since this goes to M and this is M,"},{"Start":"01:13.150 ","End":"01:16.900","Text":"this is sandwiched, and so it has to go to M,"},{"Start":"01:16.900 ","End":"01:21.015","Text":"which is maximum of x,y, as required."},{"Start":"01:21.015 ","End":"01:28.225","Text":"Now, in part b, we\u0027ll let big M be the supremum as written here."},{"Start":"01:28.225 ","End":"01:31.180","Text":"Then by the Extreme Value Theorem,"},{"Start":"01:31.180 ","End":"01:32.810","Text":"the supremum is attained,"},{"Start":"01:32.810 ","End":"01:34.805","Text":"so it\u0027s really a maximum."},{"Start":"01:34.805 ","End":"01:39.735","Text":"It\u0027s attained at some x naught in the interval a, b."},{"Start":"01:39.735 ","End":"01:44.180","Text":"The uninteresting case is when the supremum is 0,"},{"Start":"01:44.180 ","End":"01:47.360","Text":"and then the whole function must be 0,"},{"Start":"01:47.360 ","End":"01:49.925","Text":"it\u0027s non-negative and the supremum 0,"},{"Start":"01:49.925 ","End":"01:52.940","Text":"then the result is trivial, everything is 0."},{"Start":"01:52.940 ","End":"01:56.390","Text":"Let\u0027s assume that M is bigger than 0."},{"Start":"01:56.390 ","End":"01:59.705","Text":"Let\u0027s choose some Epsilon bigger than 0."},{"Start":"01:59.705 ","End":"02:07.335","Text":"Now by continuity, there are c and d in the interval such that,"},{"Start":"02:07.335 ","End":"02:10.560","Text":"well, c, d is an interval around x naught."},{"Start":"02:10.560 ","End":"02:12.870","Text":"This is the neighborhood of x naught."},{"Start":"02:12.870 ","End":"02:18.565","Text":"In this neighborhood, f of x is bigger or equal to M minus Epsilon,"},{"Start":"02:18.565 ","End":"02:21.845","Text":"this is using the Epsilon Delta for continuity."},{"Start":"02:21.845 ","End":"02:27.500","Text":"If this is so, then f of x is bigger or equal to g of x,"},{"Start":"02:27.500 ","End":"02:29.460","Text":"which I\u0027m about to define."},{"Start":"02:29.460 ","End":"02:33.800","Text":"G of x, let it be M minus Epsilon on the interval c,"},{"Start":"02:33.800 ","End":"02:37.240","Text":"d and 0 outside this interval."},{"Start":"02:37.240 ","End":"02:42.080","Text":"What we get is that the integral of f is bigger or"},{"Start":"02:42.080 ","End":"02:47.170","Text":"equal to the integral of g to the n here and here."},{"Start":"02:47.170 ","End":"02:52.880","Text":"This is equal to the integral of M minus Epsilon to the power of"},{"Start":"02:52.880 ","End":"02:58.010","Text":"n cause we define g of x to be equal to M minus Epsilon on the interval c,"},{"Start":"02:58.010 ","End":"03:01.260","Text":"d and 0 outside of that."},{"Start":"03:01.260 ","End":"03:05.490","Text":"This is a constant as far as x goes,"},{"Start":"03:05.490 ","End":"03:08.930","Text":"so it\u0027s just this constant times the width of the interval,"},{"Start":"03:08.930 ","End":"03:13.414","Text":"which is d minus c. But on the other hand,"},{"Start":"03:13.414 ","End":"03:21.035","Text":"this integral is less than or equal to M to the n dx,"},{"Start":"03:21.035 ","End":"03:25.250","Text":"because f of x is less than or equal to M on the interval."},{"Start":"03:25.250 ","End":"03:28.175","Text":"The f of x to the n less than or equal to M to the n."},{"Start":"03:28.175 ","End":"03:29.930","Text":"This is equal to,"},{"Start":"03:29.930 ","End":"03:31.790","Text":"again because it\u0027s a constant,"},{"Start":"03:31.790 ","End":"03:34.659","Text":"this M to the n times the width of this interval."},{"Start":"03:34.659 ","End":"03:36.905","Text":"Now we have 2 inequalities."},{"Start":"03:36.905 ","End":"03:41.780","Text":"This quantity is bigger or equal to something and less than or equal to something."},{"Start":"03:41.780 ","End":"03:43.370","Text":"If we combine them,"},{"Start":"03:43.370 ","End":"03:49.445","Text":"then we get this to the power of 1 over n. On the one hand,"},{"Start":"03:49.445 ","End":"03:51.710","Text":"it\u0027s going to be bigger or equal to this,"},{"Start":"03:51.710 ","End":"03:55.235","Text":"to the power of 1 over n. That\u0027s this part here."},{"Start":"03:55.235 ","End":"03:57.980","Text":"On the other hand, is going to be less than or equal to this to"},{"Start":"03:57.980 ","End":"04:01.370","Text":"the power of 1 over n, which is this."},{"Start":"04:01.370 ","End":"04:05.045","Text":"If we let n go to infinity,"},{"Start":"04:05.045 ","End":"04:07.130","Text":"on the right-hand side,"},{"Start":"04:07.130 ","End":"04:12.090","Text":"we get M. On the left-hand side,"},{"Start":"04:12.090 ","End":"04:15.585","Text":"this goes to 1, so we\u0027re left with M minus Epsilon."},{"Start":"04:15.585 ","End":"04:19.575","Text":"This thing is sandwiched between M and M minus epsilon."},{"Start":"04:19.575 ","End":"04:22.340","Text":"This is true for all Epsilon bigger than 0."},{"Start":"04:22.340 ","End":"04:26.315","Text":"In that case, this thing must actually be equal to"},{"Start":"04:26.315 ","End":"04:30.920","Text":"M. That\u0027s what we have to show because M is the supremum,"},{"Start":"04:30.920 ","End":"04:33.810","Text":"and so we are done."}],"ID":24754},{"Watched":false,"Name":"Exercise 16","Duration":"4m 24s","ChapterTopicVideoID":23813,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23813.jpeg","UploadDate":"2021-01-11T09:21:01.3830000","DurationForVideoObject":"PT4M24S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.565","Text":"In this exercise, we\u0027ll prove what is called the Cauchy-Schwarz inequality."},{"Start":"00:05.565 ","End":"00:09.270","Text":"Now, parts a and b might look very different,"},{"Start":"00:09.270 ","End":"00:14.520","Text":"but actually they\u0027re both special cases of something called an inner product space."},{"Start":"00:14.520 ","End":"00:19.920","Text":"In one case, the n-dimensional vectors over the reals is an inner product space."},{"Start":"00:19.920 ","End":"00:20.970","Text":"In the other case,"},{"Start":"00:20.970 ","End":"00:23.520","Text":"the integrable functions over an interval a,"},{"Start":"00:23.520 ","End":"00:25.305","Text":"b is an inner product space."},{"Start":"00:25.305 ","End":"00:29.880","Text":"Both of these are called the Cauchy-Schwarz inequality."},{"Start":"00:29.880 ","End":"00:33.345","Text":"Just ignore what I said if you don\u0027t care about inner product spaces."},{"Start":"00:33.345 ","End":"00:35.190","Text":"Anyway, in part a,"},{"Start":"00:35.190 ","End":"00:42.175","Text":"we have to show that if x_1 through x_n and y_1 through y_n are 2 sets of n real numbers,"},{"Start":"00:42.175 ","End":"00:46.325","Text":"then the absolute value of the sum of x_i,"},{"Start":"00:46.325 ","End":"00:51.050","Text":"y_i is always less than or equal to the square root of"},{"Start":"00:51.050 ","End":"00:56.164","Text":"the sum of x_i squared times the square root of the sum of the y_i squared."},{"Start":"00:56.164 ","End":"01:00.890","Text":"We\u0027re given a hint that this sum of things"},{"Start":"01:00.890 ","End":"01:06.194","Text":"squared is bigger or equal to 0 for all t. In part b,"},{"Start":"01:06.194 ","End":"01:09.680","Text":"f and g are 2 integrable functions on the interval a, b."},{"Start":"01:09.680 ","End":"01:12.440","Text":"We have to show the absolute value of the integral of"},{"Start":"01:12.440 ","End":"01:15.590","Text":"the product is less than or equal to the square root"},{"Start":"01:15.590 ","End":"01:18.950","Text":"of the integral of f squared times the square root of the integral of"},{"Start":"01:18.950 ","End":"01:23.330","Text":"g squared and a similar hint."},{"Start":"01:23.410 ","End":"01:30.120","Text":"In part a, what we can do is focus on t"},{"Start":"01:30.120 ","End":"01:37.670","Text":"here and this expression is actually a polynomial in t. At first sight,"},{"Start":"01:37.670 ","End":"01:43.615","Text":"it\u0027s just a function of t for a given set of x_i and y_i,"},{"Start":"01:43.615 ","End":"01:46.160","Text":"and of course this is bigger or equal to 0."},{"Start":"01:46.160 ","End":"01:50.989","Text":"Now, if we expand this and collect together powers of t,"},{"Start":"01:50.989 ","End":"01:55.420","Text":"then we get this quadratic polynomial in"},{"Start":"01:55.420 ","End":"02:01.370","Text":"t. Note that the coefficient of t squared, call it a,"},{"Start":"02:01.370 ","End":"02:06.080","Text":"is non-negative and so is the free co-efficient,"},{"Start":"02:06.080 ","End":"02:11.510","Text":"call it c. What we have here is a quadratic polynomial At squared plus"},{"Start":"02:11.510 ","End":"02:17.665","Text":"2Bt plus C. I want the 2 here variant of the general quadratic."},{"Start":"02:17.665 ","End":"02:20.320","Text":"We said that for all t,"},{"Start":"02:20.320 ","End":"02:22.815","Text":"p of t is bigger or equal to 0."},{"Start":"02:22.815 ","End":"02:28.580","Text":"Now, we know from quadratic functions that the quadratic is bigger or equal to"},{"Start":"02:28.580 ","End":"02:34.900","Text":"0 everywhere if and only if the discriminant is less than or equal to 0."},{"Start":"02:34.900 ","End":"02:36.650","Text":"In the case that we have a 2 here,"},{"Start":"02:36.650 ","End":"02:39.160","Text":"the discriminant is without the 4 here,"},{"Start":"02:39.160 ","End":"02:41.315","Text":"it\u0027s just B squared minus AC."},{"Start":"02:41.315 ","End":"02:44.690","Text":"Well, you could think of it as 4B squared minus 4AC."},{"Start":"02:44.690 ","End":"02:46.550","Text":"The 4 is a positive constant,"},{"Start":"02:46.550 ","End":"02:47.975","Text":"so we can throw it out."},{"Start":"02:47.975 ","End":"02:50.915","Text":"We get that Delta is less than or equal to 0,"},{"Start":"02:50.915 ","End":"02:53.870","Text":"meaning that this is less than or equal to 0,"},{"Start":"02:53.870 ","End":"02:56.725","Text":"so B squared is less than or equal to AC."},{"Start":"02:56.725 ","End":"02:59.090","Text":"We said that A and C are non-negative,"},{"Start":"02:59.090 ","End":"03:03.440","Text":"so we can take the square root of both sides and don\u0027t forget that"},{"Start":"03:03.440 ","End":"03:06.140","Text":"the square root of B squared is absolute value"},{"Start":"03:06.140 ","End":"03:09.120","Text":"of B less than or equal to square root of A,"},{"Start":"03:09.120 ","End":"03:15.935","Text":"square root of C. Now we just have to substitute back what B is and what A and C are,"},{"Start":"03:15.935 ","End":"03:17.735","Text":"and we get this,"},{"Start":"03:17.735 ","End":"03:20.750","Text":"which is what we had to show in the first place."},{"Start":"03:20.750 ","End":"03:23.425","Text":"So that\u0027s part a."},{"Start":"03:23.425 ","End":"03:25.505","Text":"Now on to part b."},{"Start":"03:25.505 ","End":"03:32.945","Text":"Similar idea that function p of t be the integral of tf plus g,"},{"Start":"03:32.945 ","End":"03:37.400","Text":"where now because the integrant is non-negative,"},{"Start":"03:37.400 ","End":"03:39.950","Text":"I mean it\u0027s something squared, it\u0027s always non-negative,"},{"Start":"03:39.950 ","End":"03:43.405","Text":"then the integral will also be non-negative."},{"Start":"03:43.405 ","End":"03:49.385","Text":"Expanding this and separating the coefficients of powers of t,"},{"Start":"03:49.385 ","End":"03:54.280","Text":"we get again a quadratic expression in t, like so."},{"Start":"03:54.280 ","End":"03:56.065","Text":"As before, this comes out,"},{"Start":"03:56.065 ","End":"03:58.590","Text":"call this A, call this B, call this C,"},{"Start":"03:58.590 ","End":"04:03.390","Text":"we got At squared plus 2Bt plus C. Just as before,"},{"Start":"04:03.390 ","End":"04:05.865","Text":"this is always non-negative."},{"Start":"04:05.865 ","End":"04:11.930","Text":"We get the same conclusion that absolute value of b is less than or equal to root A,"},{"Start":"04:11.930 ","End":"04:17.975","Text":"root C and substituting back A and C and B as above,"},{"Start":"04:17.975 ","End":"04:20.660","Text":"we get this inequality here,"},{"Start":"04:20.660 ","End":"04:22.415","Text":"which is what we had to show."},{"Start":"04:22.415 ","End":"04:25.560","Text":"That completes part b and we\u0027re done."}],"ID":24755},{"Watched":false,"Name":"Exercise 17","Duration":"4m 3s","ChapterTopicVideoID":23814,"CourseChapterTopicPlaylistID":257169,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23814.jpeg","UploadDate":"2021-01-11T09:22:50.8270000","DurationForVideoObject":"PT4M3S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.875","Text":"In this exercise, f is a real valued function on the interval a, b and it\u0027s integrable."},{"Start":"00:06.875 ","End":"00:15.465","Text":"Now, we make a new function by changing the values of f at a finite number of points."},{"Start":"00:15.465 ","End":"00:21.425","Text":"We have to show that the modified function is integrable."},{"Start":"00:21.425 ","End":"00:23.835","Text":"As a matter of fact, you can even say more"},{"Start":"00:23.835 ","End":"00:27.606","Text":"that the modified function has the same integral as f."},{"Start":"00:27.606 ","End":"00:30.600","Text":"Now we can actually simplify this problem."},{"Start":"00:30.600 ","End":"00:33.810","Text":"Instead of a finite number, we can change it at"},{"Start":"00:33.810 ","End":"00:38.385","Text":"just one point because we can do this by induction."},{"Start":"00:38.385 ","End":"00:42.160","Text":"If we change the points one at a time,"},{"Start":"00:42.160 ","End":"00:45.030","Text":"then we can change a finite number of points one at a time"},{"Start":"00:45.030 ","End":"00:49.520","Text":"and each time, the integrability doesn\u0027t change."},{"Start":"00:49.520 ","End":"00:51.515","Text":"We\u0027ll just do it for one point."},{"Start":"00:51.515 ","End":"00:56.100","Text":"We have a new function g which is the same as f"},{"Start":"00:56.100 ","End":"01:00.168","Text":"except it has a single point c in the interval"},{"Start":"01:00.168 ","End":"01:02.283","Text":"where g is different from f."},{"Start":"01:02.283 ","End":"01:07.390","Text":"Let\u0027s let h be the difference of g and f, g minus f."},{"Start":"01:07.390 ","End":"01:12.011","Text":"Then h is 0 except at a single point where x is c."},{"Start":"01:12.011 ","End":"01:18.605","Text":"Let\u0027s take a partition of a, b into n equal parts,"},{"Start":"01:18.605 ","End":"01:22.235","Text":"each with width b minus a over n."},{"Start":"01:22.235 ","End":"01:27.430","Text":"As usual, we\u0027ll let M_i be the supremum of h"},{"Start":"01:27.430 ","End":"01:31.310","Text":"in the interval from x_i minus 1 to x_i"},{"Start":"01:31.310 ","End":"01:36.410","Text":"and m_i is the infimum on that interval."},{"Start":"01:36.410 ","End":"01:41.720","Text":"This point c has to belong to at least one of these intervals,"},{"Start":"01:41.720 ","End":"01:43.130","Text":"x_i minus 1, x_i."},{"Start":"01:43.130 ","End":"01:48.565","Text":"I claim that it\u0027s exactly either one or two of these intervals."},{"Start":"01:48.565 ","End":"01:51.770","Text":"In other words, there are either one or two values of i"},{"Start":"01:51.770 ","End":"01:54.830","Text":"such that c belongs to x_i minus 1, x_i."},{"Start":"01:54.830 ","End":"01:57.500","Text":"Well, the way that can happen is if c falls"},{"Start":"01:57.500 ","End":"02:00.845","Text":"exactly on one of these and not one of the end ones."},{"Start":"02:00.845 ","End":"02:05.165","Text":"It\u0027s going to be in two intervals if, and only if, c is x_k,"},{"Start":"02:05.165 ","End":"02:09.315","Text":"for some k not including 0 and n."},{"Start":"02:09.315 ","End":"02:13.748","Text":"In this case, it belongs to this interval for both"},{"Start":"02:13.748 ","End":"02:16.145","Text":"i equals k plus 1 and i equals k."},{"Start":"02:16.145 ","End":"02:19.010","Text":"If i equals k plus 1, it\u0027s the left endpoint,"},{"Start":"02:19.010 ","End":"02:23.155","Text":"and if i equals k, then it\u0027s the right endpoint."},{"Start":"02:23.155 ","End":"02:27.870","Text":"Also notice that if c is in the interval x_i minus 1, x_i,"},{"Start":"02:27.870 ","End":"02:35.837","Text":"then M_i minus m_i is equal to the absolute value of h of c."},{"Start":"02:35.837 ","End":"02:42.995","Text":"This expression is either 0 minus h of c or h of c minus 0."},{"Start":"02:42.995 ","End":"02:47.405","Text":"The maximum and the minimum has to either be 0 or h of c."},{"Start":"02:47.405 ","End":"02:51.710","Text":"If c is not in the interval, then everything is 0,"},{"Start":"02:51.710 ","End":"02:55.375","Text":"so M_i minus m_i is 0."},{"Start":"02:55.375 ","End":"02:59.810","Text":"If we take the sum from 1 to n of this expression,"},{"Start":"02:59.810 ","End":"03:04.760","Text":"then this will equal 1 or 2 here times b minus a over n,"},{"Start":"03:04.760 ","End":"03:06.035","Text":"absolute value of f of c."},{"Start":"03:06.035 ","End":"03:10.370","Text":"To be safe, we\u0027ll just put less than or equal 2 and a 2."},{"Start":"03:10.370 ","End":"03:16.045","Text":"Now this will be less than epsilon, provided we choose n big enough."},{"Start":"03:16.045 ","End":"03:19.819","Text":"If this is less than an arbitrary epsilon,"},{"Start":"03:19.819 ","End":"03:22.330","Text":"that means that h is integrable."},{"Start":"03:22.330 ","End":"03:28.100","Text":"If h is integrable and we know that f is integrable by the given,"},{"Start":"03:28.100 ","End":"03:31.850","Text":"then the modified function g is equal to f plus h."},{"Start":"03:31.850 ","End":"03:34.470","Text":"The sum of two integral functions is integrable,"},{"Start":"03:34.470 ","End":"03:36.315","Text":"so g is integrable."},{"Start":"03:36.315 ","End":"03:39.980","Text":"We can actually say more, and this is not part of the question,"},{"Start":"03:39.980 ","End":"03:42.140","Text":"but it\u0027s easy to see that"},{"Start":"03:42.140 ","End":"03:45.905","Text":"if you just look at the upper sum separately and the lower sum separately"},{"Start":"03:45.905 ","End":"03:49.024","Text":"and each of them an absolute value is less than epsilon,"},{"Start":"03:49.024 ","End":"03:51.335","Text":"so the integral of h is 0,"},{"Start":"03:51.335 ","End":"03:53.760","Text":"which means that g and f have the same integral."},{"Start":"03:53.760 ","End":"03:56.875","Text":"So changing a function at a finite number of points"},{"Start":"03:56.875 ","End":"03:58.730","Text":"doesn\u0027t change the integrability"},{"Start":"03:58.730 ","End":"04:01.796","Text":"and doesn\u0027t change the value of the integral."},{"Start":"04:01.796 ","End":"04:03.840","Text":"We\u0027re done."