1 00:00:00,040 --> 00:00:01,940 FEMALE SPEAKER: The following content is provided under a 2 00:00:01,940 --> 00:00:03,690 Creative Commons License. 3 00:00:03,690 --> 00:00:06,630 Your support will help MIT OpenCourseWare continue to 4 00:00:06,630 --> 00:00:09,990 offer high quality educational resources for free. 5 00:00:09,990 --> 00:00:12,830 To make a donation or to view additional materials from 6 00:00:12,830 --> 00:00:16,760 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:16,760 --> 00:00:18,010 ocw.mit.edu. 8 00:00:31,286 --> 00:00:32,290 PROFESSOR: Hi. 9 00:00:32,290 --> 00:00:35,520 Technically speaking, I suppose, if time permitted, we 10 00:00:35,520 --> 00:00:39,030 should spend, or shall we say, give equal time to integral 11 00:00:39,030 --> 00:00:42,180 calculus, the same time that we gave to differential 12 00:00:42,180 --> 00:00:46,080 calculus, starting from scratch, building ideas slowly 13 00:00:46,080 --> 00:00:46,750 and the like. 14 00:00:46,750 --> 00:00:49,710 But there are two logistic reasons for not doing this, 15 00:00:49,710 --> 00:00:53,390 one of which is that the basic definitions would be the same 16 00:00:53,390 --> 00:00:55,620 regardless of which branch we started with. 17 00:00:55,620 --> 00:00:58,550 The second one is, of course, that time becomes preciously 18 00:00:58,550 --> 00:01:02,010 tight, and we have to make the best of what we can here. 19 00:01:02,010 --> 00:01:06,750 So what we're going to do now is try to show, in terms of 20 00:01:06,750 --> 00:01:10,700 hindsight being better than foresight, a motivation as to 21 00:01:10,700 --> 00:01:14,970 how one would have invented differential calculus had it 22 00:01:14,970 --> 00:01:18,550 been motivated by the existing integral calculus. 23 00:01:18,550 --> 00:01:22,070 In other words, what we want to do today is to show the 24 00:01:22,070 --> 00:01:24,580 beautiful interplay between differential 25 00:01:24,580 --> 00:01:26,290 and integral calculus. 26 00:01:26,290 --> 00:01:29,350 Since we have said in our last lecture that we would like to 27 00:01:29,350 --> 00:01:32,830 begin with a study of integral calculus, assuming that 28 00:01:32,830 --> 00:01:35,990 differential calculus had never been invented, I think 29 00:01:35,990 --> 00:01:39,720 that we're obligated to try to show how, then, differential 30 00:01:39,720 --> 00:01:42,860 calculus could have been invented in the environment of 31 00:01:42,860 --> 00:01:44,510 the studying of areas. 32 00:01:44,510 --> 00:01:48,030 For this reason, I have called today's lecture "The Marriage 33 00:01:48,030 --> 00:01:51,910 of Differential and Integral Calculus." And essentially, 34 00:01:51,910 --> 00:01:54,560 the idea goes something like this. 35 00:01:54,560 --> 00:01:57,370 Last time, we were talking about finding 36 00:01:57,370 --> 00:01:59,240 areas under a curve. 37 00:01:59,240 --> 00:02:02,170 And to find the area under the curve, we saw that we could do 38 00:02:02,170 --> 00:02:04,450 this as the limit of a certain sum. 39 00:02:04,450 --> 00:02:07,630 In other words, an infinite sum having a particular limit. 40 00:02:07,630 --> 00:02:11,450 Now, to motivate the concept of rate of change, one could 41 00:02:11,450 --> 00:02:13,500 ask the following question. 42 00:02:13,500 --> 00:02:16,710 A curve like 'y' equals 'f of x', and a starting point, 'x' 43 00:02:16,710 --> 00:02:17,830 equals 'a'. 44 00:02:17,830 --> 00:02:21,980 As I move out along the x-axis from 'a', I could study the 45 00:02:21,980 --> 00:02:26,670 area under the curve, say, of the curve, 'y' equals 'f of 46 00:02:26,670 --> 00:02:30,600 x', from 'x' equals 'a' to some value 'x1'. 47 00:02:30,600 --> 00:02:34,280 And the question I could then ask is, I wonder how fast this 48 00:02:34,280 --> 00:02:38,040 area is changing at the instant that 'x' equals 'x1'. 49 00:02:38,040 --> 00:02:41,860 In other words, how fast is the area changing under the 50 00:02:41,860 --> 00:02:47,150 curve as 'x' moves out along the positive x-axis this way? 51 00:02:47,150 --> 00:02:48,160 You see? 52 00:02:48,160 --> 00:02:50,960 Now, the idea is this. 53 00:02:50,960 --> 00:02:53,950 I can then mimic the same definition that we gave for 54 00:02:53,950 --> 00:02:56,890 instantaneous rate of change and differential calculus. 55 00:02:56,890 --> 00:03:00,060 Namely, I can say, why can't we define the instantaneous 56 00:03:00,060 --> 00:03:03,510 rate of change to be the average rate of change with 57 00:03:03,510 --> 00:03:04,520 respect to 'x'? 58 00:03:04,520 --> 00:03:06,350 And then we'll take the limit as 'delta x' 59 00:03:06,350 --> 00:03:08,130 approaches 0, et cetera. 60 00:03:08,130 --> 00:03:10,420 And if I do that, watch what happens over here. 61 00:03:10,420 --> 00:03:13,600 You see what I do in terms of this picture, is I say, OK, 62 00:03:13,600 --> 00:03:15,670 here's the area under the curve when 63 00:03:15,670 --> 00:03:17,300 'x' is equal to 'x1'. 64 00:03:17,300 --> 00:03:20,720 Now I'll let 'x1' change by some amount 'delta x', meaning 65 00:03:20,720 --> 00:03:23,120 this'll be the point 'x1 + delta x'. 66 00:03:23,120 --> 00:03:25,770 And this brings about a change in my area, which 67 00:03:25,770 --> 00:03:27,330 I'll call 'delta A'. 68 00:03:27,330 --> 00:03:29,790 See the change in area as 'x' goes from 'x1' 69 00:03:29,790 --> 00:03:31,840 to 'x1 + delta x'. 70 00:03:31,840 --> 00:03:35,830 Now, what I would like to find is 'delta A' divided by 'delta 71 00:03:35,830 --> 00:03:39,390 x', taking the limit as 'delta x' approaches 0. 