1 00:00:05,350 --> 00:00:10,200 We've seen what happens when we differentiate the accumulation function. 2 00:00:10,200 --> 00:00:12,550 That's the fundamental theorem of calculus. 3 00:00:12,550 --> 00:00:17,918 If I take the derivative of the integral from a to b of f of x dx, well what do I 4 00:00:17,918 --> 00:00:22,470 get? Well, it's just f. 5 00:00:22,470 --> 00:00:24,931 Of b, right? The rate of the change of the 6 00:00:24,931 --> 00:00:27,906 accumulation function is the functions value. 7 00:00:27,906 --> 00:00:31,322 But, what would happen if I were to differentiate with respect to the left 8 00:00:31,322 --> 00:00:35,910 hand end point instead of with respect to the right hand end point? 9 00:00:35,910 --> 00:00:39,59 By which I mean. What's the derivative with respect to a 10 00:00:39,59 --> 00:00:42,668 of the integral from a to b of f of x dx. Right. 11 00:00:42,668 --> 00:00:47,90 What happens if instead of diferentiating with respect to the top end point, I'm 12 00:00:47,90 --> 00:00:51,470 diferentiating with respect to the bottom end point? 13 00:00:51,470 --> 00:00:57,320 I'm asking how is this function as a function of a changing? 14 00:00:57,320 --> 00:01:01,500 Let's think about the graph. Let's draw my coordinate axes. 15 00:01:01,500 --> 00:01:05,490 And I'll draw some random looking function. 16 00:01:05,490 --> 00:01:11,312 And I'll pick some points a and b. And the integral from a to b calculates 17 00:01:11,312 --> 00:01:15,47 the area in here. And I want to know how does that integral 18 00:01:15,47 --> 00:01:18,880 change when I wiggle a? I'm asking to differentiate the integral 19 00:01:18,880 --> 00:01:22,50 with respect to this. Left-hand endpoint. 20 00:01:22,50 --> 00:01:25,792 So let's wiggle the left-hand endpoint. Let's move it over a little bit to a plus 21 00:01:25,792 --> 00:01:28,129 h. I'm imagining h is very small. 22 00:01:28,129 --> 00:01:31,342 Alright, and I want to know, how does the integral change? 23 00:01:31,342 --> 00:01:35,563 Well the quantity that calculates the absolute change in the integral is this. 24 00:01:35,563 --> 00:01:39,140 What's this thing here? This is the integral from a plus h to b. 25 00:01:39,140 --> 00:01:43,560 And this thing here is the integral from a to b, so this difference is telling me 26 00:01:43,560 --> 00:01:48,490 how the integral changes when I replace a by a plus h. 27 00:01:48,490 --> 00:01:52,648 Now if you think about it, what this is really calculating is, you know, related 28 00:01:52,648 --> 00:01:56,252 to the integral from just a plus h, right. 29 00:01:56,252 --> 00:02:01,960 This integral is calculating this area, and subtracting this larger area. 30 00:02:01,960 --> 00:02:06,160 So the difference is really just this area in here between A and A+H but it 31 00:02:06,160 --> 00:02:10,640 comes with a negative sign because I'm subtracting this smaller area and I'm 32 00:02:10,640 --> 00:02:16,945 subtracting now this larger area. So I've got this negative sign here. 33 00:02:16,945 --> 00:02:22,711 Now this region, if h is small enough, is practically a rectangle, and it's 34 00:02:22,711 --> 00:02:30,286 practically a rectangle of width h and height let's say f of a. 35 00:02:30,286 --> 00:02:35,746 So, that means that this difference is at least approximately just h times the 36 00:02:35,746 --> 00:02:39,955 function's value at a. Now how does that help? 37 00:02:39,955 --> 00:02:44,933 Remember what I'm trying to calculate. I, I'm trying to differentiate the 38 00:02:44,933 --> 00:02:49,879 interval from a to be with respect to a. That means I'm trying to take the limit 39 00:02:49,879 --> 00:02:54,534 of this difference quotient. But the numerator here, we just saw, is 40 00:02:54,534 --> 00:03:01,344 approximately, negative h times f of a. And that means in the limit I expect to 41 00:03:01,344 --> 00:03:06,84 get an answer of just negative f of a, alright, these hs will cancel in the 42 00:03:06,84 --> 00:03:12,10 limit, and that's exactly what I hope for, right? 43 00:03:12,10 --> 00:03:16,234 The derivitive of the integral from a to b with respect to the left hand endpoint 44 00:03:16,234 --> 00:03:19,910 a is negative f of a. So I can summarize this. 45 00:03:19,910 --> 00:03:24,706 So I can summarize this. The derivative with respect to a of the 46 00:03:24,706 --> 00:03:29,820 integral from a to b of f of x dx is negative f of a. 47 00:03:29,820 --> 00:03:35,20 This fact coheres with a certain convention about integration. 48 00:03:35,20 --> 00:03:39,870 The convention that we use is that if we integrate from a to b. 49 00:03:39,870 --> 00:03:46,566 the function f of x dx that's negative the interval from b to a of f of x dx. 50 00:03:46,566 --> 00:03:52,611 So if you integrate the wrong way, so to speak, we want to count that as negative 51 00:03:52,611 --> 00:03:54,755 area. 52 00:03:54,755 --> 00:03:58,140 >> Now in light of this convention, what do we know? 53 00:03:58,140 --> 00:04:02,55 >> So d da of the integral from a to b of f 54 00:04:02,55 --> 00:04:14,260 of x dx is d da of this, negative the integral from b to a of f of x dx. 55 00:04:14,260 --> 00:04:17,620 But now this is the derivative of the top end point which is just the usual 56 00:04:17,620 --> 00:04:25,610 fundamental theorem of calculus. So this is negative f of a. 57 00:04:25,610 --> 00:04:29,250 So the upshot here is that differentiating the integral from a to b, 58 00:04:29,250 --> 00:04:33,410 with respect to the left hand or bottom end point, Is negative the function's 59 00:04:33,410 --> 00:04:37,316 value. And that make sense because if you 60 00:04:37,316 --> 00:04:41,270 increase the left hand endpoint that decreases the area. 61 00:04:41,270 --> 00:04:45,68 Alright, so it seems reasonable that a negative sign should be popping up there. 62 00:04:45,68 --> 00:04:48,968 And the cool thing is that, that fact is cohering with a convention about 63 00:04:48,968 --> 00:04:56,12 integration. That compared to integrating the usual 64 00:04:56,12 --> 00:05:06,473 way, if you integrate the wrong way you introduce a negative sign.