1 00:00:00,000 --> 00:00:07,620 2 00:00:07,620 --> 00:00:09,570 PROFESSOR: Welcome back to recitation. 3 00:00:09,570 --> 00:00:12,720 In this video I want us to work on the following problem. 4 00:00:12,720 --> 00:00:14,635 So we're going to assume little f is 5 00:00:14,635 --> 00:00:16,400 a continuous function. 6 00:00:16,400 --> 00:00:19,660 And if we know that the integral from 0 to x of f of t 7 00:00:19,660 --> 00:00:24,880 dt is equal to x squared sine pi x, I want us to 8 00:00:24,880 --> 00:00:27,940 find little f of 2. 9 00:00:27,940 --> 00:00:28,930 So I'm going to give you a little while 10 00:00:28,930 --> 00:00:30,080 to work on the problem. 11 00:00:30,080 --> 00:00:33,470 When you're ready, come back and we'll take a look at how I 12 00:00:33,470 --> 00:00:34,720 do the problem. 13 00:00:34,720 --> 00:00:43,070 14 00:00:43,070 --> 00:00:44,040 Welcome back. 15 00:00:44,040 --> 00:00:47,600 So again, we're working on a problem where we know that f 16 00:00:47,600 --> 00:00:48,800 is a continuous function. 17 00:00:48,800 --> 00:00:51,850 We know that the integral from 0 to x is f of t dt is equal 18 00:00:51,850 --> 00:00:52,820 to a certain function. 19 00:00:52,820 --> 00:00:55,500 And we want to find f of 2. 20 00:00:55,500 --> 00:00:57,710 So what we're going to use, because we know little f is 21 00:00:57,710 --> 00:00:59,570 continuous, we can actually use the 22 00:00:59,570 --> 00:01:01,520 fundamental theorem of calculus. 23 00:01:01,520 --> 00:01:07,770 And we know that d dx of this whole expression here is 24 00:01:07,770 --> 00:01:09,950 actually little f of x. 25 00:01:09,950 --> 00:01:11,960 So let's just remind us-- 26 00:01:11,960 --> 00:01:13,210 I'll write it down formally. 27 00:01:13,210 --> 00:01:16,380 28 00:01:16,380 --> 00:01:26,570 FTC implies that d dx, the interval from 0 to x f of t dt 29 00:01:26,570 --> 00:01:27,186 if equal to f of x. 30 00:01:27,186 --> 00:01:27,272 Right? 31 00:01:27,272 --> 00:01:29,960 We know that. 32 00:01:29,960 --> 00:01:36,590 So, we know that d dx of the right-hand side and d dx of 33 00:01:36,590 --> 00:01:39,190 the left-hand side are the same. 34 00:01:39,190 --> 00:01:41,910 And so we know that if we take d dx of the right-hand side, 35 00:01:41,910 --> 00:01:43,040 we'll get little f of x. 36 00:01:43,040 --> 00:01:47,810 Because d dx of the left-hand side of this equation is 37 00:01:47,810 --> 00:01:48,680 little f of x. 38 00:01:48,680 --> 00:01:52,430 Do d dx of the right-hand side is also little f of x. 39 00:01:52,430 --> 00:01:55,230 So now all we have to do is apply what we know about 40 00:01:55,230 --> 00:01:59,090 taking derivatives to find little f of x. 41 00:01:59,090 --> 00:02:04,205 Then once we find little f of x we can evaluate it at f 42 00:02:04,205 --> 00:02:04,880 equal 2, at 2--sorry-- at x equal 2. 43 00:02:04,880 --> 00:02:07,210 Sorry about that. 44 00:02:07,210 --> 00:02:10,550 So let's come over to the right part of the board and 45 00:02:10,550 --> 00:02:13,530 we'll find d dx of this function here. 46 00:02:13,530 --> 00:02:21,870 47 00:02:21,870 --> 00:02:22,830 So what do we have to use? 48 00:02:22,830 --> 00:02:25,090 We have to use the product rule and then we'll have to 49 00:02:25,090 --> 00:02:28,290 use a little chain rule, a really simple chain rule in 50 00:02:28,290 --> 00:02:29,290 the second term. 51 00:02:29,290 --> 00:02:30,910 I'm going to write it down below, so we 52 00:02:30,910 --> 00:02:31,770 have some room here. 53 00:02:31,770 --> 00:02:36,550 So the first term, we get 2x sine pi x. 54 00:02:36,550 --> 00:02:38,690 The second term we get x squared. 55 00:02:38,690 --> 00:02:41,010 Derivative of sine is cosine. 56 00:02:41,010 --> 00:02:41,690 pi x. 57 00:02:41,690 --> 00:02:44,210 And then we have to pull out a pi. 58 00:02:44,210 --> 00:02:48,940 Which I'll just put, well I'll put it here for the moment. 59 00:02:48,940 --> 00:02:51,980 The pi comes from taking the derivative of pi x. 60 00:02:51,980 --> 00:02:56,670 And now all we need, this is then f of x, right? 61 00:02:56,670 --> 00:02:59,610 So I just will remind us this is f of x. 62 00:02:59,610 --> 00:03:02,920 Let me put it right here, a little f of x. 63 00:03:02,920 --> 00:03:06,290 So if I want to find f of 2, all I have to do is evaluate 64 00:03:06,290 --> 00:03:09,100 this at x equal 2. 65 00:03:09,100 --> 00:03:11,060 So f of 2 is going to equal-- 66 00:03:11,060 --> 00:03:13,450 2 times 2 is 4-- 67 00:03:13,450 --> 00:03:17,560 sine 2 pi plus 4-- 68 00:03:17,560 --> 00:03:20,050 I'll bring the pi in front-- 69 00:03:20,050 --> 00:03:23,040 4 pi cosine 2 pi. 70 00:03:23,040 --> 00:03:24,610 So what do I get? 71 00:03:24,610 --> 00:03:27,010 Sine 2 pi is 0. 72 00:03:27,010 --> 00:03:28,970 Cosine 2 pi is 1. 73 00:03:28,970 --> 00:03:32,860 So ultimately, I just get 4 pi. 74 00:03:32,860 --> 00:03:35,090 So, hopefully you were able to solve this problem and get the 75 00:03:35,090 --> 00:03:36,950 same answer. 76 00:03:36,950 --> 00:03:40,220 Now let me remind you, one more time what we did. 77 00:03:40,220 --> 00:03:41,440 Let's come over here. 78 00:03:41,440 --> 00:03:44,410 We started with a function that we knew was continuous. 79 00:03:44,410 --> 00:03:48,280 We knew the integral from 0 to x of that function was equal 80 00:03:48,280 --> 00:03:50,610 to a certain function of x. 81 00:03:50,610 --> 00:03:53,621 And we wanted to evaluate this function at a certain, at a 82 00:03:53,621 --> 00:03:55,530 certain point. 83 00:03:55,530 --> 00:03:56,840 So what we exploited was the 84 00:03:56,840 --> 00:03:59,030 fundamental theorem of calculus. 85 00:03:59,030 --> 00:04:02,700 And the fundamental theorem of calculus tells us that d dx, 86 00:04:02,700 --> 00:04:06,010 of the left-hand side is f of x. 87 00:04:06,010 --> 00:04:09,460 And so that we can take d dx of the right-hand side and 88 00:04:09,460 --> 00:04:11,070 figure out what f of x is. 89 00:04:11,070 --> 00:04:13,290 And then we just evaluate. 90 00:04:13,290 --> 00:04:15,240 So I think I'll stop there. 91 00:04:15,240 --> 00:04:15,855