1 00:00:00,040 --> 00:00:02,460 The following content is provided under a Creative 2 00:00:02,460 --> 00:00:03,870 Commons license. 3 00:00:03,870 --> 00:00:06,320 Your support will help MIT OpenCourseWare 4 00:00:06,320 --> 00:00:10,560 continue to offer high quality educational resources for free. 5 00:00:10,560 --> 00:00:13,300 To make a donation or view additional materials 6 00:00:13,300 --> 00:00:17,210 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,210 --> 00:00:32,095 at ocw.mit.edu 8 00:00:32,095 --> 00:00:33,650 HERBERT GROSS: Hi. 9 00:00:33,650 --> 00:00:37,080 Hopefully, by now you have become well experienced 10 00:00:37,080 --> 00:00:39,430 at computing limits of double sums. 11 00:00:39,430 --> 00:00:41,300 You've learned to hate me a little bit more 12 00:00:41,300 --> 00:00:43,610 because the work was quite complicated, 13 00:00:43,610 --> 00:00:47,290 and hopefully you are convinced that there must be an easier 14 00:00:47,290 --> 00:00:49,250 way to do this stuff. 15 00:00:49,250 --> 00:00:51,680 And fortunately, the answer is there 16 00:00:51,680 --> 00:00:54,640 is an easier way sometimes. 17 00:00:54,640 --> 00:00:58,340 And the concept that we're going to talk about today 18 00:00:58,340 --> 00:01:00,600 is, again, completely analogous to 19 00:01:00,600 --> 00:01:03,220 what happened in calculus of a single variable 20 00:01:03,220 --> 00:01:05,990 where we showed that the concept of definite integral 21 00:01:05,990 --> 00:01:09,360 existed independently of being able to form derivatives, 22 00:01:09,360 --> 00:01:12,990 but in the case where we knew a particular function whose 23 00:01:12,990 --> 00:01:15,180 derivative would be a given thing, 24 00:01:15,180 --> 00:01:17,830 we were able to perform the infinite sum-- 25 00:01:17,830 --> 00:01:20,470 or compute the infinite sum much more readily 26 00:01:20,470 --> 00:01:25,920 than had we had to rely on the arithmetic of infinite series. 27 00:01:25,920 --> 00:01:27,700 By the way-- and I'll mention this later 28 00:01:27,700 --> 00:01:31,140 in a different context, I hope-- that the converse problem 29 00:01:31,140 --> 00:01:32,030 was also true. 30 00:01:32,030 --> 00:01:35,180 In many cases one did not know how 31 00:01:35,180 --> 00:01:39,840 to find the required function with the given derivative, 32 00:01:39,840 --> 00:01:42,770 and in this case knowing how to find the area by the limit 33 00:01:42,770 --> 00:01:46,330 process was equivalent to how we were then 34 00:01:46,330 --> 00:01:49,620 able to invent new functions which had given derivatives. 35 00:01:49,620 --> 00:01:51,824 Essentially, the difference in point of view 36 00:01:51,824 --> 00:01:53,240 was the difference between what we 37 00:01:53,240 --> 00:01:55,840 called the first fundamental theorem of integral calculus, 38 00:01:55,840 --> 00:01:58,450 and the second fundamental theorem of integral calculus. 39 00:01:58,450 --> 00:02:00,610 As I said, the same analogy will hold here, 40 00:02:00,610 --> 00:02:03,440 and let's get into this now without further ado. 41 00:02:03,440 --> 00:02:06,170 I call today's lesson the fundamental theorem. 42 00:02:06,170 --> 00:02:08,600 And what I'm going to do now is to pretend that we never 43 00:02:08,600 --> 00:02:10,800 had the lecture of last time. 44 00:02:10,800 --> 00:02:13,110 That we have never heard of an infinite sum. 45 00:02:13,110 --> 00:02:16,090 I'm going to introduce what one calls 46 00:02:16,090 --> 00:02:18,630 the anti-derivative of a function of two 47 00:02:18,630 --> 00:02:20,860 independent variables, again keeping in mind 48 00:02:20,860 --> 00:02:23,900 the analogy that for more than two independent variables 49 00:02:23,900 --> 00:02:25,480 a similar treatment holds. 50 00:02:25,480 --> 00:02:28,080 And the exercises will include problems 51 00:02:28,080 --> 00:02:30,150 that have more than two independent variables. 52 00:02:30,150 --> 00:02:32,270 The idea is that when I write down 53 00:02:32,270 --> 00:02:34,650 this particular double integral, if I 54 00:02:34,650 --> 00:02:40,320 look at the innermost integral, notice that x appears only 55 00:02:40,320 --> 00:02:41,470 as a parameter. 56 00:02:41,470 --> 00:02:44,630 That the variable of integration is y 57 00:02:44,630 --> 00:02:47,040 notice then, that x is being treated as a constant. 58 00:02:47,040 --> 00:02:48,350 What this says is what? 59 00:02:48,350 --> 00:02:53,830 Pick a fixed value of x, compute this integral, 60 00:02:53,830 --> 00:02:56,670 evaluate this as y goes from g_1 of x 61 00:02:56,670 --> 00:02:59,310 to g_2 of x-- the resulting function is 62 00:02:59,310 --> 00:03:02,460 a function of x alone-- and integrate that function 63 00:03:02,460 --> 00:03:04,460 of x from a to b. 64 00:03:04,460 --> 00:03:06,720 And to do that for you in slow motion, 65 00:03:06,720 --> 00:03:09,440 so you see what happens here, what I'm essentially saying is, 66 00:03:09,440 --> 00:03:11,310 read this thing inside out. 67 00:03:11,310 --> 00:03:13,940 Imagine some brackets placed here. 68 00:03:13,940 --> 00:03:16,240 Not imagine them, let's put them in. 69 00:03:16,240 --> 00:03:19,000 And what we're saying is, lookit, first of all, 70 00:03:19,000 --> 00:03:20,800 fix a value of x. 71 00:03:20,800 --> 00:03:22,380 Hold x constant. 