}],"ID":24756}],"Thumbnail":null,"ID":257169},{"Name":"Advanced exercises - Riemann Sum and FTC","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"MVT for Integrals","Duration":"12m 32s","ChapterTopicVideoID":23815,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23815.jpeg","UploadDate":"2021-01-11T09:39:21.0170000","DurationForVideoObject":"PT12M32S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.800","Text":"In this clip, we\u0027ll learn about the first mean value theorem for integrals,"},{"Start":"00:04.800 ","End":"00:08.160","Text":"and there\u0027s also a second mean value theorem for integrals."},{"Start":"00:08.160 ","End":"00:12.555","Text":"But first, I want to remind you of the regular mean value theorem."},{"Start":"00:12.555 ","End":"00:16.260","Text":"The familiar normal mean value theorem"},{"Start":"00:16.260 ","End":"00:19.420","Text":"says that if f is continuous on the closed interval a,"},{"Start":"00:19.420 ","End":"00:22.215","Text":"b and it\u0027s differentiable on the open interval,"},{"Start":"00:22.215 ","End":"00:27.690","Text":"then there is some point in the open interval such that f of b minus f"},{"Start":"00:27.690 ","End":"00:33.120","Text":"of a is equal to f prime of c times b minus a,"},{"Start":"00:33.120 ","End":"00:36.345","Text":"where c is that point that\u0027s guaranteed to exist."},{"Start":"00:36.345 ","End":"00:43.250","Text":"Could be more than 1. Now, first, mean value theorem for integrals."},{"Start":"00:43.250 ","End":"00:46.820","Text":"We\u0027re given a continuous f on the closed interval a,"},{"Start":"00:46.820 ","End":"00:53.060","Text":"b, and we\u0027re guaranteed that there is some point c in the open interval,"},{"Start":"00:53.060 ","End":"00:55.325","Text":"i.e., strictly between a and b,"},{"Start":"00:55.325 ","End":"01:00.620","Text":"such that the integral of f on this interval is equal"},{"Start":"01:00.620 ","End":"01:07.790","Text":"to f at the point c times the width of the interval b minus a."},{"Start":"01:07.790 ","End":"01:14.440","Text":"The geometric meaning means that the area under the curve here,"},{"Start":"01:14.440 ","End":"01:19.685","Text":"if you make a rectangle with the same area on the same base,"},{"Start":"01:19.685 ","End":"01:29.410","Text":"then the height of this rectangle is the f of c that was guaranteed here."},{"Start":"01:30.440 ","End":"01:36.140","Text":"In fact, what seems intuitive is that f of c is"},{"Start":"01:36.140 ","End":"01:43.130","Text":"called the average value of f on the interval mean value."},{"Start":"01:43.130 ","End":"01:47.089","Text":"It\u0027s like the average of the values of the function."},{"Start":"01:47.089 ","End":"01:51.320","Text":"The proof is very short and it\u0027s like an exercise."},{"Start":"01:51.320 ","End":"01:53.840","Text":"I\u0027m going to show you that here."},{"Start":"01:53.840 ","End":"01:56.930","Text":"You can skip that if you feel you don\u0027t need it."},{"Start":"01:56.930 ","End":"02:00.455","Text":"We defined big F on the same interval."},{"Start":"02:00.455 ","End":"02:03.395","Text":"By big F of x, we often do this,"},{"Start":"02:03.395 ","End":"02:07.315","Text":"it\u0027s the integral from a to x."},{"Start":"02:07.315 ","End":"02:11.800","Text":"We need to change the variable to t of f of t dt."},{"Start":"02:13.030 ","End":"02:17.970","Text":"Of course, the big F prime"},{"Start":"02:17.970 ","End":"02:22.085","Text":"is equal to little f by the first fundamental theorem of the calculus."},{"Start":"02:22.085 ","End":"02:24.050","Text":"We\u0027ll use that in a moment."},{"Start":"02:24.050 ","End":"02:28.130","Text":"Now by the regular mean value theorem which I just showed you,"},{"Start":"02:28.130 ","End":"02:31.010","Text":"there is some point c in the open interval a,"},{"Start":"02:31.010 ","End":"02:39.060","Text":"b, such that b minus a times big F prime of c is big F of b minus big F of a."},{"Start":"02:39.620 ","End":"02:42.885","Text":"Like I said, big F prime equals"},{"Start":"02:42.885 ","End":"02:51.320","Text":"little f. We define big F on the same interval a,"},{"Start":"02:51.320 ","End":"02:58.500","Text":"b as follows; big F of x is the integral from a to x of f of t dt."},{"Start":"02:58.760 ","End":"03:01.560","Text":"By the regular mean value theorem,"},{"Start":"03:01.560 ","End":"03:03.180","Text":"which I just showed you,"},{"Start":"03:03.180 ","End":"03:05.090","Text":"but we\u0027ll use big F instead."},{"Start":"03:05.090 ","End":"03:06.860","Text":"We had little f there."},{"Start":"03:06.860 ","End":"03:09.829","Text":"There is some c in the open interval,"},{"Start":"03:09.829 ","End":"03:16.860","Text":"such that b minus a big F prime of c is f of b minus f of a."},{"Start":"03:22.880 ","End":"03:26.500","Text":"By the first fundamental theorem of the calculus,"},{"Start":"03:26.500 ","End":"03:31.010","Text":"the derivative of big F is little f. Whenever we define big F this way,"},{"Start":"03:31.010 ","End":"03:36.090","Text":"its derivative is the original integrand."},{"Start":"03:36.090 ","End":"03:39.014","Text":"What we get here,"},{"Start":"03:39.014 ","End":"03:43.575","Text":"just replace f prime by little f,"},{"Start":"03:43.575 ","End":"03:49.520","Text":"we get this, and replacing big F by its definition here and here,"},{"Start":"03:49.520 ","End":"03:52.545","Text":"so we get these 2 integrals."},{"Start":"03:52.545 ","End":"03:54.350","Text":"But the second one is 0,"},{"Start":"03:54.350 ","End":"03:56.825","Text":"it\u0027s the integral from a to a."},{"Start":"03:56.825 ","End":"03:59.675","Text":"This is just equal to this."},{"Start":"03:59.675 ","End":"04:01.655","Text":"That\u0027s all we have to show,"},{"Start":"04:01.655 ","End":"04:06.110","Text":"except that we should replace the dummy t back to x."},{"Start":"04:06.110 ","End":"04:08.645","Text":"Well, it doesn\u0027t matter what variable here."},{"Start":"04:08.645 ","End":"04:12.890","Text":"That proves the first mean value theorem for integrals."},{"Start":"04:12.890 ","End":"04:16.770","Text":"Let\u0027s also do an example exercise."},{"Start":"04:16.880 ","End":"04:20.245","Text":"Here\u0027s the example."},{"Start":"04:20.245 ","End":"04:24.795","Text":"We\u0027re asked to find the c and f of c,"},{"Start":"04:24.795 ","End":"04:27.620","Text":"although there might be more than one c. As in"},{"Start":"04:27.620 ","End":"04:30.335","Text":"the first mean value theorem for integrals above,"},{"Start":"04:30.335 ","End":"04:37.670","Text":"for the function f of x equals 4x minus x squared on the interval 1, 4."},{"Start":"04:37.670 ","End":"04:40.010","Text":"We need a function and an interval."},{"Start":"04:40.010 ","End":"04:43.220","Text":"Of course, it goes without saying that f is continuous."},{"Start":"04:43.220 ","End":"04:45.600","Text":"I mean, it\u0027s a quadratic."},{"Start":"04:45.920 ","End":"04:48.180","Text":"In case a picture helps,"},{"Start":"04:48.180 ","End":"04:51.420","Text":"this is the upside-down parabola,"},{"Start":"04:51.420 ","End":"04:53.475","Text":"4x minus x squared,"},{"Start":"04:53.475 ","End":"04:57.040","Text":"shown between 1 and 4."},{"Start":"04:58.670 ","End":"05:07.595","Text":"The mean value theorem says that f of c times b minus a is 4 minus 1,"},{"Start":"05:07.595 ","End":"05:13.025","Text":"equals the integral of f of x dx, on that interval."},{"Start":"05:13.025 ","End":"05:18.825","Text":"We can compute everything except f of c. Well,"},{"Start":"05:18.825 ","End":"05:20.355","Text":"to the right-hand side,"},{"Start":"05:20.355 ","End":"05:21.920","Text":"the integral is this,"},{"Start":"05:21.920 ","End":"05:25.790","Text":"the indefinite integral, substitute 4 and we get 30."},{"Start":"05:25.790 ","End":"05:27.275","Text":"I\u0027ll leave you to check that."},{"Start":"05:27.275 ","End":"05:30.115","Text":"Substitute 1 and we get 21."},{"Start":"05:30.115 ","End":"05:32.070","Text":"Now, 4 minus 1 is 3,"},{"Start":"05:32.070 ","End":"05:34.875","Text":"so we get 3f of c is 9,"},{"Start":"05:34.875 ","End":"05:38.200","Text":"and so f of c is 3."},{"Start":"05:38.200 ","End":"05:43.050","Text":"This is the average value of the function."},{"Start":"05:43.490 ","End":"05:46.730","Text":"Here\u0027s a picture of that."},{"Start":"05:46.730 ","End":"05:49.370","Text":"This is at height 3,"},{"Start":"05:49.370 ","End":"05:51.230","Text":"3 is the average value."},{"Start":"05:51.230 ","End":"05:53.739","Text":"What we need to do now is find c,"},{"Start":"05:53.739 ","End":"05:58.230","Text":"which will be the point below here."},{"Start":"05:58.230 ","End":"06:00.440","Text":"It\u0027s just an equation to solve."},{"Start":"06:00.440 ","End":"06:02.420","Text":"We know what f of x is,"},{"Start":"06:02.420 ","End":"06:04.190","Text":"it\u0027s 4x minus x squared,"},{"Start":"06:04.190 ","End":"06:06.255","Text":"just use c instead."},{"Start":"06:06.255 ","End":"06:08.775","Text":"We get a quadratic equation."},{"Start":"06:08.775 ","End":"06:10.845","Text":"I won\u0027t spend time-solving this."},{"Start":"06:10.845 ","End":"06:12.180","Text":"You\u0027ll probably do this in your head."},{"Start":"06:12.180 ","End":"06:16.380","Text":"There\u0027s 2 solutions where c is 1 or 3,"},{"Start":"06:16.380 ","End":"06:20.310","Text":"but the mean value theorem says"},{"Start":"06:20.310 ","End":"06:25.470","Text":"that c is strictly between the 2 endpoints between 1 and 4,"},{"Start":"06:25.470 ","End":"06:29.400","Text":"so we can rule this 1 out and just c equals 3 as a solution."},{"Start":"06:29.400 ","End":"06:31.035","Text":"We found f of c, we found c,"},{"Start":"06:31.035 ","End":"06:33.335","Text":"that\u0027s this exercise done."},{"Start":"06:33.335 ","End":"06:37.460","Text":"Now, we come to the second mean value theorem for integrals."},{"Start":"06:37.460 ","End":"06:39.890","Text":"I have to say that this is not standard."},{"Start":"06:39.890 ","End":"06:44.615","Text":"There\u0027s several variations, some of them even quite different looking,"},{"Start":"06:44.615 ","End":"06:48.095","Text":"and they\u0027re all called the second mean value theorem for integrals."},{"Start":"06:48.095 ","End":"06:50.380","Text":"It depends on the book,"},{"Start":"06:50.380 ","End":"06:52.430","Text":"the professor, or the website you visit."},{"Start":"06:52.430 ","End":"06:54.695","Text":"Anyway, this is the version we\u0027ll be using."},{"Start":"06:54.695 ","End":"06:56.975","Text":"In this, we have 2 functions, f and g,"},{"Start":"06:56.975 ","End":"06:59.330","Text":"f is required to be continuous,"},{"Start":"06:59.330 ","End":"07:04.905","Text":"and g just has to be integrable on our closed interval a, b."},{"Start":"07:04.905 ","End":"07:08.865","Text":"We want g to be non-negative."},{"Start":"07:08.865 ","End":"07:12.540","Text":"Although it will also work for non-positive."},{"Start":"07:12.540 ","End":"07:16.640","Text":"We\u0027ll just assume that it\u0027s non-negative or positive."},{"Start":"07:16.640 ","End":"07:22.580","Text":"Under these circumstances, there exists a point c in the closed interval,"},{"Start":"07:22.580 ","End":"07:25.280","Text":"a, b. Actually,"},{"Start":"07:25.280 ","End":"07:27.110","Text":"you can restrict it to the open interval."},{"Start":"07:27.110 ","End":"07:28.670","Text":"Anyway, let\u0027s leave it like this."},{"Start":"07:28.670 ","End":"07:36.590","Text":"Such that the integral of f times g is equal to f at some point times the"},{"Start":"07:36.590 ","End":"07:40.505","Text":"integral of just g. F of c"},{"Start":"07:40.505 ","End":"07:44.810","Text":"is now no longer just the average of f because it depends on g,"},{"Start":"07:44.810 ","End":"07:50.435","Text":"so it\u0027s called the g-weighted average of f on the interval."},{"Start":"07:50.435 ","End":"07:53.280","Text":"The proof is not very difficult,"},{"Start":"07:53.280 ","End":"07:54.630","Text":"so let\u0027s do it."},{"Start":"07:54.630 ","End":"07:58.340","Text":"We\u0027ll use the extreme value theorem on f. Remember,"},{"Start":"07:58.340 ","End":"08:00.575","Text":"f is continuous, so we can do that."},{"Start":"08:00.575 ","End":"08:05.935","Text":"It attains a minimum little m and a maximum big M on the interval."},{"Start":"08:05.935 ","End":"08:08.160","Text":"It attains them at some points,"},{"Start":"08:08.160 ","End":"08:11.100","Text":"I don\u0027t know, x_1 and x_2, let\u0027s say."},{"Start":"08:11.100 ","End":"08:20.515","Text":"Of course, any value of f of x in the interval is between the minimum and the maximum."},{"Start":"08:20.515 ","End":"08:25.100","Text":"Then we can multiply out by g of x."},{"Start":"08:25.100 ","End":"08:29.750","Text":"Since g of x is non-negative and preserve the direction of the inequality."},{"Start":"08:29.750 ","End":"08:32.764","Text":"Now, let\u0027s take the integral of all these."},{"Start":"08:32.764 ","End":"08:37.055","Text":"But I also took the constant out here and here."},{"Start":"08:37.055 ","End":"08:40.580","Text":"I\u0027d like to divide everything by the integral of g,"},{"Start":"08:40.580 ","End":"08:41.930","Text":"but that could be 0,"},{"Start":"08:41.930 ","End":"08:43.864","Text":"so we\u0027ll split up into cases."},{"Start":"08:43.864 ","End":"08:47.255","Text":"First, the case where this integral is 0."},{"Start":"08:47.255 ","End":"08:53.120","Text":"Then we have here that m times 0 is less than or equal to this integral,"},{"Start":"08:53.120 ","End":"08:55.070","Text":"which is less than or equal to big M times 0,"},{"Start":"08:55.070 ","End":"08:56.660","Text":"and so this is 0, and this is 0,"},{"Start":"08:56.660 ","End":"08:59.180","Text":"so this also has to be 0."},{"Start":"08:59.180 ","End":"09:03.890","Text":"But if this is 0, then we automatically get"},{"Start":"09:03.890 ","End":"09:09.350","Text":"this equality that this equals this because this is 0 times f of c is also 0."},{"Start":"09:09.350 ","End":"09:12.085","Text":"We can actually choose any c on the interval,"},{"Start":"09:12.085 ","End":"09:14.930","Text":"and it will work because 0 equals 0."},{"Start":"09:14.930 ","End":"09:20.630","Text":"Let\u0027s take the more interesting case where the integral of g is not 0."},{"Start":"09:20.630 ","End":"09:24.050","Text":"In that case, we can do the division we wanted,"},{"Start":"09:24.050 ","End":"09:28.160","Text":"and get that this integral over this integral is between little m and"},{"Start":"09:28.160 ","End":"09:32.600","Text":"big M. Since f is continuous on the interval,"},{"Start":"09:32.600 ","End":"09:35.485","Text":"we can use the intermediate value property."},{"Start":"09:35.485 ","End":"09:37.685","Text":"Should have said IVP, never mind."},{"Start":"09:37.685 ","End":"09:42.590","Text":"By this, there exists some point in the interval a, b."},{"Start":"09:42.590 ","End":"09:46.190","Text":"We can actually narrow it down to between x_1 and x_2,"},{"Start":"09:46.190 ","End":"09:48.620","Text":"that\u0027s where the minimum, maximum are achieved,"},{"Start":"09:48.620 ","End":"09:50.975","Text":"but in any event, it\u0027s in the big interval."},{"Start":"09:50.975 ","End":"09:54.950","Text":"We have this c such that f of c is equal to this value."},{"Start":"09:54.950 ","End":"10:01.940","Text":"This is just a number between m and M. Now multiply both sides by that denominator,"},{"Start":"10:01.940 ","End":"10:06.230","Text":"and we get that this integral here"},{"Start":"10:06.230 ","End":"10:11.125","Text":"is equal to f of c times integral of g. That\u0027s exactly what we had to show."},{"Start":"10:11.125 ","End":"10:13.845","Text":"So that concludes the proof."},{"Start":"10:13.845 ","End":"10:20.330","Text":"Now, I\u0027ll give an example of the use of the second MVT, and here it is."},{"Start":"10:20.330 ","End":"10:25.620","Text":"We have to show that this integral of x over natural log of x,"},{"Start":"10:25.620 ","End":"10:29.675","Text":"from 2-4, is between these 2 bounds."},{"Start":"10:29.675 ","End":"10:32.420","Text":"This is not something you could easily compute."},{"Start":"10:32.420 ","End":"10:35.605","Text":"At least we can get an estimate."},{"Start":"10:35.605 ","End":"10:38.190","Text":"We want to use the theorem,"},{"Start":"10:38.190 ","End":"10:40.684","Text":"so we need to define f and g,"},{"Start":"10:40.684 ","End":"10:43.460","Text":"and the interval\u0027s going to be 2, 4."},{"Start":"10:43.460 ","End":"10:47.490","Text":"We just have to write this as f times g in the right way to do it,"},{"Start":"10:47.490 ","End":"10:51.980","Text":"so that f be 1 over natural log of x and g of x equals x."},{"Start":"10:51.980 ","End":"10:53.525","Text":"The other way round won\u0027t work."},{"Start":"10:53.525 ","End":"10:56.510","Text":"For example, of c integral of 1 over natural log of x."},{"Start":"10:56.510 ","End":"10:59.005","Text":"Anyway, this is the way it\u0027ll work."},{"Start":"10:59.005 ","End":"11:02.485","Text":"They\u0027re both continuous and both integrable,"},{"Start":"11:02.485 ","End":"11:05.960","Text":"and g is non-negative."},{"Start":"11:05.960 ","End":"11:07.460","Text":"What is g?"},{"Start":"11:07.460 ","End":"11:11.790","Text":"It\u0027s equal to x, but x is non-negative from 2-4."},{"Start":"11:11.790 ","End":"11:16.115","Text":"It\u0027s even positive. We can apply the second mean value theorem for integrals."},{"Start":"11:16.115 ","End":"11:18.620","Text":"What we get is that there exists some point c in"},{"Start":"11:18.620 ","End":"11:22.520","Text":"the interval such that the integral is equal to,"},{"Start":"11:22.520 ","End":"11:25.675","Text":"this is f of c,"},{"Start":"11:25.675 ","End":"11:29.540","Text":"and this part is the integral of g of x, dx."},{"Start":"11:29.540 ","End":"11:33.125","Text":"We can compute the integral and we can estimate"},{"Start":"11:33.125 ","End":"11:37.980","Text":"the 1 over natural log of c. So computing."},{"Start":"11:37.980 ","End":"11:40.770","Text":"Well, I\u0027ll let you look at it. It\u0027s trivial."},{"Start":"11:40.770 ","End":"11:42.825","Text":"Comes out to be 6."},{"Start":"11:42.825 ","End":"11:48.155","Text":"The natural log is a monotonic function and so is its reciprocal."},{"Start":"11:48.155 ","End":"11:50.315","Text":"If c is between 2 and 4,"},{"Start":"11:50.315 ","End":"11:54.455","Text":"natural log of c reciprocal is between these 2."},{"Start":"11:54.455 ","End":"11:56.365","Text":"This is the biggest,"},{"Start":"11:56.365 ","End":"11:58.200","Text":"and this is the smallest."},{"Start":"11:58.200 ","End":"12:02.810","Text":"Let\u0027s just reverse the order and also rewrite"},{"Start":"12:02.810 ","End":"12:07.460","Text":"natural log of 4 as 2 natural log of 2. We get this."},{"Start":"12:07.460 ","End":"12:11.375","Text":"I want to combine this and this and insert them here."},{"Start":"12:11.375 ","End":"12:19.820","Text":"We get that 1 over natural log of 2 times the 6 is less than or equal to our integral,"},{"Start":"12:19.820 ","End":"12:24.380","Text":"which is less than or equal to 1 over natural log of 2 times 6."},{"Start":"12:24.380 ","End":"12:26.120","Text":"We can tidy it up a bit."