1 00:00:00,6 --> 00:00:04,990 [MUSIC]. 2 00:00:04,990 --> 00:00:08,528 We've seen a bunch of arguments, or perspectives, on why the fundamental 3 00:00:08,528 --> 00:00:12,124 theorem of calculus should be true, but we've also paid a high price in this 4 00:00:12,124 --> 00:00:15,270 course. We've been doing alot of things 5 00:00:15,270 --> 00:00:17,720 rigorously, and we've done the squeeze theorem. 6 00:00:17,720 --> 00:00:21,490 Well with the squeeze theorem in hand, it's possible to give a rigorous formal 7 00:00:21,490 --> 00:00:24,985 proof of the fundamental theorem of calculus. 8 00:00:24,985 --> 00:00:28,85 Well just to remind you this is actually what we're going to prove rigorously 9 00:00:28,85 --> 00:00:30,364 here. I'm going to start with some function 10 00:00:30,364 --> 00:00:32,978 from this closed interval ab to the real numbers. 11 00:00:32,978 --> 00:00:36,790 It's a continuous function, so it's actually integrable. 12 00:00:36,790 --> 00:00:40,820 then I write down the accumulation function big F which takes the integral 13 00:00:40,820 --> 00:00:46,398 from a to x of the function little f. And the big claim here, is that the 14 00:00:46,398 --> 00:00:51,300 derivative of the accumulation function is the function little f. 15 00:00:51,300 --> 00:00:54,336 The statement of the fundamental theorem of calculus is about a derivative, so 16 00:00:54,336 --> 00:00:57,170 we're really trying to do two different things. 17 00:00:57,170 --> 00:01:00,890 I'm trying to show that a limit of a difference quotient exists, that the 18 00:01:00,890 --> 00:01:04,858 derivative exists, and then second, I'm making a claim as to the value of that 19 00:01:04,858 --> 00:01:10,734 derivative. specifically, I want to show that this 20 00:01:10,734 --> 00:01:16,374 derivative, which is defined to be the limit as h approaches 0 of big F of x 21 00:01:16,374 --> 00:01:27,50 plus h minus big F of x over h, I want to show that this is equal to little f of x. 22 00:01:27,50 --> 00:01:30,869 But to make matters a little bit easier, I'm only going to consider the situation 23 00:01:30,869 --> 00:01:34,550 when h is positive. So I'll put a little plus there. 24 00:01:34,550 --> 00:01:39,138 Let's rewrite that difference of two accumulation functions, as a single 25 00:01:39,138 --> 00:01:42,270 definite integral. Well how so? 26 00:01:42,270 --> 00:01:46,997 Well let's just write this down. So, this is the limit as h approaches 0 27 00:01:46,997 --> 00:01:53,310 from the right of big F of x plus h, that's the integral from a to x plus h of 28 00:01:53,310 --> 00:01:59,302 f of t dt minus big F of x, which is the integral from a to x of f of t dt all 29 00:01:59,302 --> 00:02:08,665 over h. But now this difference of integrals, can 30 00:02:08,665 --> 00:02:14,696 be written as a single integral. This is the limit as h approaches 0 from 31 00:02:14,696 --> 00:02:20,328 the right, of, well, if I integrate from a to x plus h and subtract the integral 32 00:02:20,328 --> 00:02:28,960 from a to x, what's left over, is the integral from x to x plus h. 33 00:02:28,960 --> 00:02:34,344 Of f of t dt all over h. Let's think about how large and how small 34 00:02:34,344 --> 00:02:39,350 that integral can be. Well, first some notation. 35 00:02:39,350 --> 00:02:46,78 I'll define little m of h to be the minimum value of the function f on the 36 00:02:46,78 --> 00:02:55,58 interval between x and x plus h. And I'll define big M of h, to be the 37 00:02:55,58 --> 00:03:02,830 maximum value of the function little f on the interval, well the same interval, a 38 00:03:02,830 --> 00:03:11,477 closed interval from x to x plus h. Now you might be wondering, how do I know 39 00:03:11,477 --> 00:03:14,150 that there is a minimum and a maximum value? 40 00:03:14,150 --> 00:03:18,836 Well, little f assumed to continuous, so by the extreme value theorem there is a 41 00:03:18,836 --> 00:03:25,770 minimum and a maximum value on a closed interval, and these are closed intervals. 42 00:03:25,770 --> 00:03:31,215 This gives us a bound on the integral. Specifically let's think about the 43 00:03:31,215 --> 00:03:36,400 integral from x to x plus h of the function little f, and since the function 44 00:03:36,400 --> 00:03:42,520 little f has a maximum value of big M on this interval, this integral is no bigger 45 00:03:42,520 --> 00:03:51,924 than big M of h times h. And conversely, all right, the minimum 46 00:03:51,924 --> 00:03:58,0 value is little m of h, and that means that this integral is at least as big as 47 00:03:58,0 --> 00:04:04,440 little m of h times h. Now let's divide both sides by h, and 48 00:04:04,440 --> 00:04:08,220 that doesn't effect the inequality because I'm only really thinking about 49 00:04:08,220 --> 00:04:14,612 the case when h is positive. Well that gives me little m of h is less 50 00:04:14,612 --> 00:04:21,220 than or equal to the integral from x to x plus h, of f of t dt, over h, is less 51 00:04:21,220 --> 00:04:30,744 than or equal to big M of h. Now, we apply the squeeze theorem. 52 00:04:30,744 --> 00:04:34,940 Well how does that go? Well let's think about the limit as h 53 00:04:34,940 --> 00:04:39,282 approaches 0 from the right of little m of h. 54 00:04:39,282 --> 00:04:43,308 Now, remember what little m is, little m is the minimum value of the function f on 55 00:04:43,308 --> 00:04:49,60 the interval between x and x plus h. And as h approaches 0 from the right, 56 00:04:49,60 --> 00:04:54,906 this minimum value of f It's just the value of f at the point x. 57 00:04:54,906 --> 00:04:59,928 And for the same reasoning, the limit as h approaches 0 from the right of the 58 00:04:59,928 --> 00:05:05,193 maximum value, big M, the maximum value of f on that interval is also little f of 59 00:05:05,193 --> 00:05:11,484 x. So what I've got here is this inequality, 60 00:05:11,484 --> 00:05:16,344 and the left-hand side and the right-hand side both have equal limit as h 61 00:05:16,344 --> 00:05:21,366 approaches 0 from the right, and that means that this middle term also has a 62 00:05:21,366 --> 00:05:28,691 limit that exists, and equals f of x. Let me write that down. 63 00:05:28,691 --> 00:05:34,486 So what I can conclude, is that the limit, as h approaches 0 from the right, 64 00:05:34,486 --> 00:05:40,281 of the integral from x to x plus h, of f of t dt divided by h, well that's equal 65 00:05:40,281 --> 00:05:46,494 to f of x. And if I play the same game but think 66 00:05:46,494 --> 00:05:53,320 about h approaching 0 from the left. I can get that, that limit is also f of 67 00:05:53,320 --> 00:06:01,412 x, and consequently, the two sided limit, as h approaches 0, of this term, is equal 68 00:06:01,412 --> 00:06:09,431 to f of x. And so we've verified that the derivative 69 00:06:09,431 --> 00:06:13,487 exist, and it's equal to f of x.