72 00:03:22,380 --> 00:03:27,380 And for that fixed value of x, compute f of x,y dy from g_1 73 00:03:27,380 --> 00:03:29,280 of x to g_2 of x. 74 00:03:29,280 --> 00:03:31,810 Compute that, remember you're going to integrate 75 00:03:31,810 --> 00:03:32,820 that with respect to y. 76 00:03:32,820 --> 00:03:34,236 When you're all through with this, 77 00:03:34,236 --> 00:03:37,140 because the limits involve only x, 78 00:03:37,140 --> 00:03:39,980 the integrand that you get-- this bracketed expression will 79 00:03:39,980 --> 00:03:42,490 now be a function of x alone, call it h of x. 80 00:03:42,490 --> 00:03:45,230 Integrate h of x dx between a and b, 81 00:03:45,230 --> 00:03:47,830 and the resulting expression, assuming 82 00:03:47,830 --> 00:03:49,910 that you can carry out the integration of course, 83 00:03:49,910 --> 00:03:54,110 will be a particular number. 84 00:03:54,110 --> 00:03:57,710 That has nothing to do, as I say, with double sums so far. 85 00:03:57,710 --> 00:04:00,350 Though I suspect that you're getting a bit suspicious. 86 00:04:00,350 --> 00:04:03,420 And in the same way that the definite integral was 87 00:04:03,420 --> 00:04:08,315 a sum and not an anti-derivative, 88 00:04:08,315 --> 00:04:10,410 the notation of the definite integral 89 00:04:10,410 --> 00:04:13,727 looked enough like the notation of the indefinite integral, 90 00:04:13,727 --> 00:04:15,560 so we began to suspect there was going to be 91 00:04:15,560 --> 00:04:16,890 a connection between them. 92 00:04:16,890 --> 00:04:18,450 That connection is what's going to be 93 00:04:18,450 --> 00:04:20,890 called the fundamental theorem in our present case. 94 00:04:20,890 --> 00:04:24,890 But going on, let me just give you an example showing you how 95 00:04:24,890 --> 00:04:28,400 we compute the anti-derivative. 96 00:04:28,400 --> 00:04:31,200 Let's take, for example interval from 0 to 2, 97 00:04:31,200 --> 00:04:35,210 integral from 0 to x squared, x cubed y dy dx. 98 00:04:35,210 --> 00:04:38,340 Again we go from inside to out, the way 99 00:04:38,340 --> 00:04:41,960 we read this is that we're integrating with respect to y. 100 00:04:41,960 --> 00:04:44,270 That means we're holding x constant. 101 00:04:44,270 --> 00:04:46,760 For a fixed value of x, integrate this 102 00:04:46,760 --> 00:04:49,320 as y goes from 0 to x squared. 103 00:04:49,320 --> 00:04:52,540 So treating x as a constant, I integrate this with respect 104 00:04:52,540 --> 00:04:53,240 to y. 105 00:04:53,240 --> 00:04:56,280 The integral of y dy is 1/2 y squared. 106 00:04:56,280 --> 00:05:00,250 I evaluate that as y goes from 0 to x squared. 107 00:05:00,250 --> 00:05:01,670 That's crucial you see. 108 00:05:01,670 --> 00:05:03,940 It's y that goes from 0 to x squared. 109 00:05:03,940 --> 00:05:06,350 When I replace y by the upper limit, 110 00:05:06,350 --> 00:05:11,830 I get 1/2 x cubed times the quantity x squared, squared. 111 00:05:11,830 --> 00:05:14,530 When I replace y by the lower limit 0, 112 00:05:14,530 --> 00:05:16,450 the integrand vanishes. 113 00:05:16,450 --> 00:05:19,700 So that what I wind up with is a function of x alone. 114 00:05:19,700 --> 00:05:21,430 Specifically what function is it? 115 00:05:21,430 --> 00:05:26,710 It's 1/2 x to the seventh, and I now integrate that from 0 to 2. 116 00:05:26,710 --> 00:05:30,380 That gives me 1/8 x to the eighth over 2, 117 00:05:30,380 --> 00:05:32,350 that's x to the eighth over 16. 118 00:05:32,350 --> 00:05:34,600 Evaluated between 0 and 2. 119 00:05:34,600 --> 00:05:38,030 2 to the eighth over 16 is 2 to the eighth over 2 120 00:05:38,030 --> 00:05:41,530 to the fourth, which is 2 to the fourth, which in turn is 16. 121 00:05:41,530 --> 00:05:46,486 So again, notice, I can carry out this operation purely 122 00:05:46,486 --> 00:05:47,110 manipulatively. 123 00:05:47,110 --> 00:05:52,660 It is truly the inverse of taking the partial derivative. 124 00:05:52,660 --> 00:05:55,530 In the same way that taking the partial derivative involved 125 00:05:55,530 --> 00:05:58,190 holding all the variables but one constant, 126 00:05:58,190 --> 00:06:01,550 taking a partial integral, so to speak, an anti-derivative, 127 00:06:01,550 --> 00:06:03,870 involves simply doing what? 128 00:06:03,870 --> 00:06:06,540 Integrating, treating all the variables but the given 129 00:06:06,540 --> 00:06:08,280 one as a constant. 130 00:06:08,280 --> 00:06:10,740 Which reduces you, again, to the equivalent 131 00:06:10,740 --> 00:06:12,470 of what the anti-derivative meant 132 00:06:12,470 --> 00:06:14,700 in the calculus of a single variable. 