},{"Start":"12:26.120 ","End":"12:28.760","Text":"6 over 2 is 3, we write it like this,"},{"Start":"12:28.760 ","End":"12:29.885","Text":"and that\u0027s what we have to show,"},{"Start":"12:29.885 ","End":"12:33.030","Text":"and we\u0027re done with the whole clip."}],"ID":24757},{"Watched":false,"Name":"Exercise 1","Duration":"3m 27s","ChapterTopicVideoID":23816,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23816.jpeg","UploadDate":"2021-01-11T09:42:20.1530000","DurationForVideoObject":"PT3M27S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.160","Text":"In this exercise, we have 2 parts."},{"Start":"00:02.160 ","End":"00:08.430","Text":"Part a asks us to show that every continuous function on a closed and bounded interval,"},{"Start":"00:08.430 ","End":"00:11.580","Text":"is a derivative of some function."},{"Start":"00:11.580 ","End":"00:16.500","Text":"In part b, if we change continuous to integrable,"},{"Start":"00:16.500 ","End":"00:18.690","Text":"then it\u0027s no longer so."},{"Start":"00:18.690 ","End":"00:21.120","Text":"At least it might be so, but not necessarily,"},{"Start":"00:21.120 ","End":"00:22.695","Text":"then we\u0027ll start with part a,"},{"Start":"00:22.695 ","End":"00:24.210","Text":"where f is continuous,"},{"Start":"00:24.210 ","End":"00:27.345","Text":"close bounded interval means something like a,b,"},{"Start":"00:27.345 ","End":"00:32.235","Text":"we need to show that f is F Prime for some big F,"},{"Start":"00:32.235 ","End":"00:35.205","Text":"which is defined as a function on a, b,"},{"Start":"00:35.205 ","End":"00:38.760","Text":"f is integrable on any sub-interval of a,"},{"Start":"00:38.760 ","End":"00:40.575","Text":"b, say on a, x,"},{"Start":"00:40.575 ","End":"00:44.655","Text":"because f is continuous also on this,"},{"Start":"00:44.655 ","End":"00:49.865","Text":"and so we can define big F as the integral from a-x,"},{"Start":"00:49.865 ","End":"00:51.680","Text":"of f of tdt."},{"Start":"00:51.680 ","End":"00:54.470","Text":"Because little f is continuous,"},{"Start":"00:54.470 ","End":"00:59.675","Text":"we can say that big F from the fundamental theorem of calculus,"},{"Start":"00:59.675 ","End":"01:03.950","Text":"is differentiable and only the differentiable for the derivative of"},{"Start":"01:03.950 ","End":"01:09.005","Text":"big F is little f. Now I\u0027ll remind you of this fundamental theorem."},{"Start":"01:09.005 ","End":"01:11.990","Text":"Here it is, you read it, but basically,"},{"Start":"01:11.990 ","End":"01:15.800","Text":"we\u0027re talking about part 2 that if f is continuous,"},{"Start":"01:15.800 ","End":"01:17.675","Text":"then the function big F,"},{"Start":"01:17.675 ","End":"01:22.970","Text":"which we defined this way is differentiable and F Prime equals f. Basically,"},{"Start":"01:22.970 ","End":"01:24.855","Text":"it\u0027s exactly what we need."},{"Start":"01:24.855 ","End":"01:27.140","Text":"That\u0027s part a. Now in part b,"},{"Start":"01:27.140 ","End":"01:29.270","Text":"we need the counterexample,"},{"Start":"01:29.270 ","End":"01:34.700","Text":"and we\u0027ll take the sine function, not S-I-N-E,"},{"Start":"01:34.700 ","End":"01:41.720","Text":"but S-I-G-N, the sign meaning plus or minus depending on positive or negative."},{"Start":"01:41.720 ","End":"01:45.050","Text":"This function, which is a step function,"},{"Start":"01:45.050 ","End":"01:47.330","Text":"actually, it\u0027s not continuous,"},{"Start":"01:47.330 ","End":"01:48.695","Text":"but it is integrable,"},{"Start":"01:48.695 ","End":"01:51.440","Text":"nevertheless because well, for many reasons,"},{"Start":"01:51.440 ","End":"01:53.780","Text":"but 1 reason you could say is it\u0027s monotone."},{"Start":"01:53.780 ","End":"01:55.820","Text":"It never decreases."},{"Start":"01:55.820 ","End":"01:58.145","Text":"We know that monotone implies integrable."},{"Start":"01:58.145 ","End":"02:02.710","Text":"Now, it doesn\u0027t have the intermediate value property."},{"Start":"02:02.710 ","End":"02:06.259","Text":"I\u0027ll show you why it doesn\u0027t have the intermediate value property."},{"Start":"02:06.259 ","End":"02:07.730","Text":"Let\u0027s take for example,"},{"Start":"02:07.730 ","End":"02:13.940","Text":"x equals 0 and x equals 1/2 and apply f,at both of these in 1 case we get 0,"},{"Start":"02:13.940 ","End":"02:16.250","Text":"and the other case we get 1."},{"Start":"02:16.250 ","End":"02:21.620","Text":"Now we take a number between 0 and 1, say 2/3s."},{"Start":"02:21.620 ","End":"02:24.650","Text":"If it did have the intermediate value property,"},{"Start":"02:24.650 ","End":"02:31.815","Text":"we could find some c in this interval with f of c equals 2/3s."},{"Start":"02:31.815 ","End":"02:37.880","Text":"But there isn\u0027t, there\u0027s no such c because in this interval between 0 and 0.5,"},{"Start":"02:37.880 ","End":"02:40.250","Text":"f of c is always equal to 1."},{"Start":"02:40.250 ","End":"02:43.380","Text":"What we\u0027ve done is the proof by contradiction really,"},{"Start":"02:43.380 ","End":"02:46.250","Text":"if the function was a derivative,"},{"Start":"02:46.250 ","End":"02:51.670","Text":"then it would have the IVP because of a previous exercise."},{"Start":"02:51.670 ","End":"02:55.670","Text":"In case you missed that exercise, here it is."},{"Start":"02:55.670 ","End":"02:59.165","Text":"Say if a function f is differentiable,"},{"Start":"02:59.165 ","End":"03:03.750","Text":"then f Prime has the intermediate value property, and in our case,"},{"Start":"03:03.750 ","End":"03:09.230","Text":"F is big F and f Prime is little f. In case you want to study this,"},{"Start":"03:09.230 ","End":"03:13.210","Text":"I\u0027ll just scroll to the end and there you are."},{"Start":"03:13.210 ","End":"03:17.760","Text":"Back to where we were. There I was saying proof by contradiction,"},{"Start":"03:17.760 ","End":"03:19.070","Text":"if it did have a derivative,"},{"Start":"03:19.070 ","End":"03:20.270","Text":"it would have the IVP,"},{"Start":"03:20.270 ","End":"03:22.835","Text":"and since it doesn\u0027t have the IVP,"},{"Start":"03:22.835 ","End":"03:25.430","Text":"then it\u0027s not a derivative."},{"Start":"03:25.430 ","End":"03:28.110","Text":"We are done."}],"ID":24758},{"Watched":false,"Name":"Exercise 2","Duration":"5m 19s","ChapterTopicVideoID":23817,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23817.jpeg","UploadDate":"2021-01-11T09:46:48.6130000","DurationForVideoObject":"PT5M19S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.100","Text":"In part A of this exercise,"},{"Start":"00:02.100 ","End":"00:09.120","Text":"we\u0027re given a real-valued function f on the interval of minus 1 to 1 and it\u0027s given by"},{"Start":"00:09.120 ","End":"00:12.600","Text":"the following formula that f is 0 when x is"},{"Start":"00:12.600 ","End":"00:17.670","Text":"negative and 1 otherwise and we\u0027re defining big F of x,"},{"Start":"00:17.670 ","End":"00:21.045","Text":"in terms of little f that\u0027s the following integral."},{"Start":"00:21.045 ","End":"00:25.740","Text":"We have to sketch the graphs of little f and big F,"},{"Start":"00:25.740 ","End":"00:28.570","Text":"I\u0027ll do that right away."},{"Start":"00:28.700 ","End":"00:32.355","Text":"This 1 is little f, this 1 is big F,"},{"Start":"00:32.355 ","End":"00:36.210","Text":"and we have to observe that little f is not"},{"Start":"00:36.210 ","End":"00:41.870","Text":"continuous at least 1.0 in fact you can see that now."},{"Start":"00:41.870 ","End":"00:44.015","Text":"There\u0027s a jump at 0."},{"Start":"00:44.015 ","End":"00:47.155","Text":"But big F is continuous."},{"Start":"00:47.155 ","End":"00:53.160","Text":"Still, big F is not differential at 0 because it\u0027s not smooth, it has a corner."},{"Start":"00:53.160 ","End":"00:56.490","Text":"Okay. We have basically done part a and part b,"},{"Start":"00:56.490 ","End":"01:00.480","Text":"we have to give an example of a different f from 1 minus 1,"},{"Start":"01:00.480 ","End":"01:06.090","Text":"1 to R such that f is not continuous at 0 as before."},{"Start":"01:06.090 ","End":"01:08.265","Text":"But this time big F,"},{"Start":"01:08.265 ","End":"01:12.165","Text":"the integral, is differentiable at 0."},{"Start":"01:12.165 ","End":"01:16.620","Text":"Well, part a we\u0027ve done already except I\u0027ll give you the formula for big F,"},{"Start":"01:16.620 ","End":"01:24.495","Text":"it\u0027s just basically the integral of 0 or the integral of 1 and this is the split value,"},{"Start":"01:24.495 ","End":"01:30.480","Text":"a bit definition of big F. Part b is the harder part,"},{"Start":"01:30.480 ","End":"01:36.990","Text":"it\u0027s not intuitive how 1 would come up with such a function f. How we define"},{"Start":"01:36.990 ","End":"01:44.505","Text":"it is that it\u0027s 0 except when x is 1 over some natural number."},{"Start":"01:44.505 ","End":"01:46.635","Text":"If x is 1/2, 1/3, 1/4,"},{"Start":"01:46.635 ","End":"01:52.410","Text":"and so on then it\u0027s going to be 1 and we have to show what it says here"},{"Start":"01:52.410 ","End":"02:00.040","Text":"that little f is no continuous at 0 but big F is differentiable at 0."},{"Start":"02:00.320 ","End":"02:05.175","Text":"The not continuous part we can do with limits and sequences,"},{"Start":"02:05.175 ","End":"02:07.470","Text":"let x and be 1 over n,"},{"Start":"02:07.470 ","End":"02:15.015","Text":"then x_n turns to 0 but f of x_n which is 1 goes to 1,"},{"Start":"02:15.015 ","End":"02:17.565","Text":"which is not f of 0,"},{"Start":"02:17.565 ","End":"02:23.640","Text":"so it\u0027s not continuous a 0 and let\u0027s show that it is differentiable."},{"Start":"02:23.640 ","End":"02:30.630","Text":"We\u0027ll do this by showing that the big F of x is the constant function 0."},{"Start":"02:30.630 ","End":"02:33.960","Text":"For all x, this integral comes out 0,"},{"Start":"02:33.960 ","End":"02:37.695","Text":"and if it\u0027s always 0, it\u0027s certainly differentiable at 0."},{"Start":"02:37.695 ","End":"02:39.870","Text":"This is part we have to show."},{"Start":"02:39.870 ","End":"02:44.670","Text":"Certainly, f of x is 0 for x less than or equal to"},{"Start":"02:44.670 ","End":"02:49.530","Text":"0 because there is no 1 over n in the negative,"},{"Start":"02:49.530 ","End":"02:56.340","Text":"so it\u0027s always equal to the integral of 0 on the negative then the integral of 0 is 0."},{"Start":"02:56.340 ","End":"02:59.655","Text":"We just have to consider positive x\u0027s."},{"Start":"02:59.655 ","End":"03:02.400","Text":"The integral for minus 1 up to x,"},{"Start":"03:02.400 ","End":"03:04.320","Text":"we can break it up into 2 parts."},{"Start":"03:04.320 ","End":"03:10.140","Text":"We can break it up into minus 1 up to 0 plus 0 to x."},{"Start":"03:10.140 ","End":"03:12.990","Text":"Since the minus 1 to 0 part is 0,"},{"Start":"03:12.990 ","End":"03:16.800","Text":"the only thing that contributes is the part from 0 to x."},{"Start":"03:16.800 ","End":"03:21.945","Text":"This part is certainly less than or equal to the integral from 0 to 1"},{"Start":"03:21.945 ","End":"03:27.255","Text":"of f of t dt because it\u0027s a non-negative function and we\u0027ve just extended the interval."},{"Start":"03:27.255 ","End":"03:29.760","Text":"So we\u0027ve got that this is less than or equal to this,"},{"Start":"03:29.760 ","End":"03:32.010","Text":"and if we show that this is 0,"},{"Start":"03:32.010 ","End":"03:33.915","Text":"then we\u0027re all done."},{"Start":"03:33.915 ","End":"03:37.920","Text":"Let\u0027s say we have some positive epsilon,"},{"Start":"03:37.920 ","End":"03:42.030","Text":"we can write the integral from 0 to 1 as the integral from"},{"Start":"03:42.030 ","End":"03:46.305","Text":"0 to epsilon plus the integral from epsilon to 1."},{"Start":"03:46.305 ","End":"03:49.020","Text":"Now, let\u0027s focus on this colored part,"},{"Start":"03:49.020 ","End":"03:51.270","Text":"I claim that this is 0."},{"Start":"03:51.270 ","End":"03:54.254","Text":"On the interval from epsilon to 1,"},{"Start":"03:54.254 ","End":"03:58.080","Text":"f of t is almost always 0 when I say,"},{"Start":"03:58.080 ","End":"04:02.490","Text":"almost always I mean except for a finite number of t because"},{"Start":"04:02.490 ","End":"04:08.040","Text":"the only place it\u0027s not 0 in this interval is when t is 1 over"},{"Start":"04:08.040 ","End":"04:12.750","Text":"n and that means that n is between 1 and 1 over"},{"Start":"04:12.750 ","End":"04:15.990","Text":"epsilon and there\u0027s only a finite number of"},{"Start":"04:15.990 ","End":"04:19.640","Text":"integers between 1 and any other positive number,"},{"Start":"04:19.640 ","End":"04:22.130","Text":"so that\u0027s a finite number."},{"Start":"04:22.130 ","End":"04:25.220","Text":"This integral is indeed 0."},{"Start":"04:25.220 ","End":"04:28.760","Text":"We have the inequality that the integral from 0 to"},{"Start":"04:28.760 ","End":"04:32.970","Text":"1 is like we said 0 to epsilon plus epsilon to 1,"},{"Start":"04:32.970 ","End":"04:38.180","Text":"that part is 0 so that\u0027s less than or equal to the integral from 0 to epsilon,"},{"Start":"04:38.180 ","End":"04:40.440","Text":"and I\u0027m replacing f by 1,"},{"Start":"04:40.440 ","End":"04:43.775","Text":"it can only grow because f is either 1 or 0."},{"Start":"04:43.775 ","End":"04:46.175","Text":"That\u0027s less than or equal to this,"},{"Start":"04:46.175 ","End":"04:49.070","Text":"this integral is equal to epsilon,"},{"Start":"04:49.070 ","End":"04:51.590","Text":"it\u0027s integral of 1 from 0 to epsilon."},{"Start":"04:51.590 ","End":"04:56.310","Text":"Note what we have is that 0 is less than or equal"},{"Start":"04:56.310 ","End":"05:01.815","Text":"to this which is less than or equal to epsilon."},{"Start":"05:01.815 ","End":"05:05.955","Text":"Now, if a constant is between 0 and epsilon,"},{"Start":"05:05.955 ","End":"05:08.190","Text":"for every epsilon that\u0027s positive,"},{"Start":"05:08.190 ","End":"05:14.160","Text":"we can let epsilon go to 0 and get that this integral must be 0."},{"Start":"05:14.160 ","End":"05:17.130","Text":"That\u0027s all that remained to be shown,"},{"Start":"05:17.130 ","End":"05:20.290","Text":"and so we are done."}],"ID":24759},{"Watched":false,"Name":"Exercise 3","Duration":"1m 7s","ChapterTopicVideoID":23818,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23818.jpeg","UploadDate":"2021-01-11T09:47:22.3170000","DurationForVideoObject":"PT1M7S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.010","Text":"In this exercise, f is integrable on the interval a,b."},{"Start":"00:05.010 ","End":"00:09.270","Text":"We have to show that the integral from a to b is equal to"},{"Start":"00:09.270 ","End":"00:13.590","Text":"the limit as x approaches b from the left of the integral"},{"Start":"00:13.590 ","End":"00:19.840","Text":"from a to x of f. What we\u0027ll do is, we\u0027ll define big F"},{"Start":"00:19.840 ","End":"00:28.065","Text":"on a,b by f equals the integral of little f from a to x."},{"Start":"00:28.065 ","End":"00:31.230","Text":"Now we apply the First Fundamental Theorem of"},{"Start":"00:31.230 ","End":"00:37.470","Text":"Calculus to get that f is continuous on a,b."},{"Start":"00:37.470 ","End":"00:39.240","Text":"Here we are to show you what the"},{"Start":"00:39.240 ","End":"00:42.339","Text":"First Fundamental Theorem of Calculus says."},{"Start":"00:42.339 ","End":"00:46.080","Text":"Well, we defined f the way we did in the exercise."},{"Start":"00:46.080 ","End":"00:48.710","Text":"Part 1 says that F is continuous,"},{"Start":"00:48.710 ","End":"00:52.055","Text":"big F. Now if big F is continuous,"},{"Start":"00:52.055 ","End":"00:55.100","Text":"then the limit as x approaches b of f of x"},{"Start":"00:55.100 ","End":"00:56.690","Text":"is going to be f of b."},{"Start":"00:56.690 ","End":"00:59.510","Text":"Now translate this from big F back to"},{"Start":"00:59.510 ","End":"01:03.860","Text":"little f and you see that this interprets to being this,"},{"Start":"01:03.860 ","End":"01:07.770","Text":"which is what we had to show and so we\u0027re done."}],"ID":24760},{"Watched":false,"Name":"Exercise 4","Duration":"3m 43s","ChapterTopicVideoID":23819,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23819.jpeg","UploadDate":"2021-01-11T09:49:34.4000000","DurationForVideoObject":"PT3M43S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.415","Text":"In this exercise, we have to prove the second fundamental theorem of the calculus"},{"Start":"00:05.415 ","End":"00:10.710","Text":"with a change where instead of the integrand being integrable,"},{"Start":"00:10.710 ","End":"00:15.510","Text":"we change that word to continuous and then it makes it easier to prove."},{"Start":"00:15.510 ","End":"00:17.790","Text":"This is what it says,"},{"Start":"00:17.790 ","End":"00:19.680","Text":"and I have it here,"},{"Start":"00:19.680 ","End":"00:22.470","Text":"the original second fundamental theorem of the calculus."},{"Start":"00:22.470 ","End":"00:24.750","Text":"You see the word here is integrable."},{"Start":"00:24.750 ","End":"00:27.735","Text":"Back here, we\u0027ve changed it to continuous."},{"Start":"00:27.735 ","End":"00:30.045","Text":"This is just what it says."},{"Start":"00:30.045 ","End":"00:33.105","Text":"In essence, what we have to show is the following."},{"Start":"00:33.105 ","End":"00:37.080","Text":"We have F which is differentiable and its derivative"},{"Start":"00:37.080 ","End":"00:42.085","Text":"with f. We also know that f is continuous."},{"Start":"00:42.085 ","End":"00:48.965","Text":"We need to show that the integral from a to b of f of x dx is F of b minus F of a."},{"Start":"00:48.965 ","End":"00:53.210","Text":"What we\u0027ll do is we\u0027ll define not f of x but g of"},{"Start":"00:53.210 ","End":"00:58.475","Text":"x to be the integral from a to x of f of t dt."},{"Start":"00:58.475 ","End":"01:01.550","Text":"Now, f is continuous."},{"Start":"01:01.550 ","End":"01:05.620","Text":"By the first fundamental theorem of the calculus"},{"Start":"01:05.620 ","End":"01:10.385","Text":"and here is a reminder of the first fundamental theorem of the calculus,"},{"Start":"01:10.385 ","End":"01:17.930","Text":"we get that G prime equals f. Now, we have here that F prime"},{"Start":"01:17.930 ","End":"01:25.035","Text":"equals f. Now, we have this and we have this."},{"Start":"01:25.035 ","End":"01:29.915","Text":"From these 2, what we can do is subtract and let"},{"Start":"01:29.915 ","End":"01:36.880","Text":"F minus G be H. Then we get that H prime is 0."},{"Start":"01:36.880 ","End":"01:42.950","Text":"The claim now is that H is constant on our interval."