72 00:03:39,390 --> 00:03:41,720 Remember, last time we showed our three 73 00:03:41,720 --> 00:03:43,360 basic axioms for area. 74 00:03:43,360 --> 00:03:45,710 We're obliged to still use them. 75 00:03:45,710 --> 00:03:46,880 Let's do that here. 76 00:03:46,880 --> 00:03:49,960 For example, the way I've drawn this curve by rising 77 00:03:49,960 --> 00:03:53,540 this way, notice that the lowest point in this interval 78 00:03:53,540 --> 00:03:55,360 occurs when 'x' equals 'x1'. 79 00:03:55,360 --> 00:03:59,420 The highest point occurs when 'x' equals 'x1 + delta x'. 80 00:03:59,420 --> 00:04:03,380 Consequently, the rectangle whose base is the closed 81 00:04:03,380 --> 00:04:07,400 interval from 'x1' to 'x1 + delta x', the rectangle whose 82 00:04:07,400 --> 00:04:11,680 height corresponds to 'x' equals 'x1' is too small in 83 00:04:11,680 --> 00:04:13,670 area to be 'delta A'. 84 00:04:13,670 --> 00:04:17,160 In other words, that rectangle is inscribed in 'delta A'. 85 00:04:17,160 --> 00:04:20,760 And in the same way, the rectangle, with base 'x1' to 86 00:04:20,760 --> 00:04:25,720 'x1 + delta x', whose height is given by the x-coordinate, 87 00:04:25,720 --> 00:04:30,010 'x1 + delta x', that rectangle has an area which is too large 88 00:04:30,010 --> 00:04:31,610 to be the exact area. 89 00:04:31,610 --> 00:04:34,830 In other words, notice that the area of the rectangle, 90 00:04:34,830 --> 00:04:38,900 which is too small to be the exact area, is 'f of x1' 91 00:04:38,900 --> 00:04:40,690 times 'delta x'. 92 00:04:40,690 --> 00:04:44,310 The area of the rectangle which is too big to be the 93 00:04:44,310 --> 00:04:50,100 correct 'delta A' is 'f of 'x1 + delta x'' times 'delta x'. 94 00:04:50,100 --> 00:04:54,530 And the area, 'delta A', is caught between these two. 95 00:04:54,530 --> 00:04:58,850 In other words, we now have the inequality that 'f of x1' 96 00:04:58,850 --> 00:05:02,100 times 'delta x' is less than 'delta A', which in turn is 97 00:05:02,100 --> 00:05:06,550 less than 'f of 'x1 + delta x'' times 'delta x'. 98 00:05:06,550 --> 00:05:09,700 Now, we'll assume that 'delta x' is positive, the way we've 99 00:05:09,700 --> 00:05:11,940 drawn it in this diagram. 100 00:05:11,940 --> 00:05:15,320 Adjustments have to be made if 'delta x' is negative, in 101 00:05:15,320 --> 00:05:17,920 other words, if we let 'delta x' approach 0 102 00:05:17,920 --> 00:05:19,390 through negative values. 103 00:05:19,390 --> 00:05:22,300 This is taken care of in the textbook, as well as mentioned 104 00:05:22,300 --> 00:05:22,690 in our notes. 105 00:05:22,690 --> 00:05:25,550 But for the sake of just getting the main idea here, 106 00:05:25,550 --> 00:05:28,060 we'll assume that 'delta x' is positive. 107 00:05:28,060 --> 00:05:30,310 We divide through by 'delta x'. 108 00:05:30,310 --> 00:05:33,310 That gives us 'f of x1' is less than 'delta a' over 109 00:05:33,310 --> 00:05:35,380 'delta x', which in turn is less than 'f 110 00:05:35,380 --> 00:05:37,420 of 'x1 + delta x''. 111 00:05:37,420 --> 00:05:39,690 In other words, had 'delta x' been negative, we would have 112 00:05:39,690 --> 00:05:43,150 had to reverse the inequality signs, et cetera, but we're 113 00:05:43,150 --> 00:05:45,090 not going to worry about that right now. 114 00:05:45,090 --> 00:05:46,510 Remember, what is 'delta x'? 115 00:05:46,510 --> 00:05:50,610 It's a non-zero number which is going to be made to become 116 00:05:50,610 --> 00:05:52,380 arbitrarily close to 0. 117 00:05:52,380 --> 00:05:53,470 Well, here's the idea. 118 00:05:53,470 --> 00:05:56,500 We now have 'delta a' over 'delta x' squeezed between 119 00:05:56,500 --> 00:05:58,180 these two values. 120 00:05:58,180 --> 00:06:02,190 And we say, OK, let's let 'delta x' approach 0. 121 00:06:02,190 --> 00:06:03,490 Now, here's the key point. 122 00:06:03,490 --> 00:06:07,730 As 'delta x' approaches 0, certainly, 'x1 + delta x' 123 00:06:07,730 --> 00:06:10,110 approaches 'x1'. 124 00:06:10,110 --> 00:06:13,310 And here's where we use the fact that 'f' is continuous. 125 00:06:13,310 --> 00:06:16,460 In other words, notice that the mere fact that 'x1 + delta 126 00:06:16,460 --> 00:06:20,360 x' approaches 'x1' is not enough to say, therefore, that 127 00:06:20,360 --> 00:06:23,540 'f of 'x1 + delta x'' approaches 'f of x1'. 128 00:06:23,540 --> 00:06:27,610 As we saw before, that's only true if 'f' is continuous. 129 00:06:27,610 --> 00:06:31,660 In other words, by continuity, if 'x1 + delta x' approaches 130 00:06:31,660 --> 00:06:36,830 'x1', then 'f of 'x1 + delta x'' approaches 'f of x1'. 131 00:06:36,830 --> 00:06:40,210 So assuming, then, that 'f' is continuous, what do we get we 132 00:06:40,210 --> 00:06:41,360 put the squeeze on? 133 00:06:41,360 --> 00:06:43,910 'x1' is what's picked to be a fixed point. 134 00:06:43,910 --> 00:06:46,620 Consequently, as 'delta x' approaches 0, 'f of 135 00:06:46,620 --> 00:06:48,940 x1' stays 'f of x1'. 136 00:06:48,940 --> 00:06:52,480 'f of 'x1 + delta x'' approaches 'f of x1'. 137 00:06:52,480 --> 00:06:57,110 In other words, 'delta A' divided by 'delta x' is caught 138 00:06:57,110 --> 00:07:02,170 between two numbers, two sequences, two sets of bounds, 139 00:07:02,170 --> 00:07:05,510 whichever way you want to say this, both of which converge 140 00:07:05,510 --> 00:07:07,960 to the common limit, 'f of x1'. 141 00:07:07,960 --> 00:07:11,520 In other words, since 'delta A' over 'delta x' is squeezed 142 00:07:11,520 --> 00:07:15,320 between these two, the limit of 'delta a' over 'delta x', 143 00:07:15,320 --> 00:07:18,670 as 'delta x' approaches 0, must equal this common limit. 144 00:07:18,670 --> 00:07:23,140 In other words, 'dA/dx', the limit of 'delta A' over 'delta 145 00:07:23,140 --> 00:07:27,480 x', as 'delta x' approaches 0, evaluated at 'x' equals 'x1', 146 00:07:27,480 --> 00:07:29,130 is 'f of x1'. 147 00:07:29,130 --> 00:07:34,570 Or another way of saying this, 'dA/dx' is 'f of x'. 