133 00:06:14,700 --> 00:06:17,730 Now here's where we come to the so-called fundamental theorem. 134 00:06:17,730 --> 00:06:21,500 Let me tie what we've talked about today, 135 00:06:21,500 --> 00:06:23,420 and what we talked about last time. 136 00:06:23,420 --> 00:06:26,150 Let me see if I can't tie those together for you 137 00:06:26,150 --> 00:06:28,790 in a completely different way. 138 00:06:28,790 --> 00:06:31,230 Meaning I want you to see that conceptually, 139 00:06:31,230 --> 00:06:33,710 today's lesson and last time's lesson 140 00:06:33,710 --> 00:06:36,600 are completely different, but the punchline is 141 00:06:36,600 --> 00:06:39,730 there's a remarkable connection between the two. 142 00:06:39,730 --> 00:06:42,190 Let's suppose I take the region R, which 143 00:06:42,190 --> 00:06:45,290 is a very simple region that's bounded above by the curve y 144 00:06:45,290 --> 00:06:46,890 equals g_2 of x. 145 00:06:46,890 --> 00:06:48,950 It's bounded below by the curve y 146 00:06:48,950 --> 00:06:51,435 equals g_1 of x, and that these two curves happen 147 00:06:51,435 --> 00:06:54,620 to intersect at values corresponding to x 148 00:06:54,620 --> 00:06:56,460 equals a and x equals b. 149 00:06:56,460 --> 00:07:03,110 And what I would like to find is the mass of the region R, 150 00:07:03,110 --> 00:07:08,590 if its density is some given function f of x, y. 151 00:07:08,590 --> 00:07:12,030 Now, in terms of what we did last time, 152 00:07:12,030 --> 00:07:14,670 this involved a double sum. 153 00:07:14,670 --> 00:07:16,320 What double sum was it? 154 00:07:16,320 --> 00:07:19,340 We partitioned this into rectangles, et cetera. 155 00:07:19,340 --> 00:07:24,105 We picked a particular number c sub i, j comma d sub i, 156 00:07:24,105 --> 00:07:26,690 j in the (i, j)-th rectangle. 157 00:07:26,690 --> 00:07:32,880 We computed f of c sub i, j comma d sub i, j times delta 158 00:07:32,880 --> 00:07:35,810 a sub i, j, added these all up. 159 00:07:35,810 --> 00:07:40,160 The double sum as i went from 1 to n and j went from 1 to n, 160 00:07:40,160 --> 00:07:45,240 and took the limit as the maximum delta x_i and delta 161 00:07:45,240 --> 00:07:49,660 y_j approach 0, and that limit, if it existed-- 162 00:07:49,660 --> 00:07:52,430 and it would exist if f was piecewise continuous-- 163 00:07:52,430 --> 00:07:55,090 that limit was denoted by what? 164 00:07:55,090 --> 00:08:00,480 Double integral over the region R f of x, y dA. 165 00:08:00,480 --> 00:08:03,990 And that dA could either be viewed as being dy*dx if you 166 00:08:03,990 --> 00:08:07,536 wanted, it could be dx*dy, It could also be, say, 167 00:08:07,536 --> 00:08:10,160 in different coordinate systems, we're not going to worry about 168 00:08:10,160 --> 00:08:11,630 that part just yet. 169 00:08:11,630 --> 00:08:13,850 We'll worry about that in future lectures. 170 00:08:13,850 --> 00:08:17,820 Now, on the other hand, let's stop right here for a moment. 171 00:08:17,820 --> 00:08:20,010 Let's see how one might have been tempted 172 00:08:20,010 --> 00:08:21,810 to tackle this problem from a completely 173 00:08:21,810 --> 00:08:23,250 intuitive point of view. 174 00:08:23,250 --> 00:08:25,230 The so-called engineering approach, 175 00:08:25,230 --> 00:08:30,540 if you would never heard of the anti-derivative, 176 00:08:30,540 --> 00:08:32,390 anti-derivative for several variables, 177 00:08:32,390 --> 00:08:34,521 but had taken part one of our course. 178 00:08:34,521 --> 00:08:36,770 The engineering approach would be something like this: 179 00:08:36,770 --> 00:08:40,700 you would say, let's pick a very small infinitesimal region 180 00:08:40,700 --> 00:08:44,400 here, in which we can assume that the density is constant. 181 00:08:44,400 --> 00:08:46,300 So we'll pick a little region that's 182 00:08:46,300 --> 00:08:50,640 a very, very small rectangle of dimension what? 183 00:08:50,640 --> 00:08:52,820 We'll say it's dy by dx. 184 00:08:57,370 --> 00:09:00,065 And the density of this particular rectangle 185 00:09:00,065 --> 00:09:03,650 can be assumed to be the constant value f of x, y. 186 00:09:06,640 --> 00:09:08,940 Because that density is constant, 187 00:09:08,940 --> 00:09:12,125 the mass of this little piece should be f of x, 188 00:09:12,125 --> 00:09:15,050 y times dy*dx. 189 00:09:15,050 --> 00:09:16,240 and then we say what? 190 00:09:16,240 --> 00:09:19,700 Let's add these all up. 191 00:09:19,700 --> 00:09:21,280 Let's go down to here we say lookit, 192 00:09:21,280 --> 00:09:23,080 this is a typical element. 193 00:09:23,080 --> 00:09:25,910 And we'll add them all up for all possible y's and all 194 00:09:25,910 --> 00:09:27,130 possible x's. 195 00:09:27,130 --> 00:09:28,740 Now how would you do this intuitively? 196 00:09:28,740 --> 00:09:32,070 You'd say well, lookit, let me hold x constant. 197 00:09:32,070 --> 00:09:36,130 Well, for a constant value of x-- let's call it x sub 1. 