},{"Start":"01:42.950 ","End":"01:47.330","Text":"The way we show it\u0027s constant is take any 2 values of x,"},{"Start":"01:47.330 ","End":"01:48.530","Text":"well, we\u0027ll take them different,"},{"Start":"01:48.530 ","End":"01:56.345","Text":"otherwise, it\u0027s trivial and show that H of each of these is the same."},{"Start":"01:56.345 ","End":"01:59.360","Text":"We\u0027ll apply the mean value theorem."},{"Start":"01:59.360 ","End":"02:03.480","Text":"H is differentiable, I should say that since F and G are and"},{"Start":"02:03.480 ","End":"02:09.165","Text":"so it\u0027s H. There exists some c between x_1 and x_2,"},{"Start":"02:09.165 ","End":"02:16.605","Text":"such that H of x_2 minus H of x_1 is H prime of c times x_2 minus x_1."},{"Start":"02:16.605 ","End":"02:20.540","Text":"Now, H prime is 0 everywhere,"},{"Start":"02:20.540 ","End":"02:23.465","Text":"so H prime of c is 0,"},{"Start":"02:23.465 ","End":"02:28.070","Text":"and x_2 is not equal to x_1 from here so that"},{"Start":"02:28.070 ","End":"02:33.335","Text":"we can divide by it and get that H_2 of x_1 minus H of x_1 is 0,"},{"Start":"02:33.335 ","End":"02:38.410","Text":"which means that H of x_1 equals H of x_2."},{"Start":"02:38.410 ","End":"02:40.425","Text":"Since H is constant,"},{"Start":"02:40.425 ","End":"02:43.185","Text":"F minus G is a constant."},{"Start":"02:43.185 ","End":"02:49.985","Text":"For every x, F of x minus G of x is the same constant C. Now,"},{"Start":"02:49.985 ","End":"02:52.565","Text":"substitute x equals a here,"},{"Start":"02:52.565 ","End":"02:57.740","Text":"and we get that F of a minus G of a is c. But what is G of a?"},{"Start":"02:57.740 ","End":"03:00.860","Text":"G of a is 0,"},{"Start":"03:00.860 ","End":"03:03.080","Text":"so that\u0027s this 0 here."},{"Start":"03:03.080 ","End":"03:07.760","Text":"F of a is equal to c. Replacing that,"},{"Start":"03:07.760 ","End":"03:14.949","Text":"here, we get that F of x minus G of x is identically equal to F of a."},{"Start":"03:14.949 ","End":"03:22.235","Text":"Now, let x equals b and substitute that in here and we get F of b minus G of b, F of a."},{"Start":"03:22.235 ","End":"03:28.150","Text":"From here, we get that F of b minus F of a is G of b."},{"Start":"03:28.150 ","End":"03:31.760","Text":"But G of b, if we plug b in here,"},{"Start":"03:31.760 ","End":"03:35.585","Text":"is just the integral from a to b of F of x, dx."},{"Start":"03:35.585 ","End":"03:40.375","Text":"This is equal to this."},{"Start":"03:40.375 ","End":"03:44.400","Text":"That\u0027s what we had to show and we are done."}],"ID":24761},{"Watched":false,"Name":"Exercise 5","Duration":"3m ","ChapterTopicVideoID":23820,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23820.jpeg","UploadDate":"2021-01-11T09:50:45.6670000","DurationForVideoObject":"PT3M","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.880","Text":"In this exercise, we define a function f on the interval minus 1, 1 as follows."},{"Start":"00:05.880 ","End":"00:11.550","Text":"We let f of x be 0 for x is 0 and for x non-zero,"},{"Start":"00:11.550 ","End":"00:16.155","Text":"we define it by this formula which makes sense when x is not 0."},{"Start":"00:16.155 ","End":"00:20.625","Text":"Then we define big F on the same interval as follows."},{"Start":"00:20.625 ","End":"00:26.780","Text":"We have to show that the derivative of big F is little f. Nevertheless,"},{"Start":"00:26.780 ","End":"00:30.350","Text":"the integral of little f doesn\u0027t exist."},{"Start":"00:30.350 ","End":"00:32.795","Text":"Now for x not equal to 0,"},{"Start":"00:32.795 ","End":"00:34.490","Text":"I\u0027ll leave you to do the computation."},{"Start":"00:34.490 ","End":"00:36.905","Text":"We just use the product rule on this."},{"Start":"00:36.905 ","End":"00:41.120","Text":"It\u0027s x squared times the derivative of the sine and the derivative of x"},{"Start":"00:41.120 ","End":"00:45.735","Text":"squared which is 2x times this as is. You can check that."},{"Start":"00:45.735 ","End":"00:47.840","Text":"The question is what happens at 0,"},{"Start":"00:47.840 ","End":"00:49.970","Text":"so there we have to do it from definition."},{"Start":"00:49.970 ","End":"00:52.855","Text":"The derivative of big F at 0 is this limit,"},{"Start":"00:52.855 ","End":"00:55.710","Text":"but F of 0 is 0,"},{"Start":"00:55.710 ","End":"01:00.030","Text":"so it\u0027s just F of h over h which is,"},{"Start":"01:00.030 ","End":"01:02.460","Text":"well F of h is given by this formula."},{"Start":"01:02.460 ","End":"01:04.515","Text":"Let\u0027s put h instead of x."},{"Start":"01:04.515 ","End":"01:11.570","Text":"We limit of this quotient and then we can cancel h squared over h is h. Now look,"},{"Start":"01:11.570 ","End":"01:13.070","Text":"it\u0027s in 2 pieces."},{"Start":"01:13.070 ","End":"01:18.034","Text":"We have a product of something that tends to 0 times something which is bounded."},{"Start":"01:18.034 ","End":"01:19.565","Text":"We\u0027ve seen this frequently."},{"Start":"01:19.565 ","End":"01:26.300","Text":"It\u0027s well known that something goes to 0 times something bounded also goes to 0,"},{"Start":"01:26.300 ","End":"01:28.465","Text":"so this limit is 0."},{"Start":"01:28.465 ","End":"01:30.680","Text":"This goes to 0 clearly and the sine of"},{"Start":"01:30.680 ","End":"01:33.665","Text":"anything is between plus and minus 1 so it\u0027s bounded."},{"Start":"01:33.665 ","End":"01:38.405","Text":"Now, f of x, little f is not bounded. I have to show you this."},{"Start":"01:38.405 ","End":"01:43.940","Text":"So define a sequence x_n by x_n equals 1 over square root"},{"Start":"01:43.940 ","End":"01:50.040","Text":"of 2Pi n. 1 over x_n squared is 2Pi n,"},{"Start":"01:50.040 ","End":"01:52.110","Text":"because take the reciprocal and square it,"},{"Start":"01:52.110 ","End":"01:56.890","Text":"so what we get is that f of x_n is equal to,"},{"Start":"01:56.890 ","End":"02:00.230","Text":"here\u0027s the 2x sine of 1 over x squared."},{"Start":"02:00.230 ","End":"02:02.780","Text":"Well, I\u0027ll go back to the top so we can see it again."},{"Start":"02:02.780 ","End":"02:05.360","Text":"Yeah, 2x sine of 1 over x squared."},{"Start":"02:05.360 ","End":"02:09.425","Text":"Then we need minus 2 over x cosine 1 over x squared."},{"Start":"02:09.425 ","End":"02:15.410","Text":"It\u0027s minus 2 divided by x_n which is this goes to the numerator."},{"Start":"02:15.410 ","End":"02:18.620","Text":"Then we have the cosine of the same thing."},{"Start":"02:18.620 ","End":"02:24.450","Text":"The 1 over x_n squared is 2Pi n. Sine of 2Pi n is 0,"},{"Start":"02:24.450 ","End":"02:27.210","Text":"cosine of 2Pi n is 1,"},{"Start":"02:27.210 ","End":"02:29.280","Text":"and this is equal to,"},{"Start":"02:29.280 ","End":"02:32.300","Text":"removing this and removing this which is 1,"},{"Start":"02:32.300 ","End":"02:34.405","Text":"we get minus 2 square root of 2Pi,"},{"Start":"02:34.405 ","End":"02:37.610","Text":"square root of n, so we have a constant which is"},{"Start":"02:37.610 ","End":"02:42.820","Text":"non-zero times something which goes to infinity or at least it\u0027s unbounded."},{"Start":"02:42.820 ","End":"02:45.229","Text":"If we have an unbounded sequence,"},{"Start":"02:45.229 ","End":"02:47.185","Text":"then the function is unbounded."},{"Start":"02:47.185 ","End":"02:53.465","Text":"Then we can just say that the Riemann integral is not defined for unbounded integrand."},{"Start":"02:53.465 ","End":"02:57.080","Text":"The thing that we take the integral of if f is not bounded,"},{"Start":"02:57.080 ","End":"03:00.540","Text":"so the integral doesn\u0027t exist and we\u0027re done."}],"ID":24762},{"Watched":false,"Name":"Exercise 6","Duration":"3m 22s","ChapterTopicVideoID":23821,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23821.jpeg","UploadDate":"2021-01-11T09:52:50.5070000","DurationForVideoObject":"PT3M22S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.460","Text":"In this exercise, we have a continuous function f on the unit interval,"},{"Start":"00:05.460 ","End":"00:11.445","Text":"and it satisfies the following inequality for all x on the interval."},{"Start":"00:11.445 ","End":"00:16.079","Text":"We have to show that f is constantly 0 on this interval."},{"Start":"00:16.079 ","End":"00:18.735","Text":"Now, we\u0027re given that f is continuous,"},{"Start":"00:18.735 ","End":"00:21.480","Text":"and it\u0027s on a closed interval, 0,1,"},{"Start":"00:21.480 ","End":"00:25.784","Text":"so it must be bounded, famous theorem."},{"Start":"00:25.784 ","End":"00:31.125","Text":"Let m be the supremum of f of x on the interval,"},{"Start":"00:31.125 ","End":"00:33.120","Text":"f of x is the absolute value, that is,"},{"Start":"00:33.120 ","End":"00:37.975","Text":"and that will be bound for absolute value of x as follows."},{"Start":"00:37.975 ","End":"00:40.740","Text":"Now, this is what we\u0027re given,"},{"Start":"00:40.740 ","End":"00:47.300","Text":"and we can replace the integrand f of t by something that\u0027s bigger or equal to it,"},{"Start":"00:47.300 ","End":"00:51.720","Text":"and then the integral will only possibly grow."},{"Start":"00:52.190 ","End":"00:57.800","Text":"We can continue the inequality by replacing this by something still bigger."},{"Start":"00:57.800 ","End":"00:59.645","Text":"Because this is true,"},{"Start":"00:59.645 ","End":"01:02.690","Text":"this can be replaced by m."},{"Start":"01:02.690 ","End":"01:08.570","Text":"Now, we have absolute value of f of x is less than or equal to the integral of Mdt."},{"Start":"01:08.570 ","End":"01:10.445","Text":"We can evaluate this,"},{"Start":"01:10.445 ","End":"01:13.645","Text":"and this integral is, of course, just Mx."},{"Start":"01:13.645 ","End":"01:16.740","Text":"Now, back to this inequality."},{"Start":"01:16.740 ","End":"01:23.310","Text":"This time, we can replace the absolute value of f of t by Mt."},{"Start":"01:23.310 ","End":"01:26.750","Text":"If it works for x and x is just a dummy variable,"},{"Start":"01:26.750 ","End":"01:28.835","Text":"then it works for t also."},{"Start":"01:28.835 ","End":"01:31.090","Text":"This integrand is less than or equal to Mt,"},{"Start":"01:31.090 ","End":"01:34.495","Text":"so the integral is less than or equal to the integral."},{"Start":"01:34.495 ","End":"01:37.200","Text":"We can compute this."},{"Start":"01:37.200 ","End":"01:40.800","Text":"This time, we get Mx squared over 2."},{"Start":"01:40.800 ","End":"01:47.490","Text":"Now, once again, we can return to this blue inequality."},{"Start":"01:47.490 ","End":"01:54.365","Text":"This time we can replace absolute value of f of t by Mt squared over 2,"},{"Start":"01:54.365 ","End":"01:56.090","Text":"just as before, we have this,"},{"Start":"01:56.090 ","End":"01:58.545","Text":"but replace x by t,"},{"Start":"01:58.545 ","End":"02:05.360","Text":"and then we can do the integral of this and get Mx cubed over 6,"},{"Start":"02:05.360 ","End":"02:09.080","Text":"and so on and so on."},{"Start":"02:09.080 ","End":"02:14.420","Text":"By induction, what we\u0027ll get is that the absolute value of f of x"},{"Start":"02:14.420 ","End":"02:19.075","Text":"is less than or equal to Mx^n over n factorial."},{"Start":"02:19.075 ","End":"02:22.830","Text":"The 6 is 3 factorial because we started off with 1,"},{"Start":"02:22.830 ","End":"02:25.990","Text":"and we divided it by 2 from the square, then we divide it by 3,"},{"Start":"02:25.990 ","End":"02:29.530","Text":"so this is really 1 times 2 times 3 here."},{"Start":"02:29.530 ","End":"02:32.780","Text":"In general, we\u0027ll get 1 times 2 times 3 up to n."},{"Start":"02:32.780 ","End":"02:38.120","Text":"This is less than or equal to just M over n factorial"},{"Start":"02:38.120 ","End":"02:43.165","Text":"because the x^n part is less than or equal to 1."},{"Start":"02:43.165 ","End":"02:46.280","Text":"X is between 0 and 1."},{"Start":"02:46.280 ","End":"02:50.885","Text":"Now, we can take the limit as n goes to infinity."},{"Start":"02:50.885 ","End":"02:54.095","Text":"X is a constant as far as n goes."},{"Start":"02:54.095 ","End":"02:56.420","Text":"For a given x, this is true,"},{"Start":"02:56.420 ","End":"03:01.715","Text":"and so limit as n goes to infinity will be less than or equal to,"},{"Start":"03:01.715 ","End":"03:05.040","Text":"and n factorial goes to infinity,"},{"Start":"03:05.040 ","End":"03:07.634","Text":"so m over n factorial goes to 0,"},{"Start":"03:07.634 ","End":"03:11.030","Text":"so absolute value of f of x is less than or equal to 0,"},{"Start":"03:11.030 ","End":"03:12.980","Text":"but it\u0027s also non-negative,"},{"Start":"03:12.980 ","End":"03:17.215","Text":"and so it has to equal exactly 0."},{"Start":"03:17.215 ","End":"03:19.500","Text":"That\u0027s for any x in the interval,"},{"Start":"03:19.500 ","End":"03:20.640","Text":"and that\u0027s what we have to show,"},{"Start":"03:20.640 ","End":"03:22.990","Text":"so we are done."}],"ID":24763},{"Watched":false,"Name":"Exercise 7","Duration":"1m 53s","ChapterTopicVideoID":23822,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23822.jpeg","UploadDate":"2021-01-11T09:53:51.9730000","DurationForVideoObject":"PT1M53S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.820","Text":"In this exercise, f is real valued and defined in all the real and its continuous."},{"Start":"00:05.820 ","End":"00:11.610","Text":"We define a function g of x to be this integral here for all x."},{"Start":"00:11.610 ","End":"00:15.255","Text":"Our task is to show that the second derivative of g"},{"Start":"00:15.255 ","End":"00:19.325","Text":"is f. Just copy this definition of g here."},{"Start":"00:19.325 ","End":"00:22.315","Text":"Now, g of x, if we break it up,"},{"Start":"00:22.315 ","End":"00:24.945","Text":"we can take x out because an integral dt,"},{"Start":"00:24.945 ","End":"00:27.465","Text":"so we have this expression here."},{"Start":"00:27.465 ","End":"00:32.970","Text":"Now we want to differentiate g. This is a product of 2 functions of x."},{"Start":"00:32.970 ","End":"00:34.380","Text":"We use a product rule."},{"Start":"00:34.380 ","End":"00:37.905","Text":"Derivative of x is 1 times this as is,"},{"Start":"00:37.905 ","End":"00:45.110","Text":"plus x times the derivative of this minus the derivative of this."},{"Start":"00:45.110 ","End":"00:47.810","Text":"We\u0027re going to use the first fundamental theorem of"},{"Start":"00:47.810 ","End":"00:52.125","Text":"the calculus and here it is to remind you."},{"Start":"00:52.125 ","End":"00:56.195","Text":"We\u0027re going to use the second part that if f is continuous,"},{"Start":"00:56.195 ","End":"00:59.885","Text":"then this integral is differentiable"},{"Start":"00:59.885 ","End":"01:04.769","Text":"and its derivative is the original f. Now in our case,"},{"Start":"01:04.769 ","End":"01:08.640","Text":"we have once the original f and once we\u0027ll use"},{"Start":"01:08.640 ","End":"01:13.010","Text":"t times f. We\u0027ll use the fundamental theorem twice,"},{"Start":"01:13.010 ","End":"01:14.180","Text":"both here and here."},{"Start":"01:14.180 ","End":"01:16.900","Text":"Also notice that this is continuous."},{"Start":"01:16.900 ","End":"01:18.500","Text":"Because this is continuous,"},{"Start":"01:18.500 ","End":"01:23.000","Text":"we can apply the first ftc both here and here,"},{"Start":"01:23.000 ","End":"01:26.720","Text":"and then just replace t by x."},{"Start":"01:26.720 ","End":"01:32.060","Text":"As luck would have it, this cancels with this and we\u0027re left with just this."},{"Start":"01:32.060 ","End":"01:34.340","Text":"Now we were asked about the second derivative,"},{"Start":"01:34.340 ","End":"01:36.680","Text":"so we\u0027re going to differentiate again."},{"Start":"01:36.680 ","End":"01:39.920","Text":"It\u0027s exactly the same thing as we did from here to here."},{"Start":"01:39.920 ","End":"01:46.420","Text":"We\u0027re going to use the first fundamental theorem and that just gives us f of x."},{"Start":"01:46.420 ","End":"01:48.750","Text":"That\u0027s what we have to show. Let\u0027s see."},{"Start":"01:48.750 ","End":"01:50.880","Text":"Yeah, we have to show that g double prime is f,"},{"Start":"01:50.880 ","End":"01:54.010","Text":"and that\u0027s what we\u0027ve just done, so we\u0027re finished."}],"ID":24764},{"Watched":false,"Name":"Exercise 8","Duration":"5m 30s","ChapterTopicVideoID":23823,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23823.jpeg","UploadDate":"2021-01-11T09:57:45.7330000","DurationForVideoObject":"PT5M30S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.820","Text":"In this exercise, f is continuous and alpha is a non-zero constant."},{"Start":"00:05.820 ","End":"00:12.465","Text":"We define a second function g of x by this formula."},{"Start":"00:12.465 ","End":"00:19.340","Text":"We have to show this is actually an identity not inequality, that for all x,"},{"Start":"00:19.340 ","End":"00:23.390","Text":"f of x is g double prime of x plus alpha squared g of"},{"Start":"00:23.390 ","End":"00:29.990","Text":"x. I wrote a trigonometrical formula that we\u0027ll need."},{"Start":"00:29.990 ","End":"00:34.159","Text":"So g of x, just copied it is this."},{"Start":"00:34.159 ","End":"00:41.390","Text":"Using this formula which we apply to the sine of Alpha x minus Alpha t,"},{"Start":"00:41.390 ","End":"00:44.210","Text":"we get this after a little bit of rearranging."},{"Start":"00:44.210 ","End":"00:47.210","Text":"We break up the subtraction into"},{"Start":"00:47.210 ","End":"00:52.310","Text":"2 separate integrals and also the bits that don\u0027t involve t,"},{"Start":"00:52.310 ","End":"00:55.325","Text":"we can bring in front of the integral sign,"},{"Start":"00:55.325 ","End":"01:00.285","Text":"we get this, now we differentiate."},{"Start":"01:00.285 ","End":"01:06.875","Text":"Notice that since f is continuous and certainly cosine of Alpha t is continuous,"},{"Start":"01:06.875 ","End":"01:11.000","Text":"that this function is continuous."},{"Start":"01:11.000 ","End":"01:15.785","Text":"Take 2, now we want to differentiate this with respect to x."},{"Start":"01:15.785 ","End":"01:17.780","Text":"You\u0027ll get something which is a bit lengthy,"},{"Start":"01:17.780 ","End":"01:20.265","Text":"doesn\u0027t fit on 1 line."},{"Start":"01:20.265 ","End":"01:22.505","Text":"Here we have a product,"},{"Start":"01:22.505 ","End":"01:25.130","Text":"then we have minus, and we have another product here."},{"Start":"01:25.130 ","End":"01:27.500","Text":"We\u0027ll use the product rule on this bit."},{"Start":"01:27.500 ","End":"01:32.010","Text":"We get sine Alpha x of this times the derivative of this."},{"Start":"01:32.620 ","End":"01:35.730","Text":"Then the derivative of sine Alpha x,"},{"Start":"01:35.730 ","End":"01:39.320","Text":"which is Alpha cosine Alpha x times this integral as is,"},{"Start":"01:39.320 ","End":"01:41.585","Text":"minus and something similar."