148 00:07:34,570 --> 00:07:39,180 Now, that result is partly intuitive and partly 149 00:07:39,180 --> 00:07:39,950 remarkable. 150 00:07:39,950 --> 00:07:44,170 What it says is, roughly speaking, is that the area 151 00:07:44,170 --> 00:07:49,480 under the curve is changing at a rate equal instantaneously 152 00:07:49,480 --> 00:07:52,090 to the height corresponding to the 153 00:07:52,090 --> 00:07:53,370 x-coordinate at that point. 154 00:07:53,370 --> 00:07:56,970 In other words, to define how fast the area is changing as 155 00:07:56,970 --> 00:08:01,140 'x' moves, all you have to do is measure the height of the 156 00:08:01,140 --> 00:08:06,510 curve corresponding to that particular value of 'x'. 157 00:08:06,510 --> 00:08:09,220 Now, what does this have to do, then, with relating 158 00:08:09,220 --> 00:08:11,420 integral and differential calculus? 159 00:08:11,420 --> 00:08:14,310 Notice already we begin to get some sort of a hint. 160 00:08:14,310 --> 00:08:18,730 What we've shown is now that the area is going to be 161 00:08:18,730 --> 00:08:22,620 related somehow to the inverse derivative of 'f of x'. 162 00:08:22,620 --> 00:08:27,310 In fact, this is precisely what is meant by a rather 163 00:08:27,310 --> 00:08:28,840 important result. 164 00:08:28,840 --> 00:08:31,580 In fact, it's important enough so it's called the first 165 00:08:31,580 --> 00:08:34,890 fundamental theorem of integral calculus. 166 00:08:34,890 --> 00:08:36,390 And it simply says this. 167 00:08:36,390 --> 00:08:41,090 Suppose we know explicitly a function, capital 'G', such 168 00:08:41,090 --> 00:08:44,910 that the derivative of capital 'G' is 'f', where the 'f' now 169 00:08:44,910 --> 00:08:48,940 refers to the 'f' that we're talking about in our problem. 170 00:08:48,940 --> 00:08:52,640 Now, look it, what we already know is that 'A', the area 171 00:08:52,640 --> 00:08:55,170 function, has its derivative equal to 'f'. 172 00:08:55,170 --> 00:08:58,610 Consequently, since the area function and G have the same 173 00:08:58,610 --> 00:09:01,630 derivative, they must differ by, at most, a constant. 174 00:09:01,630 --> 00:09:05,180 In other words, the area as a function of 'x' is this 175 00:09:05,180 --> 00:09:07,310 function, 'G of x' plus 'c'. 176 00:09:07,310 --> 00:09:08,590 And what is this 'G of x'? 177 00:09:08,590 --> 00:09:10,030 It's not any old function. 178 00:09:10,030 --> 00:09:10,710 It's what? 179 00:09:10,710 --> 00:09:14,120 An inverse derivative of 'f of x'. 180 00:09:14,120 --> 00:09:21,820 Since the area evaluated at 'x' equals 'a' is 0, see, 181 00:09:21,820 --> 00:09:24,920 namely, the area under the curve, as 'x' goes from 'A' to 182 00:09:24,920 --> 00:09:26,910 'a', see, that's just a line. 183 00:09:26,910 --> 00:09:28,880 A line has no thickness, has no area. 184 00:09:28,880 --> 00:09:29,680 This is 0. 185 00:09:29,680 --> 00:09:32,690 We have, by plugging back in to here, that 'A' of 186 00:09:32,690 --> 00:09:34,450 little 'a' is 0. 187 00:09:34,450 --> 00:09:38,100 That in turn is 'G of a' plus 'c'. 188 00:09:38,100 --> 00:09:41,990 And solving this for 'c', we find that 'c' is equal to 189 00:09:41,990 --> 00:09:43,340 minus 'G of a'. 190 00:09:43,340 --> 00:09:46,820 In other words, the area under the curve 'y' equals 'f of x', 191 00:09:46,820 --> 00:09:49,670 as a function of 'x', is simply what? 192 00:09:49,670 --> 00:09:53,640 It's 'G of x' minus 'G of a', where 'G' is any function 193 00:09:53,640 --> 00:09:55,290 whose derivative is 'f'. 194 00:09:55,290 --> 00:09:59,730 If we now let 'x' equal 'b', we're back to the situation of 195 00:09:59,730 --> 00:10:03,630 finding the usual region that we're talking about here, 196 00:10:03,630 --> 00:10:05,490 namely, what is 'A sub R'? 197 00:10:05,490 --> 00:10:08,200 It's the region bounded above-- let's come back here 198 00:10:08,200 --> 00:10:11,380 and take a look at this-- it's the region bounded above by 199 00:10:11,380 --> 00:10:15,220 the curve, 'y' equals 'f of x', bounded on the left by 'x' 200 00:10:15,220 --> 00:10:19,910 equals 'a', on the right by 'x' equals 'b', and below by 201 00:10:19,910 --> 00:10:21,150 the x-axis. 202 00:10:21,150 --> 00:10:24,510 According to this notation, the area of the region 'R' is 203 00:10:24,510 --> 00:10:29,020 just the area as a function of 'x' when 'x1' is equal to 'b'. 204 00:10:29,020 --> 00:10:32,465 In other words, the area of our region 'R' is just 'a' of 205 00:10:32,465 --> 00:10:36,260 'b', which is 'G of b' minus 'G of a', where 'G' is any 206 00:10:36,260 --> 00:10:39,880 function whose derivative is 'f'. 207 00:10:39,880 --> 00:10:40,670 Now remember-- 208 00:10:40,670 --> 00:10:42,080 and here's the key point-- 209 00:10:42,080 --> 00:10:47,480 remember that the area of the region 'R' did not necessitate 210 00:10:47,480 --> 00:10:51,080 us having to know a function whose derivative was 'f'. 211 00:10:51,080 --> 00:10:53,880 In other words, remember, in our last lecture, how did we 212 00:10:53,880 --> 00:10:56,000 find areas of regions like 'R'? 213 00:10:56,000 --> 00:10:58,990 What we did was is we formed these 'U sub n's, 214 00:10:58,990 --> 00:11:00,150 these 'L sub n's. 215 00:11:00,150 --> 00:11:04,170 We inscribed and circumscribed networks of rectangles, put 216 00:11:04,170 --> 00:11:07,850 the squeeze on by evaluating this particular limit. 217 00:11:07,850 --> 00:11:10,390 And whatever that limit was, that was the area of the 218 00:11:10,390 --> 00:11:11,250 region 'R'. 219 00:11:11,250 --> 00:11:14,120 In other words, if I had never heard of the inverse 220 00:11:14,120 --> 00:11:16,350 derivative, I could still find the area of the 221 00:11:16,350 --> 00:11:18,290 region 'R' this way. 222 00:11:18,290 --> 00:11:22,510 However, if I just happen to know a function 'G', whose 223 00:11:22,510 --> 00:11:26,612 derivative is 'f', I have a much easier way of doing this. 224 00:11:26,612 --> 00:11:29,630 In fact, let me summarize that. 225 00:11:29,630 --> 00:11:33,340 You see, we can compute the limit as 226 00:11:33,340 --> 00:11:34,960 'n' approaches infinity. 