198 00:09:36,130 --> 00:09:38,590 For a constant value of x, notice 199 00:09:38,590 --> 00:09:42,190 that y can vary any place-- to be in the region R, 200 00:09:42,190 --> 00:09:46,480 for that fixed value of x, y can be any place from here to here. 201 00:09:46,480 --> 00:09:48,910 In other words, y varies continuously 202 00:09:48,910 --> 00:09:52,660 from g_1 of x_1 to g_2 of x_1. 203 00:09:52,660 --> 00:09:56,115 Or, because x_1 could have been x, for that given x, 204 00:09:56,115 --> 00:10:00,440 y varies continuously from g_1 of x to g_2 of x. 205 00:10:00,440 --> 00:10:02,470 And notice that x could have been chosen, 206 00:10:02,470 --> 00:10:06,470 if we want to be in the region R, to be any place from a to b. 207 00:10:06,470 --> 00:10:08,900 And so mechanically, we might say, 208 00:10:08,900 --> 00:10:13,110 let's just say, here is the mass of a small element, 209 00:10:13,110 --> 00:10:16,360 and we'll add these all up so that for a fixed x, 210 00:10:16,360 --> 00:10:19,040 y goes from g_1 of x to g_2 of x. 211 00:10:19,040 --> 00:10:21,750 And x could be anywhere from a to b. 212 00:10:21,750 --> 00:10:25,270 And it appears that this is a truism. 213 00:10:25,270 --> 00:10:27,220 This is not a truism. 214 00:10:27,220 --> 00:10:29,530 It's a rather remarkable result. It 215 00:10:29,530 --> 00:10:32,900 is analogous to what happened in calculus of a single variable. 216 00:10:32,900 --> 00:10:35,890 Namely, notice that when we had the definite integral from a 217 00:10:35,890 --> 00:10:39,100 to b, f of x dx, that was an infinite sum. 218 00:10:39,100 --> 00:10:43,530 It was summation f of c sub k, delta x sub k, 219 00:10:43,530 --> 00:10:46,440 as k went from 1 to n taking the limit 220 00:10:46,440 --> 00:10:49,750 as the maximum delta x sub k approached 0. 221 00:10:49,750 --> 00:10:52,850 It turned out that if we knew a function capital 222 00:10:52,850 --> 00:10:56,670 F whose derivative was little f, then this sum 223 00:10:56,670 --> 00:10:59,460 could be evaluated by computing capital 224 00:10:59,460 --> 00:11:02,350 F of b minus capital F of a. 225 00:11:02,350 --> 00:11:04,350 Again the idea being that if you knew 226 00:11:04,350 --> 00:11:07,770 a function whose derivative was f, that gave you an easy way 227 00:11:07,770 --> 00:11:09,460 to evaluate the sum. 228 00:11:09,460 --> 00:11:12,210 On the other hand, if you didn't know 229 00:11:12,210 --> 00:11:14,420 a function whose derivative was little f, 230 00:11:14,420 --> 00:11:17,260 evaluating the sum as a limit gave you 231 00:11:17,260 --> 00:11:21,620 a way of being able to find the anti-derivative. 232 00:11:21,620 --> 00:11:24,050 A particular case in point, you may recall, 233 00:11:24,050 --> 00:11:29,140 was trying to handle the problem e to the minus x squared dx, 234 00:11:29,140 --> 00:11:31,480 as x goes from 0 to 1. 235 00:11:31,480 --> 00:11:34,320 This certainly existed as an area, 236 00:11:34,320 --> 00:11:35,970 but we did not know-- in fact, there 237 00:11:35,970 --> 00:11:37,620 is no elementary function. 238 00:11:37,620 --> 00:11:39,680 We invented one, called the error function, 239 00:11:39,680 --> 00:11:43,000 whose derivative is e to the minus x squared. 240 00:11:43,000 --> 00:11:46,230 So the idea is-- and this is the key point-- 241 00:11:46,230 --> 00:11:49,790 this is what becomes known as the fundamental theorem 242 00:11:49,790 --> 00:11:52,000 for several variables. 243 00:11:52,000 --> 00:11:56,690 That there is a connection between a double infinite sum, 244 00:11:56,690 --> 00:12:00,220 and a double anti-derivative. 245 00:12:00,220 --> 00:12:02,010 That this particular expression here 246 00:12:02,010 --> 00:12:04,270 does not involve knowing anything 247 00:12:04,270 --> 00:12:06,010 about partial derivatives. 248 00:12:06,010 --> 00:12:08,080 This particular expression here does not 249 00:12:08,080 --> 00:12:11,970 involve any knowledge of knowing partial sums. 250 00:12:11,970 --> 00:12:14,500 This simply says what? 251 00:12:14,500 --> 00:12:16,470 Integrate this thing twice. 252 00:12:16,470 --> 00:12:18,910 Once holding x constant and letting y 253 00:12:18,910 --> 00:12:21,740 go from g_1 of x to g_2 of x, and integrating 254 00:12:21,740 --> 00:12:22,920 with respect to x. 255 00:12:22,920 --> 00:12:27,740 And the amazing result is that if f is continuous, 256 00:12:27,740 --> 00:12:30,450 this limit exists, and in particular, 257 00:12:30,450 --> 00:12:33,710 if we happen to know how to actually carry out 258 00:12:33,710 --> 00:12:38,480 these repeated or iterated integrations, 259 00:12:38,480 --> 00:12:41,510 we can compute this complicated sum simply 260 00:12:41,510 --> 00:12:44,950 by carrying out this anti-derivative. 261 00:12:44,950 --> 00:12:47,060 And I think the best way to emphasize that 262 00:12:47,060 --> 00:12:51,210 to you is to repeat the punchline to the homework 263 00:12:51,210 --> 00:12:54,000 exercises of last assignment. 