},{"Start":"01:41.585 ","End":"01:44.915","Text":"Cosine Alpha x times the derivative of this."},{"Start":"01:44.915 ","End":"01:50.180","Text":"The derivative of minus cosine is plus sign and the Alpha comes out here."},{"Start":"01:50.180 ","End":"01:55.655","Text":"The derivative of this bit times the integral as is."},{"Start":"01:55.655 ","End":"02:01.745","Text":"Now, these 2 functions of t are both continuous."},{"Start":"02:01.745 ","End":"02:07.225","Text":"I mentioned this because I want to apply the first fundamental theorem of the calculus."},{"Start":"02:07.225 ","End":"02:09.885","Text":"To remind you here it is."},{"Start":"02:09.885 ","End":"02:14.930","Text":"The important thing here is that when the integrand is continuous,"},{"Start":"02:14.930 ","End":"02:22.270","Text":"the derivative of the integral is the function itself, basically."},{"Start":"02:22.270 ","End":"02:30.245","Text":"Let\u0027s apply this fundamental theorem of calculus to both of these. What do we get?"},{"Start":"02:30.245 ","End":"02:35.780","Text":"The derivative of this is just whatever it says here,"},{"Start":"02:35.780 ","End":"02:40.130","Text":"but with x instead of t and this thing as is."},{"Start":"02:40.130 ","End":"02:41.990","Text":"In the second part,"},{"Start":"02:41.990 ","End":"02:47.620","Text":"we just have this with t replaced by x."},{"Start":"02:47.620 ","End":"02:51.830","Text":"This as is, some canceling we can do."},{"Start":"02:51.830 ","End":"02:56.390","Text":"First of all, this cancels with this because the same thing,"},{"Start":"02:56.390 ","End":"02:57.980","Text":"just plus and minus."},{"Start":"02:57.980 ","End":"03:00.410","Text":"I mean the sine and cosine of reverse position."},{"Start":"03:00.410 ","End":"03:03.815","Text":"The other thing we can do is there\u0027s an Alpha here and here,"},{"Start":"03:03.815 ","End":"03:05.870","Text":"and 1 over Alpha in front,"},{"Start":"03:05.870 ","End":"03:07.370","Text":"so that will cancel."},{"Start":"03:07.370 ","End":"03:09.005","Text":"After we do all that,"},{"Start":"03:09.005 ","End":"03:14.980","Text":"all we\u0027re left with is this and all this is g prime of x."},{"Start":"03:14.980 ","End":"03:18.500","Text":"We want to differentiate again."},{"Start":"03:18.500 ","End":"03:22.625","Text":"G double prime of x is equal to,"},{"Start":"03:22.625 ","End":"03:25.369","Text":"just like before, it\u0027s bit lengthy."},{"Start":"03:25.369 ","End":"03:28.760","Text":"We have a product here plus a product here."},{"Start":"03:28.760 ","End":"03:32.655","Text":"The first product is cosine Alpha x as is,"},{"Start":"03:32.655 ","End":"03:35.184","Text":"times the derivative of this integral."},{"Start":"03:35.184 ","End":"03:40.460","Text":"Then the derivative of cosine is this and the integral as is."},{"Start":"03:40.460 ","End":"03:45.430","Text":"Similarly for the second product gives us this."},{"Start":"03:45.430 ","End":"03:49.610","Text":"Now again, we\u0027re going to use the fundamental theorem of"},{"Start":"03:49.610 ","End":"03:53.060","Text":"the calculus to evaluate these derivatives,"},{"Start":"03:53.060 ","End":"03:58.645","Text":"which basically means taking the integrand and just replacing t by x."},{"Start":"03:58.645 ","End":"04:02.270","Text":"The other thing we do is to take"},{"Start":"04:02.270 ","End":"04:07.370","Text":"these 2 terms that are colored in red and combine them to this,"},{"Start":"04:07.370 ","End":"04:09.910","Text":"which is like the reverse of what we did before."},{"Start":"04:09.910 ","End":"04:20.680","Text":"I\u0027ll go back and show you what we did earlier was to replace this by this minus this."},{"Start":"04:20.680 ","End":"04:24.485","Text":"We\u0027re just doing the same thing now but in reverse."},{"Start":"04:24.485 ","End":"04:26.525","Text":"That gives us this bit."},{"Start":"04:26.525 ","End":"04:29.320","Text":"Then we can simplify and say, look,"},{"Start":"04:29.320 ","End":"04:33.445","Text":"we have cosine and cosine is cosine squared."},{"Start":"04:33.445 ","End":"04:37.960","Text":"Here we have sine and sine is sine squared."},{"Start":"04:37.960 ","End":"04:43.170","Text":"This expression here is almost the same as g of x,"},{"Start":"04:43.170 ","End":"04:48.615","Text":"except that when we define g of x equals 1 over Alpha here and not Alpha."},{"Start":"04:48.615 ","End":"04:54.130","Text":"To compensate, we make it Alpha squared times g of x."},{"Start":"05:00.320 ","End":"05:05.745","Text":"Now cosine squared plus sine squared is equal to 1."},{"Start":"05:05.745 ","End":"05:14.175","Text":"What we get is that g double prime of x is just f of x minus Alpha squared g of x."},{"Start":"05:14.175 ","End":"05:19.650","Text":"Just bring the alpha squared g of x to the other side and we have"},{"Start":"05:19.650 ","End":"05:28.250","Text":"that f of x is g double-prime plus alpha squared g. That\u0027s what we have to show."},{"Start":"05:28.250 ","End":"05:30.510","Text":"We are done."}],"ID":24765},{"Watched":false,"Name":"Exercise 9","Duration":"1m 17s","ChapterTopicVideoID":23824,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23824.jpeg","UploadDate":"2022-03-21T14:54:50.7970000","DurationForVideoObject":"PT1M17S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.925","Text":"In this exercise, the function f is differentiable on the interval 0, 1."},{"Start":"00:05.925 ","End":"00:12.090","Text":"We have to show that there is some c in the open interval 0, 1,"},{"Start":"00:12.090 ","End":"00:21.090","Text":"such that the integral of f from 0-1 is f of 0 plus a 1/2 the derivative of f at c."},{"Start":"00:21.090 ","End":"00:25.140","Text":"The way we tackle this is to define a function big F"},{"Start":"00:25.140 ","End":"00:29.380","Text":"by big F of x is the integral from 0-x of little f"},{"Start":"00:29.380 ","End":"00:31.740","Text":"and then by the first fundamental theorem,"},{"Start":"00:31.740 ","End":"00:33.660","Text":"big F prime is little f."},{"Start":"00:33.660 ","End":"00:39.030","Text":"We can use Taylor\u0027s theorem with n equals 2."},{"Start":"00:39.030 ","End":"00:41.715","Text":"Here it is with n equals 2."},{"Start":"00:41.715 ","End":"00:45.375","Text":"In our case, b is 1 and a is 0."},{"Start":"00:45.375 ","End":"00:49.245","Text":"The F of 1 is F of 0 plus F prime of 0, and so on."},{"Start":"00:49.245 ","End":"00:54.165","Text":"Just note that big F of 0 is 0."},{"Start":"00:54.165 ","End":"00:56.970","Text":"I mean the integral from 0-0 is"},{"Start":"00:56.970 ","End":"01:00.510","Text":"just 0 and big F prime is little f."},{"Start":"01:00.510 ","End":"01:04.310","Text":"What this boils down to is"},{"Start":"01:04.310 ","End":"01:06.840","Text":"that the integral from 0-1 of fx_dx"},{"Start":"01:06.840 ","End":"01:12.740","Text":"is just f of 0 plus a 1/2f prime of c"},{"Start":"01:12.740 ","End":"01:17.730","Text":"and that is exactly what we had to show and so we\u0027re done."}],"ID":24766},{"Watched":false,"Name":"Exercise 10","Duration":"1m 25s","ChapterTopicVideoID":23825,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23825.jpeg","UploadDate":"2021-01-11T09:59:18.0730000","DurationForVideoObject":"PT1M25S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.950","Text":"In this exercise, f is continuous on the interval 0,1,"},{"Start":"00:04.950 ","End":"00:08.280","Text":"and its integral on this interval happens to be 1."},{"Start":"00:08.280 ","End":"00:11.280","Text":"We have to show that there is some c in"},{"Start":"00:11.280 ","End":"00:15.690","Text":"the open interval such that f of c equals 3c squared."},{"Start":"00:15.690 ","End":"00:18.480","Text":"Now 3c squared, if you think of it as 3x squared,"},{"Start":"00:18.480 ","End":"00:21.810","Text":"reminds us of the derivative of x cubed,"},{"Start":"00:21.810 ","End":"00:25.730","Text":"and you think that this is f of c minus 3c squared is 0."},{"Start":"00:25.730 ","End":"00:34.215","Text":"If the inspiration for defining a function F as the integral of f minus x cubed."},{"Start":"00:34.215 ","End":"00:41.400","Text":"Now we can differentiate F and see that it\u0027s equal to f of x,"},{"Start":"00:41.400 ","End":"00:44.780","Text":"that\u0027s from this part by the first fundamental theorem,"},{"Start":"00:44.780 ","End":"00:47.735","Text":"minus the 3x squared that we mentioned."},{"Start":"00:47.735 ","End":"00:53.390","Text":"F of 0 is the integral from 0 to 0 minus 0 cubed, so it\u0027s 0,"},{"Start":"00:53.390 ","End":"00:58.550","Text":"and F of 1 is the integral from 0 to 1 minus 1 cubed,"},{"Start":"00:58.550 ","End":"01:00.410","Text":"and this was given to be 1,"},{"Start":"01:00.410 ","End":"01:02.240","Text":"so 1 minus 1 cubed is 0."},{"Start":"01:02.240 ","End":"01:06.375","Text":"Now we have 0 both at 0 and 1."},{"Start":"01:06.375 ","End":"01:10.489","Text":"We\u0027re going to apply Rolle\u0027s theorem to F on this interval,"},{"Start":"01:10.489 ","End":"01:12.110","Text":"and by Rolle\u0027s theorem,"},{"Start":"01:12.110 ","End":"01:15.620","Text":"there is some c in the open interval such"},{"Start":"01:15.620 ","End":"01:19.970","Text":"that F prime of c is 0 and view the computation,"},{"Start":"01:19.970 ","End":"01:25.740","Text":"this gives us exactly what we needed and we are done."}],"ID":24767},{"Watched":false,"Name":"Exercise 11","Duration":"2m 4s","ChapterTopicVideoID":23826,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23826.jpeg","UploadDate":"2021-01-11T10:00:47.8930000","DurationForVideoObject":"PT2M4S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.290","Text":"In this exercise, f is continuous on the interval from 0"},{"Start":"00:04.290 ","End":"00:09.974","Text":"to Pi/4 and we have to show that there is some constant c in this interval,"},{"Start":"00:09.974 ","End":"00:13.560","Text":"such that the following equality holds."},{"Start":"00:13.560 ","End":"00:17.715","Text":"The plan is to use the Cauchy mean value theorem."},{"Start":"00:17.715 ","End":"00:22.755","Text":"Let\u0027s define F of x as usual to be the integral of f"},{"Start":"00:22.755 ","End":"00:28.290","Text":"and we\u0027ll let G of x be the function sine 2x."},{"Start":"00:28.290 ","End":"00:32.880","Text":"Now, by the Cauchy mean value theorem for F and G on"},{"Start":"00:32.880 ","End":"00:37.950","Text":"the interval and perhaps I should remind you what this theorem is."},{"Start":"00:37.950 ","End":"00:39.990","Text":"I copied it from somewhere."},{"Start":"00:39.990 ","End":"00:46.040","Text":"What we get if we let b equals Pi over 4 and a"},{"Start":"00:46.040 ","End":"00:52.230","Text":"equals 0 and F and G replace f and g,"},{"Start":"00:52.230 ","End":"00:55.205","Text":"we get the following equality."},{"Start":"00:55.205 ","End":"00:57.925","Text":"Now if we just figure this out,"},{"Start":"00:57.925 ","End":"01:04.160","Text":"this F is the integral from 0 to Pi over 4 of f t dt by definition."},{"Start":"01:04.160 ","End":"01:05.660","Text":"F of 0, well,"},{"Start":"01:05.660 ","End":"01:08.300","Text":"that will be 0 because if we let x is 0,"},{"Start":"01:08.300 ","End":"01:14.360","Text":"then that\u0027s 0, and this is G of Pi/4,"},{"Start":"01:14.360 ","End":"01:16.730","Text":"which is actually equal to 1."},{"Start":"01:16.730 ","End":"01:18.410","Text":"Twice Pi/4 is Pi/2,"},{"Start":"01:18.410 ","End":"01:20.030","Text":"sine Pi/2 is 1,"},{"Start":"01:20.030 ","End":"01:23.570","Text":"and here its sine of 0, which is 0."},{"Start":"01:23.570 ","End":"01:31.170","Text":"On the other side, we get the derivative of this function F,"},{"Start":"01:31.170 ","End":"01:35.310","Text":"which is f by the fundamental theorem Number 1."},{"Start":"01:35.310 ","End":"01:43.700","Text":"The derivative of G is 2 cosine 2x and we need to plug in x equals c. If we do that,"},{"Start":"01:43.700 ","End":"01:47.060","Text":"all we\u0027re left with here is 1 minus 0, this is 0,"},{"Start":"01:47.060 ","End":"01:52.415","Text":"so we\u0027re just left with this integral and here we have f of c and here 2 cosine to 2c."},{"Start":"01:52.415 ","End":"01:59.900","Text":"Now, just multiply both sides by this denominator and we get this equality,"},{"Start":"01:59.900 ","End":"02:04.590","Text":"which is what we had to show and so we are done."}],"ID":24768},{"Watched":false,"Name":"Exercise 12","Duration":"2m 16s","ChapterTopicVideoID":23827,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23827.jpeg","UploadDate":"2022-03-21T14:54:58.9870000","DurationForVideoObject":"PT2M16S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.320","Text":"In this exercise, f is a function defined from"},{"Start":"00:04.320 ","End":"00:09.345","Text":"0 up to a. I guess we assume that a is a positive number."},{"Start":"00:09.345 ","End":"00:13.530","Text":"It\u0027s such that the second derivative is positive."},{"Start":"00:13.530 ","End":"00:16.395","Text":"That means that it\u0027s concave up."},{"Start":"00:16.395 ","End":"00:18.810","Text":"For every x in the interval,"},{"Start":"00:18.810 ","End":"00:23.085","Text":"we have to show that the integral from 0 to a of f"},{"Start":"00:23.085 ","End":"00:28.140","Text":"is bigger or equal to a times the function at a over 2."},{"Start":"00:28.140 ","End":"00:34.140","Text":"In this diagram, if this is 0 and this is a and this is a over 2,"},{"Start":"00:34.140 ","End":"00:40.685","Text":"what it says is that the shaded area is bigger or equal to the area of this rectangle."},{"Start":"00:40.685 ","End":"00:47.775","Text":"This height is f of a over 2 and the width of the rectangle is a."},{"Start":"00:47.775 ","End":"00:50.850","Text":"That\u0027s the area of the rectangle."},{"Start":"00:50.850 ","End":"00:58.625","Text":"First, let x-naught be any point in the interval from 0 to a, excluding the endpoints."},{"Start":"00:58.625 ","End":"01:02.300","Text":"By Taylor\u0027s theorem with n equals 2,"},{"Start":"01:02.300 ","End":"01:05.870","Text":"I should remind you what Taylor\u0027s theorem is, here it is."},{"Start":"01:05.870 ","End":"01:10.680","Text":"We get that f of x is f of x naught plus well, and so on."},{"Start":"01:11.690 ","End":"01:16.520","Text":"That means because this last term is non-negative,"},{"Start":"01:16.520 ","End":"01:21.225","Text":"that f of x is bigger or equal to this plus this."},{"Start":"01:21.225 ","End":"01:24.355","Text":"Now if we have this inequality on the interval,"},{"Start":"01:24.355 ","End":"01:29.785","Text":"we can take the integral from 0 to a of each side and that preserves the inequality."},{"Start":"01:29.785 ","End":"01:32.215","Text":"The integral, this is a constant,"},{"Start":"01:32.215 ","End":"01:36.320","Text":"so we just multiply it by x and this is the constant."},{"Start":"01:36.320 ","End":"01:39.540","Text":"The integral of x is x squared over 2,"},{"Start":"01:39.540 ","End":"01:41.775","Text":"and the integral of 1 is just x."},{"Start":"01:41.775 ","End":"01:46.795","Text":"This is the expression we get and we have to substitute the limits of integration."},{"Start":"01:46.795 ","End":"01:48.700","Text":"When x is 0, everything is 0."},{"Start":"01:48.700 ","End":"01:53.320","Text":"We just take this expression with x equals a and get this."},{"Start":"01:53.320 ","End":"01:56.560","Text":"This is true for any x naught in the interval."},{"Start":"01:56.560 ","End":"02:00.955","Text":"In particular, we could take x naught equals a over 2."},{"Start":"02:00.955 ","End":"02:02.545","Text":"If we plug that in,"},{"Start":"02:02.545 ","End":"02:05.225","Text":"then this is what we get,"},{"Start":"02:05.225 ","End":"02:08.735","Text":"and these 2 terms cancel out."},{"Start":"02:08.735 ","End":"02:12.410","Text":"All we\u0027re left with is this bit here,"},{"Start":"02:12.410 ","End":"02:16.980","Text":"and that\u0027s exactly what we had to show. We\u0027re done."}],"ID":24769},{"Watched":false,"Name":"Exercise 13","Duration":"1m 17s","ChapterTopicVideoID":23828,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23828.jpeg","UploadDate":"2021-01-11T10:02:34.4370000","DurationForVideoObject":"PT1M17S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.040","Text":"In this exercise, f is continuous on the interval a, b,"},{"Start":"00:05.040 ","End":"00:11.977","Text":"and suppose that the integral from a to x of f_(t)dt"},{"Start":"00:11.977 ","End":"00:16.860","Text":"is always equal to the integral from x to b of f_(t)dt."},{"Start":"00:16.860 ","End":"00:22.830","Text":"We have to show that f is constant function 0 on the interval."},{"Start":"00:22.830 ","End":"00:27.390","Text":"Define F to be the integral from a to x,"},{"Start":"00:27.390 ","End":"00:29.070","Text":"just like we usually do."},{"Start":"00:29.070 ","End":"00:33.090","Text":"Now notice that the integral from a to x plus the integral from x"},{"Start":"00:33.090 ","End":"00:37.695","Text":"to b is the integral all the way from a to b."},{"Start":"00:37.695 ","End":"00:42.395","Text":"So the given condition implies that, well,"},{"Start":"00:42.395 ","End":"00:44.525","Text":"this is F_(x) by the definition,"},{"Start":"00:44.525 ","End":"00:47.900","Text":"but this is equal to this by the given,"},{"Start":"00:47.900 ","End":"00:50.335","Text":"so this is also F_(x)."},{"Start":"00:50.335 ","End":"00:53.520","Text":"We get F_(x) plus F_(x) is F_(b),"},{"Start":"00:53.520 ","End":"00:57.105","Text":"which means that F_(x) is 1/2 of F_(b), and main thing is"},{"Start":"00:57.105 ","End":"01:05.310","Text":"this is a constant, then its derivative on the 1 hand is f."},{"Start":"01:05.310 ","End":"01:08.905","Text":"On the other hand, it\u0027s equal to 0,"},{"Start":"01:08.905 ","End":"01:14.220","Text":"so we get that f_(x) is equal to 0,"},{"Start":"01:14.220 ","End":"01:18.340","Text":"which is what we have to show, and so we\u0027re done."}],"ID":24770},{"Watched":false,"Name":"Exercise 14","Duration":"2m 39s","ChapterTopicVideoID":23829,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23829.jpeg","UploadDate":"2021-01-11T10:04:08.8770000","DurationForVideoObject":"PT2M39S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.030","Text":"In this exercise, f and g are both integrable functions on the interval a, b;"},{"Start":"00:06.030 ","End":"00:11.595","Text":"in addition, we\u0027re given that f is increasing and g is non-negative."},{"Start":"00:11.595 ","End":"00:16.785","Text":"In this case, we have to show that there exists some point c in the interval,"},{"Start":"00:16.785 ","End":"00:20.415","Text":"well, I won\u0027t read it out such that this equality holds."},{"Start":"00:20.415 ","End":"00:23.955","Text":"Notice that c appears here and here."