227 00:11:34,960 --> 00:11:38,700 Summation 'k' goes from 1 to 'n', 'f of 'c sub k'' 'delta 228 00:11:38,700 --> 00:11:41,850 x', by use of inverse derivatives. 229 00:11:41,850 --> 00:11:44,080 See, again, the highlight being what? 230 00:11:44,080 --> 00:11:47,590 You can still work with this the same way as we did in our 231 00:11:47,590 --> 00:11:50,110 last lecture. 232 00:11:50,110 --> 00:11:53,330 There is absolutely no need to have had to ever heard of a 233 00:11:53,330 --> 00:11:56,030 derivative to solve this type of problem, even 234 00:11:56,030 --> 00:11:57,450 though it may be messy. 235 00:11:57,450 --> 00:11:59,370 Now, what's the best proof I have of this? 236 00:11:59,370 --> 00:12:02,680 Well, I guess one of the best proofs is to specifically 237 00:12:02,680 --> 00:12:07,000 refer to one of the exercises in the previous unit. 238 00:12:07,000 --> 00:12:08,300 It was a tough exercise. 239 00:12:08,300 --> 00:12:11,050 I did it as a learning exercise because I felt it was 240 00:12:11,050 --> 00:12:13,870 something that was messy and that you had to be guided 241 00:12:13,870 --> 00:12:16,880 through in order not to become hopelessly lost. 242 00:12:16,880 --> 00:12:19,230 By way of review, the problem was this. 243 00:12:19,230 --> 00:12:23,990 We took as our region 'R' the region bounded above by 'y' 244 00:12:23,990 --> 00:12:28,850 equals 'sine x', below by the x-axis, on the right by the 245 00:12:28,850 --> 00:12:31,750 line 'x' equals pi/2. 246 00:12:31,750 --> 00:12:34,620 And the problem was, define the area of the region 'R'. 247 00:12:34,620 --> 00:12:38,420 And we solved this problem in the last unit without recourse 248 00:12:38,420 --> 00:12:41,740 to derivatives because, as of the time that we were in the 249 00:12:41,740 --> 00:12:45,780 last unit, we had no results relating derivatives 250 00:12:45,780 --> 00:12:48,890 to limits of sums. 251 00:12:48,890 --> 00:12:49,800 How did we do this? 252 00:12:49,800 --> 00:12:54,080 We partitioned this into n parts, and we formed the sum 253 00:12:54,080 --> 00:12:57,680 limit as 'n' approaches infinity. 254 00:12:57,680 --> 00:12:59,460 'Sigma k' goes from 1 to 'n'. 255 00:12:59,460 --> 00:13:02,780 We broke this thing up into 'n' equal parts, so the size 256 00:13:02,780 --> 00:13:04,020 of each piece was what? 257 00:13:04,020 --> 00:13:06,170 pi/2 divided by 'n'. 258 00:13:06,170 --> 00:13:07,440 That's 'pi/2n'. 259 00:13:07,440 --> 00:13:10,240 We computed the sine of each of these n-points, et cetera. 260 00:13:10,240 --> 00:13:12,880 And whatever that limit was, that was the area of the 261 00:13:12,880 --> 00:13:13,680 region 'R'. 262 00:13:13,680 --> 00:13:16,400 And as you recall, that homework problem, we found 263 00:13:16,400 --> 00:13:18,840 that the answer to that problem was that the area of 264 00:13:18,840 --> 00:13:21,330 the region 'R' was 1. 265 00:13:21,330 --> 00:13:24,670 Now again, as messy as that was, it proved that we could 266 00:13:24,670 --> 00:13:27,330 at least solve the problem with no knowledge of 267 00:13:27,330 --> 00:13:28,200 derivatives. 268 00:13:28,200 --> 00:13:31,230 How does the first fundamental theorem apply here? 269 00:13:31,230 --> 00:13:32,390 The way the first fundamental theorem 270 00:13:32,390 --> 00:13:34,020 applies is the following. 271 00:13:34,020 --> 00:13:35,890 The first fundamental theorem says this. 272 00:13:35,890 --> 00:13:40,060 Look it, to evaluate this particular sum, all we have to 273 00:13:40,060 --> 00:13:44,750 do is find a function whose derivative is 'sine x'. 274 00:13:44,750 --> 00:13:48,290 See, notice in this problem, the general 'a' and 'b' of the 275 00:13:48,290 --> 00:13:53,120 above is now played by 'a' equals 0 and 'b' equals pi/2. 276 00:13:53,120 --> 00:13:54,500 All we have to do is what? 277 00:13:54,500 --> 00:13:58,260 Find the function 'G' whose derivative is 'sine x', and 278 00:13:58,260 --> 00:14:02,670 the answer should then be 'G of pi/2' minus 'G of 0'. 279 00:14:02,670 --> 00:14:05,800 Well, you see this happens to be one that, with our 280 00:14:05,800 --> 00:14:08,970 knowledge of differential calculus, we 281 00:14:08,970 --> 00:14:09,970 can do rather easily. 282 00:14:09,970 --> 00:14:12,570 In other words, do we, at the tip of our tongues, have a 283 00:14:12,570 --> 00:14:15,990 function whose derivative is 'sine x'? 284 00:14:15,990 --> 00:14:17,030 And the answer is yes. 285 00:14:17,030 --> 00:14:20,130 Since the derivative of 'cosine x' is 'minus sine x', 286 00:14:20,130 --> 00:14:23,600 the derivative of 'minus cosine x' is 'sine x'. 287 00:14:23,600 --> 00:14:26,530 In other words, we can choose for our 'G of x' here 288 00:14:26,530 --> 00:14:29,020 'minus cosine x'. 289 00:14:29,020 --> 00:14:30,530 In other words, that the area of the region 290 00:14:30,530 --> 00:14:32,050 'R' should be what? 291 00:14:32,050 --> 00:14:36,460 'G of pi/2' minus 'G of 0', where 'G of x' is 292 00:14:36,460 --> 00:14:38,180 'minus cosine x'. 293 00:14:38,180 --> 00:14:41,940 Well, you see, cosine pi/2 is 0. 294 00:14:41,940 --> 00:14:49,120 Cosine of 0 is 1, so 0 minus minus 1 is equal to 1. 295 00:14:49,120 --> 00:14:52,440 And notice that, first of all, we get the same answer. 296 00:14:52,440 --> 00:14:55,510 And secondly, notice this two-line job over here. 297 00:14:55,510 --> 00:14:58,550 We not only get the correct answer, but we get the answer 298 00:14:58,550 --> 00:15:01,450 very much more rapidly than we did by the 299 00:15:01,450 --> 00:15:03,620 so-called limit process. 