264 00:12:54,000 --> 00:12:57,240 You may recall that last time we were dealing 265 00:12:57,240 --> 00:13:00,950 with the square whose vertices were (0, 0), (1, 0), (1, 1), 266 00:13:00,950 --> 00:13:02,110 and (0, 1). 267 00:13:02,110 --> 00:13:05,760 The density of the square at the point x comma y 268 00:13:05,760 --> 00:13:07,820 was x squared plus y squared. 269 00:13:07,820 --> 00:13:11,970 And I asked you as a homework problem 270 00:13:11,970 --> 00:13:16,430 to compute the mass of this plate R, to compute it exactly. 271 00:13:16,430 --> 00:13:21,439 And we said OK, by definition of-- remember, 272 00:13:21,439 --> 00:13:23,230 this thing here now, with the R under here, 273 00:13:23,230 --> 00:13:26,440 indicates a limit of a double sum. 274 00:13:26,440 --> 00:13:30,290 That this mass was by definition this particular result. 275 00:13:30,290 --> 00:13:33,870 And notice that last time we showed, 276 00:13:33,870 --> 00:13:36,790 in terms of double sums, that this came out to be 2/3. 277 00:13:36,790 --> 00:13:39,860 Let me show you in terms of our new theory 278 00:13:39,860 --> 00:13:43,410 how we can find this much more conveniently, with no sweat, 279 00:13:43,410 --> 00:13:44,550 so to speak. 280 00:13:44,550 --> 00:13:48,360 What I'm going to do now is the following-- The same analogy 281 00:13:48,360 --> 00:13:49,470 I used before. 282 00:13:49,470 --> 00:13:53,580 I look at my region here, which I'll call A, and I say what? 283 00:13:53,580 --> 00:13:56,030 For a fixed value of x. 284 00:13:56,030 --> 00:13:59,760 For a fixed value of x, notice that y 285 00:13:59,760 --> 00:14:02,470 varies continuously from 0 to 1. 286 00:14:02,470 --> 00:14:05,330 See, y varies continuously from 0 to 1. 287 00:14:05,330 --> 00:14:08,010 And then x, in turn, could have been chosen 288 00:14:08,010 --> 00:14:09,700 to be any place from 0 to 1. 289 00:14:09,700 --> 00:14:13,530 That as we run through all these strips, added up from 0 to 1, 290 00:14:13,530 --> 00:14:17,650 that covers our region R. So what we're saying is, 291 00:14:17,650 --> 00:14:22,710 all right, a little element of our mass will be x squared plus 292 00:14:22,710 --> 00:14:25,290 y squared dy*dx. 293 00:14:25,290 --> 00:14:28,930 We'll add these all up from 0 to 1, holding x constant. 294 00:14:28,930 --> 00:14:33,140 Then add them all up again, as x goes from 0 to 1. 295 00:14:33,140 --> 00:14:35,470 Again, this is the iterated integral. 296 00:14:35,470 --> 00:14:37,990 How do we evaluate this? 297 00:14:37,990 --> 00:14:42,850 Well we treat x as a constant, integrate this with respect 298 00:14:42,850 --> 00:14:43,680 to y. 299 00:14:43,680 --> 00:14:45,280 If we're treating x as a constant, 300 00:14:45,280 --> 00:14:48,980 this then will come out to be x squared y, plus 1/3 y cubed, 301 00:14:48,980 --> 00:14:53,580 and we now evaluate that as y goes from 0 to 1. 302 00:14:53,580 --> 00:14:55,329 Leaving these details out because they 303 00:14:55,329 --> 00:14:57,370 are essentially the calculus of a single variable 304 00:14:57,370 --> 00:15:00,260 all over again, this turns out to be the integral from 0 305 00:15:00,260 --> 00:15:02,990 to 1 x squared plus 1/3 dx, which 306 00:15:02,990 --> 00:15:06,690 in turn is 1/3 x cubed plus 1/3 x, evaluated 307 00:15:06,690 --> 00:15:08,390 as x goes from 0 to 1. 308 00:15:08,390 --> 00:15:11,340 The upper limit gives me a third plus a third, the lower limit's 309 00:15:11,340 --> 00:15:12,000 0. 310 00:15:12,000 --> 00:15:19,190 The mass here is 2/3, which checks with the result 311 00:15:19,190 --> 00:15:21,930 that we got last time. 312 00:15:21,930 --> 00:15:24,070 Again, I'm not going to go through the proof 313 00:15:24,070 --> 00:15:27,930 of the fundamental theorem, I don't think it's that crucial. 314 00:15:27,930 --> 00:15:31,900 It's available in textbooks on advanced calculus. 315 00:15:31,900 --> 00:15:34,660 Some of the exercises may possibly 316 00:15:34,660 --> 00:15:38,170 give you hints as to how these results come about. 317 00:15:38,170 --> 00:15:39,960 But by and large, I'm more interested 318 00:15:39,960 --> 00:15:43,980 now in you seeing the overview, and at this stage of the game 319 00:15:43,980 --> 00:15:47,230 letting you get whatever specific theoretical details 320 00:15:47,230 --> 00:15:50,030 you desire on your own. 321 00:15:50,030 --> 00:15:53,950 At any rate, let's continue on with examples. 322 00:15:53,950 --> 00:15:57,710 I think a very nice counterpart to example one-- 323 00:15:57,710 --> 00:16:00,170 and to refresh your memories without looking back at it, 324 00:16:00,170 --> 00:16:06,370 example one asked us to compute this anti-derivative. 325 00:16:06,370 --> 00:16:08,900 And what I'd like to do now is to emphasize 326 00:16:08,900 --> 00:16:12,500 the fundamental theorem by wording this a different way. 327 00:16:12,500 --> 00:16:16,900 What I want to do now is to describe the plate R. 328 00:16:16,900 --> 00:16:22,090 If its mass is given by the double integral over the region 329 00:16:22,090 --> 00:16:26,170 R, rho of x, y, dA and that turns out to be integral from 0 330 00:16:26,170 --> 00:16:30,070 to 2, 0 to x squared, x cubed y, dy*dx. 331 00:16:30,070 --> 00:16:33,000 What is the region R, whose-- what 332 00:16:33,000 --> 00:16:36,900 is this region R, if this is how its mass is given? 