},{"Start":"00:23.955 ","End":"00:30.105","Text":"What we\u0027ll do is we\u0027ll define a function h according to the right-hand side of this,"},{"Start":"00:30.105 ","End":"00:32.700","Text":"just replace c by x,"},{"Start":"00:32.700 ","End":"00:35.070","Text":"and we also have to replace x by t."},{"Start":"00:35.070 ","End":"00:40.020","Text":"Now, h of a is equal to f of a times this."},{"Start":"00:40.020 ","End":"00:45.064","Text":"The reason for that is because if we replace x by a,"},{"Start":"00:45.064 ","End":"00:47.550","Text":"this part becomes 0,"},{"Start":"00:47.550 ","End":"00:51.295","Text":"so we just get this part where x is a."},{"Start":"00:51.295 ","End":"00:55.459","Text":"We can throw the f of a inside the integral,"},{"Start":"00:55.459 ","End":"00:56.930","Text":"because it\u0027s a constant."},{"Start":"00:56.930 ","End":"01:02.239","Text":"I claim that this integral is less than or equal to this integral,"},{"Start":"01:02.239 ","End":"01:04.625","Text":"which is less than or equal to this integral."},{"Start":"01:04.625 ","End":"01:06.680","Text":"The difference is here I have f of a,"},{"Start":"01:06.680 ","End":"01:09.860","Text":"here I have f of x, and here I have f of b."},{"Start":"01:09.860 ","End":"01:12.010","Text":"Let me explain why."},{"Start":"01:12.010 ","End":"01:15.690","Text":"First of all, g of x is non-negative,"},{"Start":"01:15.690 ","End":"01:17.990","Text":"so if I increase this part,"},{"Start":"01:17.990 ","End":"01:19.670","Text":"it only can get bigger,"},{"Start":"01:19.670 ","End":"01:23.285","Text":"f of a is less than or equal to f of x is less than or equal to f of b."},{"Start":"01:23.285 ","End":"01:29.880","Text":"That is so because f is increasing in the sense of non-decreasing,"},{"Start":"01:29.880 ","End":"01:32.135","Text":"so it could be less than or equal to."},{"Start":"01:32.135 ","End":"01:35.465","Text":"Well, we can take the f of b outside the brackets,"},{"Start":"01:35.465 ","End":"01:38.915","Text":"and just like we got that h of a is this,"},{"Start":"01:38.915 ","End":"01:42.410","Text":"similarly, this is equal to h of b."},{"Start":"01:42.410 ","End":"01:43.730","Text":"When you plug in x equals b,"},{"Start":"01:43.730 ","End":"01:45.510","Text":"this part goes to 0,"},{"Start":"01:45.510 ","End":"01:47.525","Text":"we\u0027re left with this, which is this."},{"Start":"01:47.525 ","End":"01:51.915","Text":"The summary of this is that h of a, I\u0027ll highlight it,"},{"Start":"01:51.915 ","End":"01:57.830","Text":"this is less than or equal to this,"},{"Start":"01:57.830 ","End":"02:04.045","Text":"which is less than or equal to h of b."},{"Start":"02:04.045 ","End":"02:06.250","Text":"This what I\u0027ve written here,"},{"Start":"02:06.250 ","End":"02:09.470","Text":"except to call Alpha equals this integral,"},{"Start":"02:09.470 ","End":"02:11.825","Text":"so we have this inequality."},{"Start":"02:11.825 ","End":"02:13.415","Text":"I labeled it so,"},{"Start":"02:13.415 ","End":"02:17.410","Text":"because this will remind us of the intermediate value property."},{"Start":"02:17.410 ","End":"02:21.080","Text":"By this property since Alpha\u0027s between this and this,"},{"Start":"02:21.080 ","End":"02:25.190","Text":"there are some c between a and b inclusive such that"},{"Start":"02:25.190 ","End":"02:27.680","Text":"Alpha equals h of c."},{"Start":"02:27.680 ","End":"02:30.350","Text":"That means if we plug it in,"},{"Start":"02:30.350 ","End":"02:34.100","Text":"well, this here is exactly what we had to prove."},{"Start":"02:34.100 ","End":"02:35.330","Text":"It\u0027s just scrolled off."},{"Start":"02:35.330 ","End":"02:39.390","Text":"This is it. Basically, we\u0027re done."}],"ID":24771},{"Watched":false,"Name":"Exercise 15","Duration":"3m 18s","ChapterTopicVideoID":23830,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23830.jpeg","UploadDate":"2021-01-11T10:06:22.9600000","DurationForVideoObject":"PT3M18S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.180","Text":"In this exercise, we have to show that the Mean Value Theorem implies,"},{"Start":"00:06.180 ","End":"00:09.676","Text":"what is known as, the First Mean Value Theorem for Integrals,"},{"Start":"00:09.676 ","End":"00:12.105","Text":"and also the other way around."},{"Start":"00:12.105 ","End":"00:15.585","Text":"If we add in an extra little condition that"},{"Start":"00:15.585 ","End":"00:19.110","Text":"we\u0027re talking about functions with continuous first derivative,"},{"Start":"00:19.110 ","End":"00:22.755","Text":"I better remind you of these theorems."},{"Start":"00:22.755 ","End":"00:27.945","Text":"Here\u0027s the First Mean Value Theorem for Integrals in case you aren\u0027t familiar with it."},{"Start":"00:27.945 ","End":"00:33.765","Text":"There\u0027s a diagram here that if f is continuous,"},{"Start":"00:33.765 ","End":"00:38.130","Text":"then the some point between a and b here such"},{"Start":"00:38.130 ","End":"00:43.429","Text":"that the area of this rectangle equals the area under the graph."},{"Start":"00:43.429 ","End":"00:45.965","Text":"That\u0027s basically what this says."},{"Start":"00:45.965 ","End":"00:49.340","Text":"Regular Mean Value Theorem, you\u0027re familiar with, of course,"},{"Start":"00:49.340 ","End":"00:51.970","Text":"but here it is in case you need it."},{"Start":"00:51.970 ","End":"00:55.195","Text":"Let\u0027s start with one direction."},{"Start":"00:55.195 ","End":"00:58.115","Text":"We\u0027re assuming that the Mean Value Theorem is true."},{"Start":"00:58.115 ","End":"01:00.350","Text":"We have to prove this."},{"Start":"01:00.350 ","End":"01:04.453","Text":"Let f be continuous on the interval a, b"},{"Start":"01:04.453 ","End":"01:09.572","Text":"and define big F, as usual, the integral of little f."},{"Start":"01:09.572 ","End":"01:11.330","Text":"Now we\u0027re going to use"},{"Start":"01:11.330 ","End":"01:15.012","Text":"the first fundamental theorem of the calculus"},{"Start":"01:15.012 ","End":"01:18.320","Text":"and that gives us that big F is differentiable."},{"Start":"01:18.320 ","End":"01:22.370","Text":"Not only that, but the derivative of big F is little f."},{"Start":"01:22.370 ","End":"01:28.745","Text":"Now we\u0027re applying the regular Mean Value Theorem because we\u0027re assuming that in part a."},{"Start":"01:28.745 ","End":"01:32.660","Text":"There exist c such that this is true."},{"Start":"01:32.660 ","End":"01:36.938","Text":"Note that where it says big F prime, that\u0027s the same as little f."},{"Start":"01:36.938 ","End":"01:40.625","Text":"Now we can substitute both of these, F of b."},{"Start":"01:40.625 ","End":"01:49.675","Text":"Just replace x by b gives us the integral of f from a to b and F of a is, of course, 0."},{"Start":"01:49.675 ","End":"01:54.290","Text":"What that gives us is that this left-hand side is"},{"Start":"01:54.290 ","End":"01:57.690","Text":"the integral of f from the interval"},{"Start":"01:57.690 ","End":"02:01.925","Text":"and the right-hand side is little f of c times b minus a."},{"Start":"02:01.925 ","End":"02:05.465","Text":"That\u0027s what we had to show in part a."},{"Start":"02:05.465 ","End":"02:07.930","Text":"Now, part b."},{"Start":"02:07.930 ","End":"02:12.230","Text":"Let\u0027s take little f to be differentiable"},{"Start":"02:12.230 ","End":"02:17.410","Text":"and with the extra condition that f prime is continuous."},{"Start":"02:17.410 ","End":"02:21.650","Text":"We have to prove the regular Mean Value Theorem for f."},{"Start":"02:21.650 ","End":"02:28.035","Text":"Apply the First Mean Value Theorem for Integrals to f prime."},{"Start":"02:28.035 ","End":"02:32.940","Text":"What that gives us is that there is some c in the interval"},{"Start":"02:32.940 ","End":"02:38.060","Text":"such that the integral of this f prime"},{"Start":"02:38.060 ","End":"02:44.240","Text":"is this function f prime at some c times b minus a."},{"Start":"02:44.240 ","End":"02:50.435","Text":"Now, by the Second Fundamental Theorem of the Calculus,"},{"Start":"02:50.435 ","End":"02:55.250","Text":"this integral here is just f of b minus f of a."},{"Start":"02:55.250 ","End":"02:58.140","Text":"That\u0027s why we needed the continuity of f prime"},{"Start":"02:58.140 ","End":"03:02.405","Text":"because the Second FTC needs continuity of f prime."},{"Start":"03:02.405 ","End":"03:05.480","Text":"Now, this is equal to both this and this,"},{"Start":"03:05.480 ","End":"03:09.900","Text":"so we can compare these 2 quantities and that will give us"},{"Start":"03:09.900 ","End":"03:15.125","Text":"that f of b minus f of a is f prime of c times b minus a."},{"Start":"03:15.125 ","End":"03:17.266","Text":"That concludes part b."},{"Start":"03:17.266 ","End":"03:19.500","Text":"We\u0027re done."}],"ID":24772},{"Watched":false,"Name":"Exercise 16","Duration":"59s","ChapterTopicVideoID":23831,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23831.jpeg","UploadDate":"2021-01-11T10:06:54.7000000","DurationForVideoObject":"PT59S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.390","Text":"In this exercise, we have to use the first mean value theorem for"},{"Start":"00:03.390 ","End":"00:07.260","Text":"integrals to show that the integral from n to n plus 1 or 1"},{"Start":"00:07.260 ","End":"00:12.300","Text":"over x dx is less than 1 over n for any integer n."},{"Start":"00:12.300 ","End":"00:15.570","Text":"I\u0027ve put the first Mean Value Theorem for"},{"Start":"00:15.570 ","End":"00:18.910","Text":"integrals over here for you to see for reference."},{"Start":"00:18.910 ","End":"00:25.300","Text":"What we\u0027re going to do, is we\u0027re going to let f of x be 1 over x. I mean, that\u0027s clear."},{"Start":"00:25.300 ","End":"00:30.425","Text":"We\u0027re going to do it on the interval from n to n plus 1."},{"Start":"00:30.425 ","End":"00:33.410","Text":"What we get, is that there is some c in"},{"Start":"00:33.410 ","End":"00:37.790","Text":"this interval such that this integral is equal to,"},{"Start":"00:37.790 ","End":"00:42.120","Text":"now f of c is just 1 over c and b"},{"Start":"00:42.120 ","End":"00:47.865","Text":"minus a is m plus 1 minus n. This simplifies to just 1 over c,"},{"Start":"00:47.865 ","End":"00:50.899","Text":"and since c is in this interval,"},{"Start":"00:50.899 ","End":"00:52.505","Text":"c is bigger than n,"},{"Start":"00:52.505 ","End":"00:55.640","Text":"and so this is less than 1 over n. This is less than this,"},{"Start":"00:55.640 ","End":"00:58.920","Text":"and that\u0027s what we had to show, so we\u0027re done."}],"ID":24773},{"Watched":false,"Name":"Exercise 17","Duration":"1m 22s","ChapterTopicVideoID":23832,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23832.jpeg","UploadDate":"2021-01-11T10:07:48.8570000","DurationForVideoObject":"PT1M22S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.105","Text":"In this exercise, f and g are continuous on the closed interval a,"},{"Start":"00:04.105 ","End":"00:07.750","Text":"b, and their integrals are equal on this interval."},{"Start":"00:07.750 ","End":"00:14.330","Text":"We have to show that there is a point c in the interval where f and g agree."},{"Start":"00:14.330 ","End":"00:20.800","Text":"We\u0027ll start by letting h equal f minus g. h is also continuous,"},{"Start":"00:20.800 ","End":"00:22.600","Text":"difference of continuous functions."},{"Start":"00:22.600 ","End":"00:25.060","Text":"Its integral on the interval is 0 because it\u0027s"},{"Start":"00:25.060 ","End":"00:27.815","Text":"the difference of these 2 equal quantities."},{"Start":"00:27.815 ","End":"00:32.020","Text":"Now we can apply the first mean value theorem for integrals."},{"Start":"00:32.020 ","End":"00:33.490","Text":"You may have forgotten it,"},{"Start":"00:33.490 ","End":"00:35.980","Text":"so I\u0027ll show you, I prepared it."},{"Start":"00:35.980 ","End":"00:38.885","Text":"Here it is for reference."},{"Start":"00:38.885 ","End":"00:45.050","Text":"What we get is that there is some point c it should be the closed interval"},{"Start":"00:45.050 ","End":"00:53.000","Text":"such that the integral of h on the interval is equal to h of c times b minus a."},{"Start":"00:53.000 ","End":"01:00.365","Text":"Now this integral is 0 and so h of c times b minus a is 0."},{"Start":"01:00.365 ","End":"01:02.990","Text":"I guess we\u0027re assuming that a is less than b,"},{"Start":"01:02.990 ","End":"01:05.495","Text":"that is not a point interval."},{"Start":"01:05.495 ","End":"01:08.915","Text":"In that case, it\u0027s easy to prove it\u0027s trivial."},{"Start":"01:08.915 ","End":"01:13.250","Text":"h of c has got to be 0 because b minus a is not 0."},{"Start":"01:13.250 ","End":"01:22.350","Text":"That means that f of c equals g of c because h is f minus g, and we\u0027re done."}],"ID":24774},{"Watched":false,"Name":"Exercise 18","Duration":"2m 9s","ChapterTopicVideoID":23833,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23833.jpeg","UploadDate":"2021-01-11T10:09:08.2100000","DurationForVideoObject":"PT2M9S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.980","Text":"For this exercise, we\u0027re going to need the second mean value theorem for integrals."},{"Start":"00:04.980 ","End":"00:07.965","Text":"There are several versions of these going round."},{"Start":"00:07.965 ","End":"00:11.520","Text":"So this is the version that we\u0027ll be using,"},{"Start":"00:11.520 ","End":"00:13.965","Text":"and we\u0027ll need it for this exercise."},{"Start":"00:13.965 ","End":"00:16.680","Text":"Let\u0027s just get to the exercise."},{"Start":"00:16.680 ","End":"00:21.780","Text":"We have to show that this integral from Pi over 6 to Pi over 2 of x"},{"Start":"00:21.780 ","End":"00:26.860","Text":"over sine x dx is sandwiched between these 2 quantities,"},{"Start":"00:26.860 ","End":"00:29.795","Text":"Pi squared over 9, 2 Pi squared over 9."},{"Start":"00:29.795 ","End":"00:33.380","Text":"What we\u0027ll do is we\u0027ll define f and g,"},{"Start":"00:33.380 ","End":"00:35.465","Text":"and these are the f and g here."},{"Start":"00:35.465 ","End":"00:43.015","Text":"We\u0027ll take f to be the 1 over sine x part and g of x to be the x part of this."},{"Start":"00:43.015 ","End":"00:45.980","Text":"Of course, it satisfies the condition"},{"Start":"00:45.980 ","End":"00:50.030","Text":"because f and g are both continuous and both integrable."},{"Start":"00:50.030 ","End":"00:57.790","Text":"G is always positive on this interval, g of x is x."},{"Start":"00:57.790 ","End":"01:02.269","Text":"Applying this second mean value theorem for integrals,"},{"Start":"01:02.269 ","End":"01:10.745","Text":"what we get is that there is some point C on the interval such that this integral equals,"},{"Start":"01:10.745 ","End":"01:16.955","Text":"f of c is 1 over sine C because f of x is 1 over sine x."},{"Start":"01:16.955 ","End":"01:22.030","Text":"Then we need the integral of g of x dx, which is x dx."},{"Start":"01:22.030 ","End":"01:25.545","Text":"The rest of the exercise is just computations."},{"Start":"01:25.545 ","End":"01:30.125","Text":"This integral comes out to be Pi squared over 9."},{"Start":"01:30.125 ","End":"01:38.225","Text":"Now we want to estimate 1 over sine C. C is between Pi over 2 and Pi over 6."},{"Start":"01:38.225 ","End":"01:42.725","Text":"Sine is increasing in this interval,"},{"Start":"01:42.725 ","End":"01:46.235","Text":"and so 1 over sine is decreasing."},{"Start":"01:46.235 ","End":"01:48.565","Text":"In short, we get this."},{"Start":"01:48.565 ","End":"01:53.530","Text":"Therefore, 1 over sine C is between 1 and 2."},{"Start":"01:53.530 ","End":"01:56.660","Text":"Now just combining these 2,"},{"Start":"01:56.660 ","End":"01:59.180","Text":"we get that Pi squared over 9 is fixed,"},{"Start":"01:59.180 ","End":"02:04.760","Text":"and then this is between 1 times Pi squared over 9 and 2 times Pi squared over 9."},{"Start":"02:04.760 ","End":"02:07.620","Text":"That\u0027s what we had to show up here."},{"Start":"02:07.620 ","End":"02:09.970","Text":"So we are done."}],"ID":24775},{"Watched":false,"Name":"Exercise 19","Duration":"1m 56s","ChapterTopicVideoID":23834,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23834.jpeg","UploadDate":"2021-01-11T10:10:18.2700000","DurationForVideoObject":"PT1M56S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.980","Text":"In this exercise, we will need the"},{"Start":"00:01.980 ","End":"00:04.620","Text":"Second Mean Value Theorem for integrals,"},{"Start":"00:04.620 ","End":"00:07.375","Text":"and there\u0027s quite a few variations of this."},{"Start":"00:07.375 ","End":"00:11.505","Text":"I\u0027ve put here the variation that I am going to use."},{"Start":"00:11.505 ","End":"00:16.755","Text":"F is continuous function on the interval 0,1,"},{"Start":"00:16.755 ","End":"00:22.920","Text":"and we have to show that the limit of the integral of x^n"},{"Start":"00:22.920 ","End":"00:27.120","Text":"f of x is 0 as n goes to infinity."},{"Start":"00:27.120 ","End":"00:31.155","Text":"What we\u0027ll do is we have f from the question,"},{"Start":"00:31.155 ","End":"00:35.895","Text":"and we\u0027ll define g as g of x equals x^n,"},{"Start":"00:35.895 ","End":"00:39.795","Text":"so g is certainly integrable, it\u0027s continuous."},{"Start":"00:39.795 ","End":"00:43.700","Text":"g of x is also non-negative as required,"},{"Start":"00:43.700 ","End":"00:48.680","Text":"so we can apply this theorem and get that the integral of"},{"Start":"00:48.680 ","End":"00:53.960","Text":"f times g is f of c times the integral of g. Now,"},{"Start":"00:53.960 ","End":"00:57.424","Text":"the integral from 0 to 1 of x^n dx,"},{"Start":"00:57.424 ","End":"00:59.735","Text":"it\u0027s 1 over n plus 1,"},{"Start":"00:59.735 ","End":"01:01.745","Text":"so we end up with this."},{"Start":"01:01.745 ","End":"01:05.490","Text":"Because f is continuous, it\u0027s bounded,"},{"Start":"01:05.490 ","End":"01:09.080","Text":"and so the sum M such that"},{"Start":"01:09.080 ","End":"01:11.570","Text":"absolute value of f of x is always less than"},{"Start":"01:11.570 ","End":"01:13.750","Text":"or equal to M on our interval,"},{"Start":"01:13.750 ","End":"01:19.685","Text":"and that gives us that the absolute value of the integral,"},{"Start":"01:19.685 ","End":"01:24.710","Text":"which is the absolute value of f of c"},{"Start":"01:24.710 ","End":"01:26.900","Text":"over n plus 1 is positive already,"},{"Start":"01:26.900 ","End":"01:29.450","Text":"and that\u0027s less than or equal to M over n plus 1."},{"Start":"01:29.450 ","End":"01:32.315","Text":"On the other side, it\u0027s bigger or equal to 0."},{"Start":"01:32.315 ","End":"01:35.720","Text":"This goes to 0 as n goes to infinity,"},{"Start":"01:35.720 ","End":"01:39.500","Text":"so our integral is sandwiched between 0 and 0,"},{"Start":"01:39.500 ","End":"01:43.400","Text":"and so it also must be 0."