1 00:00:00,12 --> 00:00:05,324 [MUSIC]. 2 00:00:05,324 --> 00:00:09,580 I want to begin with a statement of he fundamental theorem of calculus. 3 00:00:09,580 --> 00:00:12,935 Suppose the function little f, from the closed interval a to b, to the real 4 00:00:12,935 --> 00:00:17,394 numbers, is a continuous function. And let's define a function big F, to be 5 00:00:17,394 --> 00:00:20,535 the accumulation function. Right, that's the integral. 6 00:00:20,535 --> 00:00:24,882 From a to x of the function on the left well then the function big F turns out to 7 00:00:24,882 --> 00:00:29,505 be continuous in the closed interval a to b differentiable on the open interval a 8 00:00:29,505 --> 00:00:36,490 to b and this is the big point the derivative of big F is little f. 9 00:00:36,490 --> 00:00:39,665 We've seen a bit of evidence for this already. 10 00:00:39,665 --> 00:00:41,680 And suppose I got the accumulation function. 11 00:00:41,680 --> 00:00:45,110 So big F of x is the integral from a to x of f of tdt. 12 00:00:45,110 --> 00:00:49,10 And now I can ask, how is the accumulation function changing? 13 00:00:49,10 --> 00:00:53,234 Well, the accumulation function is increasing, meaning it's derivative is 14 00:00:53,234 --> 00:00:57,770 positive if the function that I'm integrating is positive. 15 00:00:57,770 --> 00:01:01,552 And the accumulation function is decreasing, meaning the derivative is 16 00:01:01,552 --> 00:01:05,442 negative. If the function that I'm integrating is 17 00:01:05,442 --> 00:01:08,664 negative. And this suggests that the derivative of 18 00:01:08,664 --> 00:01:12,136 the accumulation function and the function that I'm integrating might be 19 00:01:12,136 --> 00:01:16,570 related in some way. Here's a graph that's particularly 20 00:01:16,570 --> 00:01:20,180 compelling. So, I'll draw the coordinate plane, 21 00:01:20,180 --> 00:01:25,220 here's the y axis, here's the t axis in this case and I'll draw some Random 22 00:01:25,220 --> 00:01:32,30 looking graph. That'll be the graph y equals f of t. 23 00:01:32,30 --> 00:01:37,80 And then I'll pick some point. I'll call it a and some other point x. 24 00:01:37,80 --> 00:01:42,580 And then I'll look at the area under the graph between a and x, rigth? 25 00:01:42,580 --> 00:01:47,368 And that's the integral from a to x of f of t d t. 26 00:01:47,368 --> 00:01:52,680 Right that area is the accumulation function evaluated x. 27 00:01:52,680 --> 00:01:56,700 Now what's the derivative of the accumulation function. 28 00:01:56,700 --> 00:02:01,254 What got the accumulation function, right f(x) equals the interval from a to x that 29 00:02:01,254 --> 00:02:05,970 f(t)dt and I don't know how that changes when x changes. 30 00:02:05,970 --> 00:02:09,340 So lets make a small change in x. Lets go from x. 31 00:02:09,340 --> 00:02:14,758 To say x plus h and then ask how does the area in this region compare to the area 32 00:02:14,758 --> 00:02:21,920 in this much larger region. What's big F of x plus h minus big F of 33 00:02:21,920 --> 00:02:25,726 x? Yeah, so what's F of x plus h right 34 00:02:25,726 --> 00:02:31,887 that's the area. From a to x plus h, minus F of x, big F 35 00:02:31,887 --> 00:02:36,629 of x. That's the area from a to x, and that 36 00:02:36,629 --> 00:02:41,929 difference, right. That difference is just the area inside 37 00:02:41,929 --> 00:02:46,68 here. That little leftover piece is practically 38 00:02:46,68 --> 00:02:51,335 a rectangle of height, f of x, little f of x and width h. 39 00:02:51,335 --> 00:02:53,710 I mean, this region is not quite a rectangle. 40 00:02:53,710 --> 00:02:58,60 But it's awfully close to a rectangle of, width h. 41 00:02:58,60 --> 00:03:03,602 And area f of x plus h minus f of x. Now, remember, by goal was to 42 00:03:03,602 --> 00:03:08,598 differentiate the accumulation function. And all I've got is this rectangle. 43 00:03:08,598 --> 00:03:16,330 Whose width is about h, and whose height, you think about it, it's about f of x. 44 00:03:16,330 --> 00:03:18,990 So what's up with this rectangle? Alright? 45 00:03:18,990 --> 00:03:23,828 I mean what I've really got here is a rectangle and its area is about f of x 46 00:03:23,828 --> 00:03:29,310 plus h minus f of x, this accumulation function. 47 00:03:29,310 --> 00:03:32,436 It's width is h, and it's height is about f of x. 48 00:03:32,436 --> 00:03:38,270 Well I've got a rectangle with this area and this width and, and this height. 49 00:03:38,270 --> 00:03:41,62 Alright, what do I know? I know its height times its width is its 50 00:03:41,62 --> 00:03:46,770 area, so if I take its area and divide by its width, I should get its height. 51 00:03:46,770 --> 00:03:49,930 Now all this is just approximate, but let's write that down. 52 00:03:49,930 --> 00:03:56,930 I The area of this you know, rectangle is f of x plus h minus f of x. 53 00:03:56,930 --> 00:04:01,990 And if I divide by the width, I should get, the height. 54 00:04:01,990 --> 00:04:07,42 What does that mean after I take a limit? Oh of course this is only approximately 55 00:04:07,42 --> 00:04:11,784 true, but if I take a limit, right. If I take a limit as h approaches 0. 56 00:04:11,784 --> 00:04:16,992 This gives me the derivative of the accumulation function and that will in 57 00:04:16,992 --> 00:04:23,800 fact be equal to the function's value. And that is the whole game. 58 00:04:23,800 --> 00:04:29,385 The accumulation function F is an antiderivative of f. 59 00:04:29,385 --> 00:04:32,740 Well this, I mean this is really remarkable. 60 00:04:32,740 --> 00:04:36,221 For instance you might be wondering what's an anti-derivative of this 61 00:04:36,221 --> 00:04:40,450 function, little f of x equals e to the negative x squared? 62 00:04:40,450 --> 00:04:45,509 Well if a functions continuous then we can find an anti-derivative. 63 00:04:45,509 --> 00:04:48,370 The anti-dirivitive is the accumulation function. 64 00:04:48,370 --> 00:04:51,898 So an anti-derivative of either the negative x squared is the accumulation 65 00:04:51,898 --> 00:04:54,920 function. I can define a function big f of x to be 66 00:04:54,920 --> 00:05:00,390 the intregal from a to x, for some number a, of e to the negative t squared d t. 67 00:05:00,390 --> 00:05:05,13 And if I differentiate this accumulation function, I get back e to the negative x 68 00:05:05,13 --> 00:05:08,846 squared. So I found an anti-derivative. 69 00:05:08,846 --> 00:05:12,202 For this function. Well I can't write it down any more 70 00:05:12,202 --> 00:05:14,304 nicely. Because I can't write down an 71 00:05:14,304 --> 00:05:18,0 anti-derivative for e to the negative x squared using the functions that I have 72 00:05:18,0 --> 00:05:26,300 at hand. But nevertheless, the anti-derivative is 73 00:05:26,300 --> 00:05:32,33 the accumulation function.