333 00:16:36,900 --> 00:16:39,420 The first thing I want you to observe 334 00:16:39,420 --> 00:16:42,990 is that this part here is identified with the density 335 00:16:42,990 --> 00:16:45,110 part of the problem. 336 00:16:45,110 --> 00:16:47,210 And that these limits of integration 337 00:16:47,210 --> 00:16:50,190 determine the region R. For example, what this says 338 00:16:50,190 --> 00:16:55,760 is, if you hold x constant, y varies from 0 to x squared. 339 00:16:55,760 --> 00:16:57,300 Let's see what that means. 340 00:16:57,300 --> 00:17:01,620 If you hold x constant, y varies from 0-- well y 341 00:17:01,620 --> 00:17:04,640 equals 0 is the x-axis-- to y equals 342 00:17:04,640 --> 00:17:07,240 x squared-- that's this particular parabola. 343 00:17:07,240 --> 00:17:12,800 What this says is, for a fixed value of x, say x equals x_1, 344 00:17:12,800 --> 00:17:14,920 to be in the region R, you can be 345 00:17:14,920 --> 00:17:19,859 any place from the x-axis along this strip up to y 346 00:17:19,859 --> 00:17:21,560 equals x_1 squared. 347 00:17:21,560 --> 00:17:23,440 So your strip is like this. 348 00:17:23,440 --> 00:17:26,540 Then we're told that in turn x could be chosen 349 00:17:26,540 --> 00:17:28,730 to be any place from 0 to 2. 350 00:17:28,730 --> 00:17:31,600 So now what we know is that this strip 351 00:17:31,600 --> 00:17:35,080 could have been chosen for any value of x between 0 and 2. 352 00:17:35,080 --> 00:17:37,070 And what that tells us therefore, 353 00:17:37,070 --> 00:17:40,130 is that the shape of our region R 354 00:17:40,130 --> 00:17:43,280 is the curve-- the region that's bounded 355 00:17:43,280 --> 00:17:48,650 above by the curved y equals x squared, below by the x-axis, 356 00:17:48,650 --> 00:17:52,940 and on the right by the line x equals 2. 357 00:17:52,940 --> 00:17:56,470 This is our region R. You see, the region R 358 00:17:56,470 --> 00:17:58,650 is determined by the limits of integration, 359 00:17:58,650 --> 00:18:03,210 and the density of R is given by rho of x, y 360 00:18:03,210 --> 00:18:06,263 equals x cubed y at the point x comma y. 361 00:18:06,263 --> 00:18:06,762 OK? 362 00:18:10,150 --> 00:18:13,700 We'd mentioned before that why do you have to write dy*dx? 363 00:18:13,700 --> 00:18:15,740 Couldn't you have written dx*dy? 364 00:18:15,740 --> 00:18:19,190 What I thought I'd like to do, for my next example for you, 365 00:18:19,190 --> 00:18:22,830 is to show you how one inverts-- or changes-- 366 00:18:22,830 --> 00:18:24,920 the order of integration. 367 00:18:24,920 --> 00:18:28,360 For example, given the same integral, the same region R, 368 00:18:28,360 --> 00:18:31,850 suppose we now want to express the mass in the form double 369 00:18:31,850 --> 00:18:34,720 integral x cubed y dx*dy? 370 00:18:34,720 --> 00:18:37,830 You see the region R is still the same as it was before. 371 00:18:37,830 --> 00:18:41,590 But now you see what we want to do, how would we read this? 372 00:18:41,590 --> 00:18:44,730 This says we're integrating with respect to x. x is varying. 373 00:18:44,730 --> 00:18:49,030 So this says, for a fixed value of y, integrate this thing. 374 00:18:49,030 --> 00:18:52,100 See, for a fixed value of y, integrate this. 375 00:18:52,100 --> 00:18:53,880 Evaluated between the appropriate limits 376 00:18:53,880 --> 00:18:56,060 of x, in terms of y. 377 00:18:56,060 --> 00:18:58,230 And then integrate this with respect to y. 378 00:18:58,230 --> 00:19:01,450 Well, what this says is let's pick a fixed value of y. 379 00:19:01,450 --> 00:19:03,490 Let's say y equals y_1. 380 00:19:03,490 --> 00:19:07,890 For that fixed value of y notice that the curve y equals 381 00:19:07,890 --> 00:19:11,480 x squared, for x positive, in inverted form 382 00:19:11,480 --> 00:19:15,050 has the form x equals the square root of y. 383 00:19:15,050 --> 00:19:18,250 So for a fixed value of y, notice 384 00:19:18,250 --> 00:19:25,340 that x varies from the square root of y_1 up to x equals 2. 385 00:19:25,340 --> 00:19:28,310 See, x goes from the square root of y to 2, 386 00:19:28,310 --> 00:19:31,250 for a fixed value of y. 387 00:19:31,250 --> 00:19:33,350 Now where can y be? 388 00:19:33,350 --> 00:19:34,840 This is our limits here. 389 00:19:34,840 --> 00:19:36,780 To be in the region R, y could have 390 00:19:36,780 --> 00:19:42,210 been chosen as low as this, or as high as this. 391 00:19:42,210 --> 00:19:46,660 In other words y varies continuously from 0 to 4. 