},{"Start":"01:43.400 ","End":"01:45.949","Text":"If the absolute value goes to 0,"},{"Start":"01:45.949 ","End":"01:49.895","Text":"then so does the integral without the absolute value,"},{"Start":"01:49.895 ","End":"01:52.190","Text":"it\u0027s if and only if something goes to 0,"},{"Start":"01:52.190 ","End":"01:57.210","Text":"if and only if its absolute value goes to 0. We\u0027re done."}],"ID":24776},{"Watched":false,"Name":"Exercise 20","Duration":"1m 18s","ChapterTopicVideoID":23835,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23835.jpeg","UploadDate":"2021-01-11T10:11:06.3800000","DurationForVideoObject":"PT1M18S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.910","Text":"In this exercise, we\u0027re going to use the first mean value theorem for integrals,"},{"Start":"00:03.910 ","End":"00:06.330","Text":"so I\u0027ll just put it up here for reference."},{"Start":"00:06.330 ","End":"00:10.125","Text":"We\u0027re given a continuous function f on the unit interval."},{"Start":"00:10.125 ","End":"00:14.250","Text":"We have to show that the limit as n goes to infinity"},{"Start":"00:14.250 ","End":"00:19.840","Text":"of the integral of f of x^n is f of 0."},{"Start":"00:19.840 ","End":"00:23.280","Text":"By the first MVT for integrals,"},{"Start":"00:23.280 ","End":"00:26.955","Text":"we have c strictly between 0 and 1."},{"Start":"00:26.955 ","End":"00:28.499","Text":"It\u0027s in the open interval."},{"Start":"00:28.499 ","End":"00:29.895","Text":"That\u0027s important here."},{"Start":"00:29.895 ","End":"00:35.610","Text":"Such that the integral of f of x^n dx is f"},{"Start":"00:35.610 ","End":"00:43.665","Text":"of c^n because our function is f of x^n times b minus a is 1 minus 0."},{"Start":"00:43.665 ","End":"00:46.785","Text":"That\u0027s just equal to f of c^n."},{"Start":"00:46.785 ","End":"00:51.110","Text":"If we take the limit as n goes to infinity, then the limit,"},{"Start":"00:51.110 ","End":"00:53.870","Text":"this thing, is equal to f of c^n,"},{"Start":"00:53.870 ","End":"00:55.815","Text":"so let\u0027s just copy that,"},{"Start":"00:55.815 ","End":"00:58.080","Text":"and because of continuity of f,"},{"Start":"00:58.080 ","End":"01:01.415","Text":"we can put the limit inside the function,"},{"Start":"01:01.415 ","End":"01:06.335","Text":"and here\u0027s where we use the fact that c is strictly between 0 and 1."},{"Start":"01:06.335 ","End":"01:08.480","Text":"This is equal to f of 0."},{"Start":"01:08.480 ","End":"01:13.685","Text":"The limit of c^n is 0 because c is smaller than 1."},{"Start":"01:13.685 ","End":"01:15.890","Text":"This is what we have to show,"},{"Start":"01:15.890 ","End":"01:18.690","Text":"and so we are done."}],"ID":24777},{"Watched":false,"Name":"Exercise 21","Duration":"3m 55s","ChapterTopicVideoID":23836,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23836.jpeg","UploadDate":"2021-01-11T10:13:48.2530000","DurationForVideoObject":"PT3M55S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.750","Text":"This exercise relates to Riemann sums."},{"Start":"00:03.750 ","End":"00:09.510","Text":"We have a continuous function f on the interval ab, closed interval."},{"Start":"00:09.510 ","End":"00:12.135","Text":"We have to show the following."},{"Start":"00:12.135 ","End":"00:13.515","Text":"I won\u0027t read it out,"},{"Start":"00:13.515 ","End":"00:18.690","Text":"but I will say this means the norm of p turns to 0,"},{"Start":"00:18.690 ","End":"00:25.965","Text":"and this S is a Riemann sum and P is a partition of the interval ab."},{"Start":"00:25.965 ","End":"00:28.230","Text":"I have put the notation here,"},{"Start":"00:28.230 ","End":"00:30.120","Text":"it\u0027s not entirely standard,"},{"Start":"00:30.120 ","End":"00:34.095","Text":"so I\u0027ve written it to what each of these things means."},{"Start":"00:34.095 ","End":"00:37.215","Text":"Correction, this should be ci not ck."},{"Start":"00:37.215 ","End":"00:39.030","Text":"To rephrase what it is,"},{"Start":"00:39.030 ","End":"00:44.670","Text":"we have to show with Epsilon Delta is that for all Epsilon bigger than 0,"},{"Start":"00:44.670 ","End":"00:47.030","Text":"there exists some Delta bigger than 0,"},{"Start":"00:47.030 ","End":"00:51.140","Text":"such that if the norm of P is less than Delta,"},{"Start":"00:51.140 ","End":"00:59.430","Text":"then this Riemann sum is close to the definite integral to within Epsilon."},{"Start":"00:59.600 ","End":"01:06.140","Text":"The key step in the proof is to note that since f is continuous on a closed interval,"},{"Start":"01:06.140 ","End":"01:08.690","Text":"it must be uniformly continuous."},{"Start":"01:08.690 ","End":"01:10.880","Text":"That\u0027s a well-known theorem."},{"Start":"01:10.880 ","End":"01:14.300","Text":"If we\u0027re given an Epsilon bigger than 0,"},{"Start":"01:14.300 ","End":"01:21.935","Text":"then there exists a Delta such that if 2 points in the interval are close within Delta,"},{"Start":"01:21.935 ","End":"01:26.620","Text":"then the f of the points is there close"},{"Start":"01:26.620 ","End":"01:31.945","Text":"within any given Epsilon and we\u0027ll choose Epsilon over b minus a,"},{"Start":"01:31.945 ","End":"01:34.955","Text":"where Epsilon is the original Epsilon."},{"Start":"01:34.955 ","End":"01:39.115","Text":"If our partition has norm less than Delta,"},{"Start":"01:39.115 ","End":"01:42.325","Text":"then the supremum minus"},{"Start":"01:42.325 ","End":"01:47.620","Text":"the infimum on any interval is less than this Epsilon over b minus a."},{"Start":"01:47.620 ","End":"01:50.845","Text":"Since because of the continuity,"},{"Start":"01:50.845 ","End":"01:56.035","Text":"these supremum and infimum are achieved at some points,"},{"Start":"01:56.035 ","End":"02:00.820","Text":"x1 and x2, and these are in Delta xi,"},{"Start":"02:00.820 ","End":"02:05.670","Text":"so the absolute value of the difference is less than Delta."},{"Start":"02:05.670 ","End":"02:11.170","Text":"I guess I should have written that x1 and x2 are in this interval, the ith interval."},{"Start":"02:11.170 ","End":"02:15.380","Text":"The norm of p is less than Delta, just repeating that,"},{"Start":"02:15.380 ","End":"02:21.875","Text":"then the upper sum minus the lower sum is equal to this sum,"},{"Start":"02:21.875 ","End":"02:24.620","Text":"where each interval we take the width of"},{"Start":"02:24.620 ","End":"02:28.640","Text":"the interval times the difference between the supremum and infimum."},{"Start":"02:28.640 ","End":"02:30.740","Text":"Now, this we\u0027ve already said,"},{"Start":"02:30.740 ","End":"02:34.420","Text":"is less than Epsilon over b minus a."},{"Start":"02:34.420 ","End":"02:38.420","Text":"The sum of all the Delta xi is equal to"},{"Start":"02:38.420 ","End":"02:42.455","Text":"b minus a because altogether they span the interval."},{"Start":"02:42.455 ","End":"02:45.230","Text":"What we get altogether is Epsilon."},{"Start":"02:45.230 ","End":"02:49.345","Text":"Now in general, we know that l is less than or equal to s,"},{"Start":"02:49.345 ","End":"02:51.010","Text":"is less than or equal to u."},{"Start":"02:51.010 ","End":"02:58.385","Text":"This is because ci is bigger or equal to m_i and it\u0027s less than or equal to M_i,"},{"Start":"02:58.385 ","End":"03:00.335","Text":"so we get this."},{"Start":"03:00.335 ","End":"03:04.970","Text":"Similarly, the integral is between L and U,"},{"Start":"03:04.970 ","End":"03:10.625","Text":"because the integral is the infimum of these and the supremum of these."},{"Start":"03:10.625 ","End":"03:12.440","Text":"Now, a little bit of algebra,"},{"Start":"03:12.440 ","End":"03:15.560","Text":"if both of these are sandwiched between the same 2 quantities,"},{"Start":"03:15.560 ","End":"03:18.810","Text":"we can say like C is between A and B,"},{"Start":"03:18.810 ","End":"03:20.855","Text":"and D is between A and B,"},{"Start":"03:20.855 ","End":"03:26.915","Text":"then it follows that the distance from C to B is at most B minus A."},{"Start":"03:26.915 ","End":"03:31.880","Text":"In our case, what we get is that the integral minus"},{"Start":"03:31.880 ","End":"03:34.880","Text":"the Riemann sum in absolute value is less than or"},{"Start":"03:34.880 ","End":"03:38.360","Text":"equal to the upper sum minus the lower sum,"},{"Start":"03:38.360 ","End":"03:40.925","Text":"and this is less than Epsilon."},{"Start":"03:40.925 ","End":"03:45.875","Text":"What we\u0027ve shown is that if the norm of p is less than Delta,"},{"Start":"03:45.875 ","End":"03:49.055","Text":"then this difference is less than Epsilon."},{"Start":"03:49.055 ","End":"03:53.150","Text":"That\u0027s what we had to show off-screen."},{"Start":"03:53.150 ","End":"03:55.830","Text":"We are done."}],"ID":24778},{"Watched":false,"Name":"Exercise 22","Duration":"4m 23s","ChapterTopicVideoID":23837,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23837.jpeg","UploadDate":"2021-01-11T10:16:35.7300000","DurationForVideoObject":"PT4M23S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.835","Text":"In this exercise, we have a sequence a_n,"},{"Start":"00:02.835 ","End":"00:05.235","Text":"where a_n is given by this formula,"},{"Start":"00:05.235 ","End":"00:10.725","Text":"for all n. We have to convert the general term a_n to a Riemann sum,"},{"Start":"00:10.725 ","End":"00:15.240","Text":"and using this, find the limit as n goes to infinity of a_n."},{"Start":"00:15.240 ","End":"00:18.465","Text":"So we have several ingredients in a Riemann sum."},{"Start":"00:18.465 ","End":"00:21.015","Text":"1 of them is the function,"},{"Start":"00:21.015 ","End":"00:26.310","Text":"and we\u0027ll let the function be natural log of x on the interval 0, 1,"},{"Start":"00:26.310 ","End":"00:28.995","Text":"except at x equals 0,"},{"Start":"00:28.995 ","End":"00:31.170","Text":"because natural log is not defined there,"},{"Start":"00:31.170 ","End":"00:33.840","Text":"so choose f of 0 to be whatever you want,"},{"Start":"00:33.840 ","End":"00:36.825","Text":"could be 0, could be 7, anything."},{"Start":"00:36.825 ","End":"00:43.220","Text":"We\u0027ll choose a partition p_n like so to make it equally spaced into n sub-intervals."},{"Start":"00:43.220 ","End":"00:45.560","Text":"As an intermediate point in each interval,"},{"Start":"00:45.560 ","End":"00:47.330","Text":"we choose the right end-point,"},{"Start":"00:47.330 ","End":"00:51.960","Text":"and the formula comes out c_i equals i over n. Also"},{"Start":"00:51.960 ","End":"00:56.580","Text":"note that all the Delta x_i are equal, the way we\u0027ve chosen it,"},{"Start":"00:56.580 ","End":"01:00.575","Text":"they\u0027re all 1 over n. The maximum of these,"},{"Start":"01:00.575 ","End":"01:01.940","Text":"which is the norm of p,"},{"Start":"01:01.940 ","End":"01:07.055","Text":"is also 1 over n. This tends to 0 when n goes to infinity."},{"Start":"01:07.055 ","End":"01:11.300","Text":"Now the Riemann sum for this partition and the function"},{"Start":"01:11.300 ","End":"01:16.080","Text":"and the collection of c_i is given by this sum,"},{"Start":"01:16.080 ","End":"01:19.365","Text":"the sum of f of c_i times delta x_i,"},{"Start":"01:19.365 ","End":"01:23.280","Text":"but all the delta x_i or 1 over n,"},{"Start":"01:23.280 ","End":"01:28.470","Text":"and that comes out f of i over n times 1 over n because that\u0027s what c_i is."},{"Start":"01:28.470 ","End":"01:31.800","Text":"Then we pull the 1 over n out in front,"},{"Start":"01:31.800 ","End":"01:35.465","Text":"and we get this, replacing f by the natural logarithm."},{"Start":"01:35.465 ","End":"01:39.980","Text":"We\u0027ve avoided the 0 here because we\u0027ve taken the right end-point,"},{"Start":"01:39.980 ","End":"01:43.045","Text":"so we don\u0027t have to use this, whatever."},{"Start":"01:43.045 ","End":"01:46.610","Text":"The sum, if we write it out with dot dot dot,"},{"Start":"01:46.610 ","End":"01:49.205","Text":"comes out to be this."},{"Start":"01:49.205 ","End":"01:53.660","Text":"Using the property of the logarithm,"},{"Start":"01:53.660 ","End":"01:57.650","Text":"we can get that this is equal to this,"},{"Start":"01:57.650 ","End":"02:01.655","Text":"because what we do is we take the logarithm of the product of these."},{"Start":"02:01.655 ","End":"02:07.160","Text":"Multiply 1 over n times 2 over n and so on up to 1 over n. In the numerator,"},{"Start":"02:07.160 ","End":"02:08.420","Text":"we get n factorial,"},{"Start":"02:08.420 ","End":"02:12.465","Text":"and then denominator we get n to the power of n. Now,"},{"Start":"02:12.465 ","End":"02:14.435","Text":"again using the properties of the logarithm,"},{"Start":"02:14.435 ","End":"02:19.805","Text":"we can put the power of 1 over n in the numerator and then denominator."},{"Start":"02:19.805 ","End":"02:25.250","Text":"Denominator leaves it as just n. Here we have n factorial to the 1 over n,"},{"Start":"02:25.250 ","End":"02:31.115","Text":"and so the limit as n goes to infinity of a_n,"},{"Start":"02:31.115 ","End":"02:36.005","Text":"I should have said, yeah, this is just equal to a_n."},{"Start":"02:36.005 ","End":"02:40.730","Text":"The limit is the limit as n goes to infinity,"},{"Start":"02:40.730 ","End":"02:43.820","Text":"which is norm of p goes to 0."},{"Start":"02:43.820 ","End":"02:50.000","Text":"This we get using the Riemann sum definition of the integral."},{"Start":"02:50.000 ","End":"02:56.740","Text":"We get the limit is the integral from 0-1 of natural log of xdx."},{"Start":"02:56.740 ","End":"02:59.720","Text":"This actually comes out to be an improper integral,"},{"Start":"02:59.720 ","End":"03:02.390","Text":"since the natural logarithm is not defined at 0,"},{"Start":"03:02.390 ","End":"03:04.179","Text":"it goes to minus infinity."},{"Start":"03:04.179 ","End":"03:06.650","Text":"Well, we could leave the answer like that,"},{"Start":"03:06.650 ","End":"03:08.930","Text":"but if you want a numerical result,"},{"Start":"03:08.930 ","End":"03:11.480","Text":"and maybe you\u0027re curious to know what it comes out,"},{"Start":"03:11.480 ","End":"03:13.700","Text":"I\u0027ll do the computations."},{"Start":"03:13.700 ","End":"03:16.325","Text":"It\u0027s optional."},{"Start":"03:16.325 ","End":"03:19.430","Text":"The integral of this being improper,"},{"Start":"03:19.430 ","End":"03:23.585","Text":"we take Epsilon tending to 0 from the right."},{"Start":"03:23.585 ","End":"03:26.480","Text":"And then the integral is from Epsilon to 1."},{"Start":"03:26.480 ","End":"03:32.299","Text":"The indefinite integral, that\u0027s a quick integration by parts, gives us this."},{"Start":"03:32.299 ","End":"03:35.845","Text":"You can differentiate this to verify that you get this,"},{"Start":"03:35.845 ","End":"03:39.135","Text":"and then the limit as Epsilon goes to 0 from the right,"},{"Start":"03:39.135 ","End":"03:41.690","Text":"plug in 1, and we\u0027ve got this,"},{"Start":"03:41.690 ","End":"03:43.945","Text":"plug-in epsilon, we have this."},{"Start":"03:43.945 ","End":"03:46.590","Text":"What we get is, this is minus 1,"},{"Start":"03:46.590 ","End":"03:51.180","Text":"and this is minus the limit as Epsilon goes to 0 of Epsilon,"},{"Start":"03:51.180 ","End":"03:52.575","Text":"natural log of Epsilon."},{"Start":"03:52.575 ","End":"03:55.534","Text":"The limit of this comes out to be 0."},{"Start":"03:55.534 ","End":"04:00.100","Text":"This limit is a well-known limit and it\u0027s equal to 0."},{"Start":"04:00.100 ","End":"04:02.385","Text":"The answer is minus 1."},{"Start":"04:02.385 ","End":"04:06.290","Text":"Again, you might want me to do some more computations,"},{"Start":"04:06.290 ","End":"04:08.180","Text":"show you that this is equal to 0,"},{"Start":"04:08.180 ","End":"04:10.430","Text":"and I\u0027ve prepared that as well."},{"Start":"04:10.430 ","End":"04:16.220","Text":"You know what, I think, I\u0027ll just leave it up here for you to verify."},{"Start":"04:16.220 ","End":"04:18.470","Text":"We use L\u0027Hopital\u0027s rule here,"},{"Start":"04:18.470 ","End":"04:21.755","Text":"and this shows why this limit is equal to 0."},{"Start":"04:21.755 ","End":"04:24.270","Text":"Okay. We\u0027re done."}],"ID":24779},{"Watched":false,"Name":"Exercise 23","Duration":"1m 37s","ChapterTopicVideoID":23838,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23838.jpeg","UploadDate":"2021-01-11T10:17:16.1470000","DurationForVideoObject":"PT1M37S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.000","Text":"In this exercise, we\u0027ll use the fundamental theorem of"},{"Start":"00:03.000 ","End":"00:06.750","Text":"calculus to prove the integration by parts formula."},{"Start":"00:06.750 ","End":"00:09.390","Text":"This is 1 way of displaying it,"},{"Start":"00:09.390 ","End":"00:12.750","Text":"sometimes we reverse the order of g and f prime,"},{"Start":"00:12.750 ","End":"00:18.495","Text":"and sometimes we use u and v as in the integral of"},{"Start":"00:18.495 ","End":"00:25.200","Text":"u dv equals uv minus the integral of v du."},{"Start":"00:25.200 ","End":"00:30.600","Text":"In the exercise, f and g are continuously differentiable on the interval a,"},{"Start":"00:30.600 ","End":"00:34.270","Text":"b, and we have to show that this formula holds."},{"Start":"00:34.270 ","End":"00:38.390","Text":"What we\u0027ll do is we\u0027ll let h equal the product f"},{"Start":"00:38.390 ","End":"00:42.785","Text":"times g. By the product rule for derivatives,"},{"Start":"00:42.785 ","End":"00:47.630","Text":"we get that h prime equals f prime g plus f g prime."},{"Start":"00:47.630 ","End":"00:53.600","Text":"Also h prime is continuous since all these 4 pieces are f prime,"},{"Start":"00:53.600 ","End":"00:55.220","Text":"g f and g prime."},{"Start":"00:55.220 ","End":"01:01.160","Text":"We can apply the second fundamental theorem of the calculus and get"},{"Start":"01:01.160 ","End":"01:09.095","Text":"that the integral of h prime on the interval a b is h of b minus h of a,"},{"Start":"01:09.095 ","End":"01:14.090","Text":"which I\u0027d like to write as h of x evaluated from a to b."},{"Start":"01:14.090 ","End":"01:19.855","Text":"Now here we\u0027ll replace h of x by what it\u0027s equal to from here,"},{"Start":"01:19.855 ","End":"01:25.220","Text":"this part here goes instead of h prime of x here,"},{"Start":"01:25.220 ","End":"01:27.940","Text":"and we get this."},{"Start":"01:27.940 ","End":"01:29.865","Text":"On the right-hand side,"},{"Start":"01:29.865 ","End":"01:33.470","Text":"h is just f times g. Then we just have to rearrange"},{"Start":"01:33.470 ","End":"01:38.370","Text":"the order and we get what we\u0027re trying to prove and we\u0027re done."}],"ID":24780},{"Watched":false,"Name":"Exercise 24","Duration":"3m 28s","ChapterTopicVideoID":23839,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23839.jpeg","UploadDate":"2021-01-11T10:19:03.6430000","DurationForVideoObject":"PT3M28S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.760","Text":"In this exercise, we\u0027ll prove the integration by substitution formula."