300 00:15:03,620 --> 00:15:06,640 Now, I'll come back to this in a minute, because you may be a 301 00:15:06,640 --> 00:15:08,190 little bit angry at me now. 302 00:15:08,190 --> 00:15:10,810 Namely, you may be asking, after having to do this 303 00:15:10,810 --> 00:15:14,220 problem the hard way, why did I have to do the usual 304 00:15:14,220 --> 00:15:17,510 teacher's trick here and show you the hard way of doing this 305 00:15:17,510 --> 00:15:20,000 and then waiting until the next unit before I show you 306 00:15:20,000 --> 00:15:21,120 the easy way? 307 00:15:21,120 --> 00:15:23,780 Well, there happens to be a catch here that I'll come back 308 00:15:23,780 --> 00:15:25,490 to in just a moment. 309 00:15:25,490 --> 00:15:29,340 But before I do that, I'd like to make an aside, an aside 310 00:15:29,340 --> 00:15:30,410 that's rather important. 311 00:15:30,410 --> 00:15:34,220 Namely, you may recall, as I started to work over here, 312 00:15:34,220 --> 00:15:38,290 that this reminded you of a notation that we were using 313 00:15:38,290 --> 00:15:42,280 back when we introduced the concept of the inverse of 314 00:15:42,280 --> 00:15:43,700 differentiation. 315 00:15:43,700 --> 00:15:48,070 Namely, we earlier used a notation, integral from 'a' to 316 00:15:48,070 --> 00:15:53,160 'b', 'f of x' 'dx' to denote 'G of b' minus 'G of a', where 317 00:15:53,160 --> 00:15:55,600 'G prime' equals 'f'. 318 00:15:55,600 --> 00:15:58,300 What we have shown by the first fundamental theorem, 319 00:15:58,300 --> 00:16:02,770 then, using this notation, is that the area of the region 320 00:16:02,770 --> 00:16:06,990 'R' is integral from 'a' to 'b', 'f of x' 'dx'. 321 00:16:06,990 --> 00:16:08,480 In other words, this means what? 322 00:16:08,480 --> 00:16:14,910 It means 'G of b' minus 'G of a', where 'G 323 00:16:14,910 --> 00:16:16,880 prime' equals 'f'. 324 00:16:16,880 --> 00:16:19,220 OK, let's pause here for a moment. 325 00:16:19,220 --> 00:16:22,010 See, as we developed our course, this is 326 00:16:22,010 --> 00:16:23,930 how this would evolve. 327 00:16:23,930 --> 00:16:28,470 The interesting thing is that, historically, this notation 328 00:16:28,470 --> 00:16:33,400 was not used to define 'G of b' minus 'G of a', where 'G 329 00:16:33,400 --> 00:16:34,750 prime' equaled 'f'. 330 00:16:34,750 --> 00:16:38,400 Historically, what happened was that this notation, called 331 00:16:38,400 --> 00:16:40,970 the definite integral, the integral from 'a' to 'b', 'f 332 00:16:40,970 --> 00:16:42,910 of x' 'dx', was-- 333 00:16:42,910 --> 00:16:45,620 I don't know if it's proper to say invented, but let me just 334 00:16:45,620 --> 00:16:47,650 say it in quotation marks to play it safe-- 335 00:16:47,650 --> 00:16:51,490 it was "invented" to denote the limit of this sum. 336 00:16:51,490 --> 00:16:55,840 In fact, notice how much more meaningful the symbol looks in 337 00:16:55,840 --> 00:16:56,930 this connotation. 338 00:16:56,930 --> 00:16:59,320 In other words, this is the limit of a sum. 339 00:16:59,320 --> 00:17:02,100 You can think of sum as beginning with 's', and the 340 00:17:02,100 --> 00:17:05,180 integral of sine as an elongated 's'. 341 00:17:05,180 --> 00:17:08,690 In other words, when one first wrote this symbol, the 342 00:17:08,690 --> 00:17:13,040 definite integral, it was meant to denote this limit. 343 00:17:13,040 --> 00:17:16,160 The important point is that, by the first fundamental 344 00:17:16,160 --> 00:17:19,355 theorem, the definite integral, whether it's a limit 345 00:17:19,355 --> 00:17:23,230 or not, turns out to be 'G of b' minus 'G of a', where 'G 346 00:17:23,230 --> 00:17:24,589 prime' equals 'f'. 347 00:17:24,589 --> 00:17:28,089 In other words, it really makes no difference in terms 348 00:17:28,089 --> 00:17:30,540 of what answer you get, whether you think of this 349 00:17:30,540 --> 00:17:34,010 definite integral as meaning this, or whether you think of 350 00:17:34,010 --> 00:17:36,940 the definite integral as meaning this. 351 00:17:36,940 --> 00:17:38,800 See, numerically, they'll be the same. 352 00:17:38,800 --> 00:17:41,820 But conceptually, they're quite different. 353 00:17:41,820 --> 00:17:46,260 And that's precisely the point that leads to the most, I 354 00:17:46,260 --> 00:17:49,720 think, probably the most difficult part all the course, 355 00:17:49,720 --> 00:17:51,090 at least up until now. 356 00:17:51,090 --> 00:17:53,670 It's something that more people cause and have 357 00:17:53,670 --> 00:17:55,210 misinterpretation over. 358 00:17:55,210 --> 00:17:57,090 And so I'd like to present this in a very, 359 00:17:57,090 --> 00:17:58,580 very gradual way. 360 00:17:58,580 --> 00:18:01,540 See, the result will ultimately be known as the 361 00:18:01,540 --> 00:18:04,490 second fundamental theorem of integral calculus. 362 00:18:04,490 --> 00:18:07,510 I've taken the liberty of writing that over here. 363 00:18:07,510 --> 00:18:10,620 But it's going to be quite a while before I actually say 364 00:18:10,620 --> 00:18:12,760 what the theorem is explicitly. 365 00:18:12,760 --> 00:18:16,280 What I'd like to do is to pick up where I left off when I 366 00:18:16,280 --> 00:18:18,120 made my aside about the notation 367 00:18:18,120 --> 00:18:19,770 of a definite integral. 368 00:18:19,770 --> 00:18:21,820 And what I'd like to mention now is, for the person who 369 00:18:21,820 --> 00:18:23,760 says, look it, why did you make us find the 370 00:18:23,760 --> 00:18:24,900 areas the hard way? 371 00:18:24,900 --> 00:18:29,250 Why didn't you just say let us compute 'G of b' minus 'G of 372 00:18:29,250 --> 00:18:32,200 a', where 'G prime' equals 'f', and the heck with all 373 00:18:32,200 --> 00:18:34,090 this summation process? 