392 00:19:46,660 --> 00:19:49,570 And again, because I don't want to have our lecture 393 00:19:49,570 --> 00:19:52,010 obscured by computational detail, 394 00:19:52,010 --> 00:19:55,580 I simply urge you to-- on your own-- 395 00:19:55,580 --> 00:19:58,830 compute this double integral, compute the double integral 396 00:19:58,830 --> 00:20:01,720 that we obtained in example number three, 397 00:20:01,720 --> 00:20:04,150 and show that the answers are the same. 398 00:20:04,150 --> 00:20:09,080 What I do want you to observe is how different the limits look. 399 00:20:09,080 --> 00:20:11,440 Notice that in the other integral it was from 0 to 2 400 00:20:11,440 --> 00:20:14,010 outside, now it's from 0 to 4. 401 00:20:14,010 --> 00:20:16,190 Notice also on the inside integral 402 00:20:16,190 --> 00:20:18,180 it was from 0 to x squared. 403 00:20:18,180 --> 00:20:20,570 Now it's from the square root of y to 2. 404 00:20:20,570 --> 00:20:23,530 There is no mechanical way of doing this, at least 405 00:20:23,530 --> 00:20:25,370 in the two-dimensional case, we can 406 00:20:25,370 --> 00:20:27,110 see from the picture what's happening. 407 00:20:27,110 --> 00:20:29,190 In the multi-dimensional case, we 408 00:20:29,190 --> 00:20:32,100 have to resort to the theory of inverse functions, 409 00:20:32,100 --> 00:20:35,260 and this becomes, at best, a very messy procedure. 410 00:20:35,260 --> 00:20:38,070 Of course, the answer is, if it's such a messy procedure, 411 00:20:38,070 --> 00:20:39,790 why bother with it? 412 00:20:39,790 --> 00:20:42,060 And so in terms of another example that I would like 413 00:20:42,060 --> 00:20:47,470 to show you, let me give you an illustration in which it 414 00:20:47,470 --> 00:20:51,870 may be possible to find the required answer if we 415 00:20:51,870 --> 00:20:55,060 integrated one order, but not if we integrate with respect 416 00:20:55,060 --> 00:20:56,060 to the other order. 417 00:20:56,060 --> 00:21:01,470 Let me evaluate the double integral 0 to 2, x to 2, 418 00:21:01,470 --> 00:21:04,290 e to the y squared, dy*dx. 419 00:21:04,290 --> 00:21:07,070 Again, let me see if I can at least write down 420 00:21:07,070 --> 00:21:08,950 what this thing means geometrically 421 00:21:08,950 --> 00:21:10,490 before I even begin. 422 00:21:10,490 --> 00:21:14,190 Notice that I can think of this as a plate R, 423 00:21:14,190 --> 00:21:18,340 whose density at the point x comma y is e to the y squared, 424 00:21:18,340 --> 00:21:21,250 and what is the shape of this plate? 425 00:21:21,250 --> 00:21:22,740 Integrate with respect to y first. 426 00:21:22,740 --> 00:21:27,050 For a fixed value of x, y goes from y 427 00:21:27,050 --> 00:21:31,310 equals x-- well that's this line here-- to y equals 2. 428 00:21:31,310 --> 00:21:32,830 That's this line here. 429 00:21:32,830 --> 00:21:37,280 So for a fixed x, y varies continuously from here to here. 430 00:21:37,280 --> 00:21:40,340 x can be any place from 0 to 2. 431 00:21:40,340 --> 00:21:45,970 That happens to be the point 2 comma 2 where 432 00:21:45,970 --> 00:21:47,310 these two lines intersect. 433 00:21:47,310 --> 00:21:50,870 So the region R is this rectangular region here. 434 00:21:50,870 --> 00:21:53,160 OK, this is our region R. And what we're saying 435 00:21:53,160 --> 00:21:55,180 is find the mass of the region R, 436 00:21:55,180 --> 00:21:59,119 if its density at the point x comma y is e to the y squared. 437 00:21:59,119 --> 00:22:01,410 Now, notice that this density exists even if I've never 438 00:22:01,410 --> 00:22:03,190 heard of the anti-derivative. 439 00:22:03,190 --> 00:22:06,530 If, however I elect to use the fundamental theorem, 440 00:22:06,530 --> 00:22:11,290 and I say OK, what I'll do is I'll compute this. 441 00:22:11,290 --> 00:22:13,660 Notice I'm back to an old bugaboo. 442 00:22:13,660 --> 00:22:17,720 I don't know how to integrate e to the y squared dy, 443 00:22:17,720 --> 00:22:19,340 other than by an approximation. 444 00:22:19,340 --> 00:22:21,050 In terms of elementary functions, 445 00:22:21,050 --> 00:22:24,210 there is no function whose derivative with respect to y 446 00:22:24,210 --> 00:22:26,240 is e to the y squared. 447 00:22:26,240 --> 00:22:28,590 So what I do is I elect to change 448 00:22:28,590 --> 00:22:30,490 the order of integration. 449 00:22:30,490 --> 00:22:31,630 Why is that? 450 00:22:31,630 --> 00:22:34,820 Well, because if I change the order of integration, 451 00:22:34,820 --> 00:22:37,160 sure I'm integrating e to the y squared, 452 00:22:37,160 --> 00:22:40,520 but now with respect to x, which means that e to the y squared 453 00:22:40,520 --> 00:22:41,860 is a constant. 454 00:22:41,860 --> 00:22:43,660 This is trivial to integrate. 455 00:22:43,660 --> 00:22:47,200 It's just going to be x times e to the y squared. 456 00:22:47,200 --> 00:22:49,000 The problem is what? 457 00:22:49,000 --> 00:22:51,070 If I'm going to change the order here, 458 00:22:51,070 --> 00:22:53,230 I must make sure that I take care 459 00:22:53,230 --> 00:22:56,260 of the limits of integration accordingly. 