},{"Start":"00:05.760 ","End":"00:08.475","Text":"Now, one way of phrasing this is as follows."},{"Start":"00:08.475 ","End":"00:12.480","Text":"We have a function Phi from an interval Alpha,"},{"Start":"00:12.480 ","End":"00:17.070","Text":"Beta to R, and it\u0027s continuously differentiable."},{"Start":"00:17.070 ","End":"00:18.525","Text":"We have another function f,"},{"Start":"00:18.525 ","End":"00:22.185","Text":"which is continuous on the range of Phi."},{"Start":"00:22.185 ","End":"00:30.750","Text":"In that case, the integral from Alpha to Beta of the composition f composed with Phi,"},{"Start":"00:30.750 ","End":"00:38.355","Text":"which is defined because f is continuous on the range of Phi times Phi prime of t,"},{"Start":"00:38.355 ","End":"00:40.230","Text":"dt is equal to,"},{"Start":"00:40.230 ","End":"00:41.935","Text":"and this is a substitution."},{"Start":"00:41.935 ","End":"00:46.600","Text":"This is like saying that x equals Phi of t,"},{"Start":"00:46.600 ","End":"00:49.935","Text":"and so dx is Phi prime of t, dt,"},{"Start":"00:49.935 ","End":"00:55.335","Text":"so we have here f of x dx and also we have to substitute the limits,"},{"Start":"00:55.335 ","End":"00:57.705","Text":"and if t goes from Alpha to Beta,"},{"Start":"00:57.705 ","End":"01:01.545","Text":"then x goes from Phi of Alpha to Phi of Beta."},{"Start":"01:01.545 ","End":"01:04.895","Text":"Start by defining a function big F of x,"},{"Start":"01:04.895 ","End":"01:08.720","Text":"which will be the integral from Phi of Alpha,"},{"Start":"01:08.720 ","End":"01:11.795","Text":"which is a constant to the variable x,"},{"Start":"01:11.795 ","End":"01:13.895","Text":"and we denote that letter u,"},{"Start":"01:13.895 ","End":"01:16.160","Text":"this will be f of u du."},{"Start":"01:16.160 ","End":"01:18.770","Text":"Now we apply the first fundamental theorem of"},{"Start":"01:18.770 ","End":"01:23.585","Text":"the calculus and we get that the derivative of F is"},{"Start":"01:23.585 ","End":"01:29.330","Text":"f. Now let\u0027s apply the chain rule and differentiate f of"},{"Start":"01:29.330 ","End":"01:35.300","Text":"Phi of t with respect to t. The chain rule says we take the outer derivative first,"},{"Start":"01:35.300 ","End":"01:38.450","Text":"that\u0027s f prime over the same Phi of t,"},{"Start":"01:38.450 ","End":"01:44.630","Text":"and then the inner derivative Phi prime of t. We can write this as f of Phi of t,"},{"Start":"01:44.630 ","End":"01:48.305","Text":"Phi prime of t, because big F prime is little f,"},{"Start":"01:48.305 ","End":"01:51.440","Text":"both f and Phi prime are continuous."},{"Start":"01:51.440 ","End":"01:55.250","Text":"Phi prime is continuous because Phi was continuously differentiable."},{"Start":"01:55.250 ","End":"01:59.405","Text":"So this d by dt of f of Phi of t is also"},{"Start":"01:59.405 ","End":"02:02.630","Text":"continuous because it\u0027s just a composition"},{"Start":"02:02.630 ","End":"02:06.580","Text":"of continuous functions, composition and product."},{"Start":"02:06.580 ","End":"02:13.325","Text":"The integral from Alpha to Beta of the derivative of f composed with Phi prime,"},{"Start":"02:13.325 ","End":"02:15.680","Text":"the prime is the same as the d by dt,"},{"Start":"02:15.680 ","End":"02:17.570","Text":"I forget which 1 is Newton,"},{"Start":"02:17.570 ","End":"02:20.430","Text":"which 1 is Leibniz notation."},{"Start":"02:20.530 ","End":"02:26.650","Text":"I forgot to say what we\u0027re doing now is using the second fundamental theorem of calculus,"},{"Start":"02:26.650 ","End":"02:33.280","Text":"so this is equal to f composed with Phi at Beta minus its value at Alpha."},{"Start":"02:33.280 ","End":"02:35.795","Text":"What we get is that this,"},{"Start":"02:35.795 ","End":"02:38.885","Text":"which is equal to this, because of this,"},{"Start":"02:38.885 ","End":"02:41.300","Text":"is equal to this minus this,"},{"Start":"02:41.300 ","End":"02:46.230","Text":"just rewrite it as that of composition as follows."},{"Start":"02:46.780 ","End":"02:50.010","Text":"We\u0027ll use this formula here."},{"Start":"02:50.010 ","End":"02:59.055","Text":"At 1 time, we have to plug in Phi of b and we get F of Phi of b,"},{"Start":"02:59.055 ","End":"03:02.340","Text":"which is just putting Phi of b here."},{"Start":"03:02.340 ","End":"03:03.660","Text":"That\u0027s what we have here,"},{"Start":"03:03.660 ","End":"03:07.005","Text":"replacing the x here by Phi of b."},{"Start":"03:07.005 ","End":"03:10.840","Text":"In the other 1, we replace x by Phi of Alpha,"},{"Start":"03:10.840 ","End":"03:13.520","Text":"but the integral of something to itself is 0,"},{"Start":"03:13.520 ","End":"03:15.590","Text":"so this term drops off."},{"Start":"03:15.590 ","End":"03:19.815","Text":"What we\u0027re left with is just this,"},{"Start":"03:19.815 ","End":"03:23.760","Text":"but we can replace the dummy variable u by the variable x"},{"Start":"03:23.760 ","End":"03:28.680","Text":"and then it\u0027s exactly what we needed and we are done."}],"ID":24781},{"Watched":false,"Name":"Exercise 25","Duration":"3m 16s","ChapterTopicVideoID":23840,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23840.jpeg","UploadDate":"2021-01-11T10:21:26.8170000","DurationForVideoObject":"PT3M16S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.555","Text":"In this exercise, we\u0027re going to prove the Leibniz rule."},{"Start":"00:04.555 ","End":"00:09.580","Text":"We have a continuous function f on the closed interval a, b,"},{"Start":"00:09.580 ","End":"00:13.570","Text":"and we have 2 differentiable functions: u and v."},{"Start":"00:13.570 ","End":"00:22.550","Text":"We are assuming that the ranges of u and v are contained in the interval a, b."},{"Start":"00:22.550 ","End":"00:28.105","Text":"What we have to show is that the derivative of this expression,"},{"Start":"00:28.105 ","End":"00:33.850","Text":"it\u0027s an integral where both the upper and lower bounds are functions of x"},{"Start":"00:33.850 ","End":"00:35.740","Text":"so that we get this formula."},{"Start":"00:35.740 ","End":"00:37.030","Text":"I won\u0027t read it out."},{"Start":"00:37.030 ","End":"00:38.870","Text":"I\u0027ll just point out a special case,"},{"Start":"00:38.870 ","End":"00:45.140","Text":"if u_(x) is a constant and v_(x) is identically equal to x,"},{"Start":"00:45.140 ","End":"00:48.020","Text":"then d_u by d_x is 0,"},{"Start":"00:48.020 ","End":"00:50.030","Text":"d_v by d_x is 1,"},{"Start":"00:50.030 ","End":"00:51.350","Text":"and v_(x) is x,"},{"Start":"00:51.350 ","End":"00:53.240","Text":"so we just get f_(x),"},{"Start":"00:53.240 ","End":"00:57.215","Text":"and that\u0027s the first fundamental theorem of the calculus."},{"Start":"00:57.215 ","End":"01:03.270","Text":"This is a generalization where the limits of integration are variable."},{"Start":"01:03.410 ","End":"01:07.625","Text":"Let\u0027s do part of it first."},{"Start":"01:07.625 ","End":"01:11.645","Text":"We\u0027ll take the lower limit to be a constant a,"},{"Start":"01:11.645 ","End":"01:13.730","Text":"and we\u0027ll just vary the upper limit."},{"Start":"01:13.730 ","End":"01:17.300","Text":"With the help of this, we\u0027ll go to the full general case."},{"Start":"01:17.300 ","End":"01:24.470","Text":"We start by defining a capital F_(x) to be the integral from a to x of f,"},{"Start":"01:24.470 ","End":"01:27.379","Text":"and by the first fundamental theorem of the calculus,"},{"Start":"01:27.379 ","End":"01:33.500","Text":"the derivative of big F is little f. This expression where we replaced x by"},{"Start":"01:33.500 ","End":"01:40.220","Text":"v_(x) is just F of v_(x). Differentiate both sides with respect to x,"},{"Start":"01:40.220 ","End":"01:47.505","Text":"and the right-hand side by the chain rule is the derivative of big F of v_(x),"},{"Start":"01:47.505 ","End":"01:50.010","Text":"and then timed d_v by d_x."},{"Start":"01:50.010 ","End":"01:55.140","Text":"But F prime is little f, we wrote here."},{"Start":"01:55.140 ","End":"01:58.410","Text":"So we\u0027ve got to this point here."},{"Start":"01:58.410 ","End":"02:03.530","Text":"Now, we can use this result which is like"},{"Start":"02:03.530 ","End":"02:09.050","Text":"halfway there to get the full result where we replace the lower limit by u of x."},{"Start":"02:09.050 ","End":"02:12.815","Text":"Note that if I take the integral from u to v_(x),"},{"Start":"02:12.815 ","End":"02:20.360","Text":"it\u0027s the same as the integral from a to v minus the integral from a to u."},{"Start":"02:20.360 ","End":"02:21.410","Text":"If you don\u0027t see this right away,"},{"Start":"02:21.410 ","End":"02:26.780","Text":"just put a plus here and switch the a with the u."},{"Start":"02:26.780 ","End":"02:31.880","Text":"What we get is from u to a plus from a to v,"},{"Start":"02:31.880 ","End":"02:36.620","Text":"that means all the way from u to v. So the derivative of"},{"Start":"02:36.620 ","End":"02:43.100","Text":"this is the derivative of this minus the derivative of this."},{"Start":"02:43.100 ","End":"02:50.750","Text":"Above, we\u0027ve shown that this is equal to f of v_(x) times d_v by d_x."},{"Start":"02:50.750 ","End":"02:54.290","Text":"There\u0027s no essential difference between v and u."},{"Start":"02:54.290 ","End":"03:00.740","Text":"They are both given to be differentiable on the interval a, b."},{"Start":"03:00.740 ","End":"03:04.580","Text":"Same formula for v will apply to u,"},{"Start":"03:04.580 ","End":"03:06.580","Text":"and we get this expression."},{"Start":"03:06.580 ","End":"03:07.955","Text":"If you go back,"},{"Start":"03:07.955 ","End":"03:13.715","Text":"you\u0027ll see that what we had to show was precisely that this equals this,"},{"Start":"03:13.715 ","End":"03:16.920","Text":"and so we are done.f"}],"ID":24782},{"Watched":false,"Name":"Exercise 26","Duration":"3m 7s","ChapterTopicVideoID":23841,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/23841.jpeg","UploadDate":"2021-01-11T10:23:25.2930000","DurationForVideoObject":"PT3M7S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.660","Text":"In this exercise, we have the function f defined for x bigger or equal to 1,"},{"Start":"00:06.660 ","End":"00:09.085","Text":"and it\u0027s given by the formula f of x equals the"},{"Start":"00:09.085 ","End":"00:13.570","Text":"integral from 1 to x natural log of t over 1 plus t dt."},{"Start":"00:13.570 ","End":"00:15.960","Text":"We have to solve the equation"},{"Start":"00:15.960 ","End":"00:19.680","Text":"f of x plus f of 1/x equals 2."},{"Start":"00:19.680 ","End":"00:22.905","Text":"The thing is that it\u0027s not easy to integrate"},{"Start":"00:22.905 ","End":"00:25.119","Text":"natural log of t over 1 plus t."},{"Start":"00:25.119 ","End":"00:28.695","Text":"We\u0027re going to use some trick."},{"Start":"00:28.695 ","End":"00:32.040","Text":"Let\u0027s see what f of 1/x is."},{"Start":"00:32.040 ","End":"00:34.583","Text":"Well, it\u0027s the same as this except that we put"},{"Start":"00:34.583 ","End":"00:37.815","Text":"1/x in the upper limit of integration."},{"Start":"00:37.815 ","End":"00:40.395","Text":"Let\u0027s make a substitution now."},{"Start":"00:40.395 ","End":"00:47.735","Text":"Let\u0027s say that tone is 1/y and we get dt is minus 1/y^2 dy."},{"Start":"00:47.735 ","End":"00:50.423","Text":"Then we have to substitute the limits also."},{"Start":"00:50.423 ","End":"00:52.472","Text":"When t is 1, y is 1,"},{"Start":"00:52.472 ","End":"00:56.605","Text":"and when t is 1/x, then y is x."},{"Start":"00:56.605 ","End":"01:02.105","Text":"What we get is that f of 1/x is the integral from 1 to x."},{"Start":"01:02.105 ","End":"01:06.485","Text":"Here, we replace by the substitution and we get this,"},{"Start":"01:06.485 ","End":"01:08.810","Text":"and we simplify it a bit."},{"Start":"01:08.810 ","End":"01:12.440","Text":"Natural log of 1/y is minus log y,"},{"Start":"01:12.440 ","End":"01:16.085","Text":"which absorbs this minus 1 and makes it plus."},{"Start":"01:16.085 ","End":"01:20.810","Text":"The denominator, we just multiply out and we get y^2 plus y,"},{"Start":"01:20.810 ","End":"01:22.910","Text":"which is y, y plus 1."},{"Start":"01:22.910 ","End":"01:26.930","Text":"We have f of x, which is this that was given."},{"Start":"01:26.930 ","End":"01:31.430","Text":"We now have f of 1/x, which is this."},{"Start":"01:31.430 ","End":"01:35.720","Text":"Now we can add them together and get this plus this."},{"Start":"01:35.720 ","End":"01:37.440","Text":"We can combine these,"},{"Start":"01:37.440 ","End":"01:41.570","Text":"and these are mostly the same except that this integrant"},{"Start":"01:41.570 ","End":"01:44.570","Text":"is like this one multiplied by 1/t,"},{"Start":"01:44.570 ","End":"01:47.110","Text":"so we get this expression."},{"Start":"01:47.110 ","End":"01:50.300","Text":"If we put this in a common denominator,"},{"Start":"01:50.300 ","End":"01:53.720","Text":"we get 1 plus t over t."},{"Start":"01:53.720 ","End":"01:58.730","Text":"The 1 plus t cancels and that gives us t in the denominator instead."},{"Start":"01:58.730 ","End":"02:04.070","Text":"This is easy to integrate because the derivative of log t is 1 over t."},{"Start":"02:04.070 ","End":"02:06.200","Text":"A simple substitution,"},{"Start":"02:06.200 ","End":"02:09.875","Text":"we see that this is 1/2 log t squared."},{"Start":"02:09.875 ","End":"02:12.805","Text":"Then we have to put in the limits of integration."},{"Start":"02:12.805 ","End":"02:16.005","Text":"This should be a 1, not a 0 of course."},{"Start":"02:16.005 ","End":"02:21.965","Text":"This comes out to be just 1/2 natural log of x squared."},{"Start":"02:21.965 ","End":"02:23.945","Text":"We have this equality,"},{"Start":"02:23.945 ","End":"02:26.480","Text":"but we\u0027re given the equation,"},{"Start":"02:26.480 ","End":"02:29.270","Text":"this plus this equals 2."},{"Start":"02:29.270 ","End":"02:35.180","Text":"We can compare the right-hand sides and get 1/2 log of x squared is 2."},{"Start":"02:35.180 ","End":"02:36.380","Text":"That\u0027s easy to solve."},{"Start":"02:36.380 ","End":"02:40.655","Text":"We get natural log of x is plus or minus the square root of 4"},{"Start":"02:40.655 ","End":"02:42.909","Text":"because we bring the 2 to the other side,"},{"Start":"02:42.909 ","End":"02:44.900","Text":"and that\u0027s plus or minus 2."},{"Start":"02:44.900 ","End":"02:50.870","Text":"That means that x will either be e to the power of 2 or e to the power of minus 2,"},{"Start":"02:50.870 ","End":"02:54.535","Text":"but x is bigger or equal to 1,"},{"Start":"02:54.535 ","End":"02:59.280","Text":"and so it can\u0027t be e to the minus 2 because that\u0027s less than 1."},{"Start":"02:59.280 ","End":"03:01.490","Text":"It has to be equal to e^2."},{"Start":"03:01.490 ","End":"03:05.503","Text":"The answer is that x equals e^2."},{"Start":"03:05.503 ","End":"03:08.040","Text":"We\u0027re done."}],"ID":24783},{"Watched":false,"Name":"Exercise 27","Duration":"2m 27s","ChapterTopicVideoID":24876,"CourseChapterTopicPlaylistID":257170,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/24876.jpeg","UploadDate":"2021-02-23T16:33:34.4170000","DurationForVideoObject":"PT2M27S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.870","Text":"In this exercise, we\u0027re given 2 functions, f and g,"},{"Start":"00:03.870 ","End":"00:08.670","Text":"continuous on the interval from a to infinity and differentiable."},{"Start":"00:08.670 ","End":"00:12.240","Text":"Now suppose f of a equals g of a,"},{"Start":"00:12.240 ","End":"00:19.125","Text":"but f prime is less than or equal to g prime for all x to the right of a,"},{"Start":"00:19.125 ","End":"00:22.680","Text":"we have to prove that f of x is less than or equal to g"},{"Start":"00:22.680 ","End":"00:27.060","Text":"of x for all x bigger or equal to a."},{"Start":"00:27.060 ","End":"00:30.315","Text":"For the solution, we take the difference,"},{"Start":"00:30.315 ","End":"00:34.110","Text":"let h be g minus f,"},{"Start":"00:34.110 ","End":"00:38.260","Text":"and let\u0027s take x bigger than a."},{"Start":"00:38.260 ","End":"00:43.355","Text":"Then because f and g are continuous and differentiable,"},{"Start":"00:43.355 ","End":"00:48.650","Text":"so is h. Also if f of a equals g of a,"},{"Start":"00:48.650 ","End":"00:50.180","Text":"then h of a is 0,"},{"Start":"00:50.180 ","End":"00:53.720","Text":"it\u0027s the difference and also g minus f prime is"},{"Start":"00:53.720 ","End":"00:57.670","Text":"h prime that will be bigger or equal to 0."},{"Start":"00:57.670 ","End":"00:59.280","Text":"Yeah, h of a is 0,"},{"Start":"00:59.280 ","End":"01:02.355","Text":"h prime is bigger than 0."},{"Start":"01:02.355 ","End":"01:08.209","Text":"We can use the mean value theorem on the interval a x."},{"Start":"01:08.209 ","End":"01:10.205","Text":"You have continuous on the closed interval,"},{"Start":"01:10.205 ","End":"01:12.190","Text":"differentiable on the open interval,"},{"Start":"01:12.190 ","End":"01:15.770","Text":"it meets all the conditions for the Lagrange mean value theorem,"},{"Start":"01:15.770 ","End":"01:20.930","Text":"so h of x minus h a is x minus a times the"},{"Start":"01:20.930 ","End":"01:26.959","Text":"derivative at some point c in the open interval from a to x."},{"Start":"01:26.959 ","End":"01:32.030","Text":"Now, h prime is bigger than 0 or bigger or"},{"Start":"01:32.030 ","End":"01:38.014","Text":"equal to 0 rather and x minus a is bigger than 0."},{"Start":"01:38.014 ","End":"01:46.720","Text":"In any event, this is bigger or equal to 0 so h of x is bigger or equal to h of a."},{"Start":"01:46.720 ","End":"01:48.350","Text":"If the differences bigger or equal to 0,"},{"Start":"01:48.350 ","End":"01:54.150","Text":"this is bigger or equal to this and h of a is 0."},{"Start":"01:54.150 ","End":"01:57.890","Text":"H of x is bigger or equal to 0,"},{"Start":"01:57.890 ","End":"01:59.815","Text":"for x bigger than a."},{"Start":"01:59.815 ","End":"02:02.810","Text":"In fact, we can extend the bigger than a,"},{"Start":"02:02.810 ","End":"02:04.610","Text":"to bigger or equal to,"},{"Start":"02:04.610 ","End":"02:07.550","Text":"because at a itself, h is 0,"},{"Start":"02:07.550 ","End":"02:14.240","Text":"so it\u0027s still bigger or equal to 0 and that means that for x bigger or equal to a,"},{"Start":"02:14.240 ","End":"02:19.070","Text":"we have h of x bigger or equal to 0, just summarizing this."},{"Start":"02:19.070 ","End":"02:22.325","Text":"This means that g is bigger or equal to f,"},{"Start":"02:22.325 ","End":"02:28.080","Text":"because h is g minus f. This is what we had to show and we\u0027re done."}],"ID":25789}],"Thumbnail":null,"ID":257170}]

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