374 00:18:34,090 --> 00:18:36,690 Now, I think the best way to answer that question-- 375 00:18:36,690 --> 00:18:38,510 and this is probably the ultimate in teaching 376 00:18:38,510 --> 00:18:41,270 technique, when somebody thinks that he has an easier 377 00:18:41,270 --> 00:18:44,410 way than the way that he was taught-- 378 00:18:44,410 --> 00:18:47,980 give them a problem that can't be done by that easier way. 379 00:18:47,980 --> 00:18:51,040 In fact, as a sarcastic aside, if you can't invent that kind 380 00:18:51,040 --> 00:18:53,990 of a problem, maybe the so-called easier way is the 381 00:18:53,990 --> 00:18:55,330 better way of doing a thing. 382 00:18:55,330 --> 00:18:57,640 But let me just show you what I'm driving at over here. 383 00:18:57,640 --> 00:19:00,620 Suppose I say to you, let's find the area of the region 384 00:19:00,620 --> 00:19:03,220 'R' where 'R' is now the following. 385 00:19:03,220 --> 00:19:07,130 It's bounded above by the curve, 'y' equals '1/x'. 386 00:19:07,130 --> 00:19:09,950 It's bounded below by the x-axis. 387 00:19:09,950 --> 00:19:13,220 It's bounded on the left by the line, 'x' equals 1, and on 388 00:19:13,220 --> 00:19:15,630 the right by the line, 'x' equals 2. 389 00:19:15,630 --> 00:19:18,870 I would like to find the area of this region 'R'. 390 00:19:18,870 --> 00:19:24,520 Now look it, from a purely conceptual point of view, what 391 00:19:24,520 --> 00:19:27,810 difference is there between this problem and the problem 392 00:19:27,810 --> 00:19:31,620 where the upper curve was 'y' equals 'x squared'? 393 00:19:31,620 --> 00:19:33,840 Conceptually, what's going on is the same thing. 394 00:19:33,840 --> 00:19:37,280 We have a bounded region and we want to compute the area. 395 00:19:37,280 --> 00:19:41,490 To this end, notice that, if I use the precise definition of 396 00:19:41,490 --> 00:19:44,890 a definite integral, the area of the region 'R' is what? 397 00:19:44,890 --> 00:19:49,370 The definite integral from 1 to 2, 'dx/x'. 398 00:19:49,370 --> 00:19:50,970 What does that mean? 399 00:19:50,970 --> 00:19:52,900 It means a particular limit. 400 00:19:52,900 --> 00:19:53,890 What limit? 401 00:19:53,890 --> 00:19:56,990 Well, you partition this integral from 1 to 2 into n 402 00:19:56,990 --> 00:20:02,200 equal parts, pick a point, 'c sub k' in each partition, then 403 00:20:02,200 --> 00:20:05,880 'f of 'c sub k'' in this problem is just '1/'c sub k''. 404 00:20:05,880 --> 00:20:08,230 In other words, what we want to do is to compute this 405 00:20:08,230 --> 00:20:12,870 limit. 'Sigma k' goes from 1 to 'n', '1/'c sub k'', 406 00:20:12,870 --> 00:20:14,430 times 'delta x'. 407 00:20:14,430 --> 00:20:17,570 Now, the thing to notice is, this thing may be a mess but 408 00:20:17,570 --> 00:20:18,990 it's computable. 409 00:20:18,990 --> 00:20:24,320 We could use specially designed graph paper or 410 00:20:24,320 --> 00:20:27,650 measuring devices to count the units of area under here. 411 00:20:27,650 --> 00:20:31,720 We can find all sorts of ways of getting estimates, even the 412 00:20:31,720 --> 00:20:35,520 long hard way of the 'U sub n's and the 'L sub n's to 413 00:20:35,520 --> 00:20:39,080 pinpoint the area of the region 'R' to as close a 414 00:20:39,080 --> 00:20:41,340 degree of accuracy as we want, et cetera. 415 00:20:41,340 --> 00:20:44,650 The point is that, if we have never heard of a derivative, 416 00:20:44,650 --> 00:20:50,070 the area of the region 'R' is given precisely by this sum. 417 00:20:50,070 --> 00:20:53,970 And admittedly, the sum is a mess. 418 00:20:53,970 --> 00:20:57,200 So let's try to do it the so-called easier way. 419 00:20:57,200 --> 00:20:58,530 The skeptic looks at this thing and 420 00:20:58,530 --> 00:21:00,630 says, who needs this? 421 00:21:00,630 --> 00:21:03,460 The area of the region 'R', we just saw by the first 422 00:21:03,460 --> 00:21:08,370 fundamental theorem, is 'G of 2', 'G of 2', minus 'G of 1', 423 00:21:08,370 --> 00:21:11,360 where 'G' is any function of 'x' whose derivative with 424 00:21:11,360 --> 00:21:14,010 respect to 'x' is '1/x'. 425 00:21:14,010 --> 00:21:15,860 And the answer to that is yes. 426 00:21:15,860 --> 00:21:17,770 So far, so good. 427 00:21:17,770 --> 00:21:21,840 But do we know a function 'G' whose derivative with respect 428 00:21:21,840 --> 00:21:23,610 to 'x' is '1/x'? 429 00:21:23,610 --> 00:21:25,870 Remember, this is calculus revisited. 430 00:21:25,870 --> 00:21:27,790 For those of you who remember calculus from 431 00:21:27,790 --> 00:21:29,100 the first time around-- 432 00:21:29,100 --> 00:21:30,770 and I'll talk about this later-- 433 00:21:30,770 --> 00:21:33,570 it's going to turn out that the required 'G' is a 434 00:21:33,570 --> 00:21:35,110 logarithmic function. 435 00:21:35,110 --> 00:21:37,570 For those of you don't remember that, there's no harm 436 00:21:37,570 --> 00:21:38,770 in not remembering that. 437 00:21:38,770 --> 00:21:40,850 All I'm trying to bring out is that, whether you remember it 438 00:21:40,850 --> 00:21:44,000 or you don't, as far as we're concerned so far in this 439 00:21:44,000 --> 00:21:47,090 course, as far as what we've developed in this course, we 440 00:21:47,090 --> 00:21:50,720 do not know a function 'G' whose derivative with respect 441 00:21:50,720 --> 00:21:52,710 to 'x' is '1/x'. 442 00:21:52,710 --> 00:21:56,450 That's what we mean by saying we can't exhibit 'G' 443 00:21:56,450 --> 00:21:58,340 explicitly. 444 00:21:58,340 --> 00:21:59,490 What do I mean by that? 445 00:21:59,490 --> 00:22:01,150 Somebody says, I am thinking of a function whose 446 00:22:01,150 --> 00:22:02,660 derivative is '1/x'. 447 00:22:02,660 --> 00:22:04,310 I say, well, that's simple. 448 00:22:04,310 --> 00:22:05,150 It's 'G of x'. 