460 00:22:56,260 --> 00:22:57,472 Now what this says is what? 461 00:22:57,472 --> 00:22:58,930 I'm going to integrate with respect 462 00:22:58,930 --> 00:23:01,260 to x first. x is going to vary, so I'm 463 00:23:01,260 --> 00:23:02,900 treating y as a constant. 464 00:23:02,900 --> 00:23:05,510 For that constant value of y, notice 465 00:23:05,510 --> 00:23:11,030 that to be in the region R, for that constant value of y, 466 00:23:11,030 --> 00:23:14,310 our x varies from what? 467 00:23:14,310 --> 00:23:20,000 x varies from 0 to the line y equals x, or accordingly x 468 00:23:20,000 --> 00:23:21,050 equals y. 469 00:23:21,050 --> 00:23:26,710 So for that fixed value of y, x varies 470 00:23:26,710 --> 00:23:29,120 from x equals 0 to x equal y. 471 00:23:29,120 --> 00:23:32,270 And correspondingly, y can be chosen 472 00:23:32,270 --> 00:23:35,300 to be any fixed number between 0 and 2, 473 00:23:35,300 --> 00:23:37,840 and that gives me this particular double integral. 474 00:23:37,840 --> 00:23:41,720 And by the way, from this point on, the rest is child's play. 475 00:23:41,720 --> 00:23:43,860 Because when I integrate this, you see, 476 00:23:43,860 --> 00:23:46,005 this integrand is x e to the y squared. 477 00:23:53,400 --> 00:23:56,760 When x is equal to y, this becomes y e to the y squared. 478 00:23:56,760 --> 00:23:59,340 When x is 0 the lower limit drops out. 479 00:23:59,340 --> 00:24:03,040 So what I want to integrate now is 0 to 2 y 480 00:24:03,040 --> 00:24:04,750 e to the y squared dy. 481 00:24:04,750 --> 00:24:08,460 But this is beautiful for me, because the factor of y in here 482 00:24:08,460 --> 00:24:11,080 is exactly what I need to be able to handle this. 483 00:24:11,080 --> 00:24:13,940 In other words, this is nothing more than 1/2 484 00:24:13,940 --> 00:24:18,340 e to the y squared evaluated between 0 and 2. 485 00:24:18,340 --> 00:24:21,810 Replacing y by two gives me 1/2 e to the fourth. 486 00:24:21,810 --> 00:24:25,070 Replacing y by 0-- remember e to the 0 is one-- 487 00:24:25,070 --> 00:24:29,130 gives me minus 1/2, because I'm subtracting the lower limit. 488 00:24:29,130 --> 00:24:32,690 The answer to evaluating this integral-- which 489 00:24:32,690 --> 00:24:35,526 was impossible to do in this form, because of the fact 490 00:24:35,526 --> 00:24:37,900 that there was no elementary function whose derivative is 491 00:24:37,900 --> 00:24:41,280 e to the y squared-- turns out quite nicely 492 00:24:41,280 --> 00:24:45,710 to be given by e to the fourth minus 1 over 2. 493 00:24:45,710 --> 00:24:48,090 To summarize today's lecture, to show you 494 00:24:48,090 --> 00:24:50,950 what the fundamental theorem really means, all we're saying 495 00:24:50,950 --> 00:24:55,110 is that we often compute the infinite double sum-- 496 00:24:55,110 --> 00:25:01,050 double integral R, f of x, y, dA by an appropriate iterated 497 00:25:01,050 --> 00:25:01,560 integral. 498 00:25:01,560 --> 00:25:02,930 That's-- what we mean is what? 499 00:25:02,930 --> 00:25:04,340 The anti-derivative. 500 00:25:04,340 --> 00:25:07,850 Integrating-- taking the anti-derivative successively. 501 00:25:07,850 --> 00:25:09,270 Iterated integral. 502 00:25:09,270 --> 00:25:10,040 See? 503 00:25:10,040 --> 00:25:12,820 Iterated integral. 504 00:25:12,820 --> 00:25:15,230 And conversely. 505 00:25:15,230 --> 00:25:18,400 In other words, the gist of this whole thing 506 00:25:18,400 --> 00:25:21,860 is that we now have two entirely different topics that 507 00:25:21,860 --> 00:25:25,600 are related by a fantastic unifying thread that 508 00:25:25,600 --> 00:25:28,900 allows us to solve one of the problems 509 00:25:28,900 --> 00:25:31,320 in terms of the other, and vice versa. 510 00:25:31,320 --> 00:25:35,010 You see, again, the analogy being complete 511 00:25:35,010 --> 00:25:39,420 with what happened in the calculus of a single variable. 512 00:25:39,420 --> 00:25:41,430 That we can evaluate double sums, 513 00:25:41,430 --> 00:25:45,640 which was the aim of the previous lecture, 514 00:25:45,640 --> 00:25:47,920 by means of an iterated integral. 515 00:25:47,920 --> 00:25:52,060 We can evaluate iterated integrals, sometimes, 516 00:25:52,060 --> 00:25:56,270 by appropriately knowing how to find the limit 517 00:25:56,270 --> 00:25:59,040 of an appropriate double sum. 518 00:25:59,040 --> 00:26:03,250 And again we shall make use of this in the exercises. 519 00:26:03,250 --> 00:26:06,670 We will go into this in more detail from other aspects, 520 00:26:06,670 --> 00:26:08,400 and other points of view next time. 521 00:26:08,400 --> 00:26:10,200 At any rate, until next time, goodbye. 522 00:26:13,230 --> 00:26:15,600 Funding for the publication of this video 523 00:26:15,600 --> 00:26:20,480 was provided by Gabriella and Paul Rosenbaum Foundation. 524 00:26:20,480 --> 00:26:24,650 Help OCW continue to provide free and open access to MIT 525 00:26:24,650 --> 00:26:32,360 courses by making a donation at ocw.mit.edu/donate.