449 00:22:05,150 --> 00:22:06,470 He says, well, what's 'G of x'? 450 00:22:06,470 --> 00:22:08,600 I say, that's the function whose derivative with respect 451 00:22:08,600 --> 00:22:10,340 to 'x' is '1/x'. 452 00:22:10,340 --> 00:22:13,210 Well, you see, that implicitly tells me what 'G' is like. 453 00:22:13,210 --> 00:22:15,680 But in terms of concrete measurements, I don't know 454 00:22:15,680 --> 00:22:16,630 anything about 'G'. 455 00:22:16,630 --> 00:22:18,390 I can't express it in terms of 456 00:22:18,390 --> 00:22:20,060 well-known, familiar functions. 457 00:22:20,060 --> 00:22:21,610 You see, I'm hung up now. 458 00:22:21,610 --> 00:22:23,620 Namely, this is precise. 459 00:22:23,620 --> 00:22:26,970 The answer to this problem will be 'G of 2' minus 'G of 460 00:22:26,970 --> 00:22:29,790 1', where 'G prime' is '1/x'. 461 00:22:29,790 --> 00:22:33,850 But I don't know explicitly-- 462 00:22:33,850 --> 00:22:35,020 yikes-- 463 00:22:35,020 --> 00:22:37,470 I don't know, explicitly such a 'G'. 464 00:22:37,470 --> 00:22:40,290 And I put the exclamation point out here to emphasize 465 00:22:40,290 --> 00:22:42,690 that particular fact. 466 00:22:42,690 --> 00:22:43,980 That's the hang-up. 467 00:22:43,980 --> 00:22:47,130 The statement that says that the area is 'G of b' minus 'G 468 00:22:47,130 --> 00:22:50,960 of a', where 'G prime' equals 'f' hinges on the fact that 469 00:22:50,960 --> 00:22:54,120 you can explicitly exhibit such as 'G'. 470 00:22:54,120 --> 00:22:56,710 Certainly, if you can't exhibit that 'G', you can 471 00:22:56,710 --> 00:22:58,910 still compute this area as a limit. 472 00:22:58,910 --> 00:23:03,650 It would be tragic to say, oh, the area doesn't exist because 473 00:23:03,650 --> 00:23:06,560 I don't know a 'G' whose derivative is '1/x'. 474 00:23:06,560 --> 00:23:09,320 Certainly, this region 'R' has an area. 475 00:23:09,320 --> 00:23:13,550 In fact, because we can pinpoint the area of the 476 00:23:13,550 --> 00:23:17,920 region 'R', we can actually construct the 'G' such that 'G 477 00:23:17,920 --> 00:23:20,220 prime of x' equals '1/x'. 478 00:23:20,220 --> 00:23:22,580 And by the way, there's plenty of drill on this. 479 00:23:22,580 --> 00:23:24,210 This is a hard concept. 480 00:23:24,210 --> 00:23:26,690 And as a result, you'll notice that the exercises in this 481 00:23:26,690 --> 00:23:30,250 section hammer home on this point, because it's a point 482 00:23:30,250 --> 00:23:33,090 that I'm positive that, if you're having trouble at all 483 00:23:33,090 --> 00:23:35,580 with integral calculus, this is certainly the most 484 00:23:35,580 --> 00:23:38,910 sophisticated part of what we're doing right now. 485 00:23:38,910 --> 00:23:41,730 You see, look it, suppose I want to construct a function 486 00:23:41,730 --> 00:23:45,770 'G' whose derivative is '1/x'. 487 00:23:45,770 --> 00:23:47,470 The idea looks something like this. 488 00:23:47,470 --> 00:23:48,990 I'm on the interval from 1 to 2 here. 489 00:23:48,990 --> 00:23:52,970 What I do is, I plot the curve, 'y' equals '1/x'. 490 00:23:52,970 --> 00:23:57,450 And I compute the area of the region formed by the curve, 491 00:23:57,450 --> 00:23:59,410 'y' equals '1/x'. 492 00:23:59,410 --> 00:24:00,740 'x' equals 1. 493 00:24:00,740 --> 00:24:06,500 The x-axis and the line, 'x' equals 'x1'. 494 00:24:06,500 --> 00:24:09,160 This is my region 'R'. 495 00:24:09,160 --> 00:24:11,210 And what I say is, look it, what do I 496 00:24:11,210 --> 00:24:13,220 know about this area? 497 00:24:13,220 --> 00:24:16,120 Remember, what I started this lecture with was the knowledge 498 00:24:16,120 --> 00:24:19,870 that a prime of 'x' is 'f of x'. 'f of x', in 499 00:24:19,870 --> 00:24:21,360 this case, is '1/x'. 500 00:24:21,360 --> 00:24:24,460 In other words, what property does this area function have? 501 00:24:24,460 --> 00:24:27,970 It has the property that its derivative with respect to 'x' 502 00:24:27,970 --> 00:24:29,890 is going to be '1/x'. 503 00:24:29,890 --> 00:24:33,440 In other words, if I now say, look it, boys, I can compute 504 00:24:33,440 --> 00:24:37,700 this area, if not exactly, at least to as many decimal place 505 00:24:37,700 --> 00:24:41,500 accuracy as I wish, so that the area function is something 506 00:24:41,500 --> 00:24:42,640 I can construct. 507 00:24:42,640 --> 00:24:48,100 Let me define 'G of x1' simply to be the area under this 508 00:24:48,100 --> 00:24:50,150 curve when 'x' is equal to 'x1'. 509 00:24:50,150 --> 00:24:51,940 In other words, written now as a limit-- 510 00:24:51,940 --> 00:24:53,600 see how this definite integral comes in here-- 511 00:24:53,600 --> 00:24:58,980 written as integral from 1 to 'x sub 1', 'dx/x', which I now 512 00:24:58,980 --> 00:25:01,460 can compute as closely as I want as a limit. 513 00:25:01,460 --> 00:25:02,840 The beauty is what? 514 00:25:02,840 --> 00:25:07,270 That 'G prime of x1' is, by definition, 'A prime of 'x1'. 515 00:25:07,270 --> 00:25:11,120 And we have already seen that the derivative of 'A' is 'f', 516 00:25:11,120 --> 00:25:13,350 where 'f' is the top curve. 517 00:25:13,350 --> 00:25:16,440 In this case, the derivative of 'A' is '1/x'. 518 00:25:16,440 --> 00:25:20,090 In other words, the area under the curve explicitly can be 519 00:25:20,090 --> 00:25:23,020 computed as a function of 'x'. 520 00:25:23,020 --> 00:25:25,930 And that function has the property, but its derivative 521 00:25:25,930 --> 00:25:28,380 with respect to 'x', in this case, is '1/x'. 522 00:25:30,910 --> 00:25:34,670 And by the way, let me make just a quick aside over here. 523 00:25:34,670 --> 00:25:37,010 I happened to throw in the word "logarithms" before. 524 00:25:37,010 --> 00:25:39,340 It's going to turn out, as we'll talk about in our next 525 00:25:39,340 --> 00:25:43,030 block of material, that 'G of x' turns out to be the natural 526 00:25:43,030 --> 00:25:46,480 log of 'x', which means that the area of the region 'R' is 527 00:25:46,480 --> 00:25:49,760 precisely natural log 2. 528 00:25:49,760 --> 00:25:51,540 Now, the idea is-- 529 00:25:51,540 --> 00:25:53,460 and this is exactly what's going to happen in our last 530 00:25:53,460 --> 00:25:54,890 block in our course-- 531 00:25:54,890 --> 00:25:56,790 that this is exactly how the log tables are 532 00:25:56,790 --> 00:25:58,450 constructed, in fact. 533 00:25:58,450 --> 00:26:00,820 I mean, how do you think they find the log of 2 to eight 534 00:26:00,820 --> 00:26:01,630 decimal places? 535 00:26:01,630 --> 00:26:04,070 Did they strand somebody on a desert island and tell them to 536 00:26:04,070 --> 00:26:08,330 compute these things, measure it with laser beams, what? 537 00:26:08,330 --> 00:26:08,740 No. 538 00:26:08,740 --> 00:26:11,490 The way we do it is, we know it's an area, and knowing how 539 00:26:11,490 --> 00:26:14,710 to find the area as a limit to as many decimal place accuracy 540 00:26:14,710 --> 00:26:17,910 as we wish, give or take a few tricks of the trade, this is 541 00:26:17,910 --> 00:26:21,080 how we find logs, and later on, trigonometric 542 00:26:21,080 --> 00:26:22,330 tables, and the like. 543 00:26:22,330 --> 00:26:25,050 But I just mention this as an aside to show you how 544 00:26:25,050 --> 00:26:28,260 important knowing area under a curve is. 545 00:26:28,260 --> 00:26:31,750 At any rate, to summarize what we're saying over here, we 546 00:26:31,750 --> 00:26:32,670 have the following. 547 00:26:32,670 --> 00:26:36,520 In general, if 'f' is any continuous function on the 548 00:26:36,520 --> 00:26:40,080 closed interval from 'a' to 'b', we define a function 'G' 549 00:26:40,080 --> 00:26:41,320 as follows. 550 00:26:41,320 --> 00:26:46,180 'G of x1' is the definite integral from 'a' to 'x1', 'f 551 00:26:46,180 --> 00:26:49,510 of x' 'dx', where 'x1' is any point in the closed interval 552 00:26:49,510 --> 00:26:50,430 for 'a' to 'b'. 553 00:26:50,430 --> 00:26:52,670 What does this thing mean, geometrically? 554 00:26:52,670 --> 00:26:57,200 It's the area under the curve, 'y' equals 'f of x' on top. 555 00:26:57,200 --> 00:27:00,940 'x' equals the x-axis on the bottom, the line 'x' equals 556 00:27:00,940 --> 00:27:05,190 'a' on the left, the line 'x' equals 'x1' on the right. 557 00:27:05,190 --> 00:27:06,800 And that area is what? 558 00:27:06,800 --> 00:27:08,750 It's the limit as 'n' approaches infinity. 559 00:27:08,750 --> 00:27:12,730 'Sigma k' goes from 1 to 'n', 'f of 'c sub k'' 'delta x', et 560 00:27:12,730 --> 00:27:16,060 cetera, et cetera, et cetera, meaning we can go through this 561 00:27:16,060 --> 00:27:19,740 the same way as we did in our previous lecture. 562 00:27:19,740 --> 00:27:21,370 This function, 'G of x1'-- 563 00:27:21,370 --> 00:27:25,060 which we can now compute explicitly as an area, because 564 00:27:25,060 --> 00:27:27,710 we have the curve 'f' drawn, we can approximate it to as 565 00:27:27,710 --> 00:27:29,510 close a degree of accuracy as we want-- 566 00:27:29,510 --> 00:27:33,420 whatever that function 'G' is, what property does it have? 567 00:27:33,420 --> 00:27:38,380 It has the property that 'G prime' is equal to 'f'. 568 00:27:38,380 --> 00:27:41,090 And that's exactly what the second fundamental theorem 569 00:27:41,090 --> 00:27:42,020 really means. 570 00:27:42,020 --> 00:27:45,430 The second fundamental theorem says, look it, if you can 571 00:27:45,430 --> 00:27:49,480 compute the area, you can now reverse this procedure that 572 00:27:49,480 --> 00:27:52,150 we're talking about in the first fundamental theorem. 573 00:27:52,150 --> 00:27:54,890 See, in the first fundamental theorem, what did we do? 574 00:27:54,890 --> 00:27:57,510 In the first fundamental theorem, we found a quick way 575 00:27:57,510 --> 00:28:00,100 of computing the area as a limit by knowing the 576 00:28:00,100 --> 00:28:01,980 appropriate inverse derivative. 577 00:28:01,980 --> 00:28:05,060 The second fundamental theorem, in a sense, is the 578 00:28:05,060 --> 00:28:07,980 inverse of the first fundamental theorem. 579 00:28:07,980 --> 00:28:11,110 Namely, it switches the roles. 580 00:28:11,110 --> 00:28:12,950 Namely, with the second fundamental theorem, what we 581 00:28:12,950 --> 00:28:13,800 say is this. 582 00:28:13,800 --> 00:28:16,760 If we know how to find the area under the curve, 'y' 583 00:28:16,760 --> 00:28:21,640 equals 'f of x', that gives us a way of computing a function 584 00:28:21,640 --> 00:28:23,910 'G' whose derivative is 'f'. 585 00:28:23,910 --> 00:28:26,100 In fact, to summarize this-- 586 00:28:26,100 --> 00:28:27,340 and I think this is very important. 587 00:28:27,340 --> 00:28:28,860 Let me just summarize this, then. 588 00:28:28,860 --> 00:28:32,560 First of all, the first fundamental theorem allows us 589 00:28:32,560 --> 00:28:37,480 to compute this particular limit, provided we can find a 590 00:28:37,480 --> 00:28:40,320 'G', such that 'G prime' equals 'f'. 591 00:28:40,320 --> 00:28:43,690 In fact, in this case, the limit, as 'n' approaches 592 00:28:43,690 --> 00:28:47,990 infinity, summation 'k' goes from 1 to 'n', 'f of 'c sub 593 00:28:47,990 --> 00:28:52,660 k'' 'delta x', is given very, very ingeniously, beautifully, 594 00:28:52,660 --> 00:28:57,110 and concisely by 'G of b' minus 'G of a'. 595 00:28:57,110 --> 00:29:01,400 Secondly, the second fundamental theorem allows us, 596 00:29:01,400 --> 00:29:05,010 given 'f', to construct 'G', such that 'G 597 00:29:05,010 --> 00:29:06,480 prime' equals 'f'. 598 00:29:06,480 --> 00:29:10,430 Namely, the required 'G' is just the definite integral 599 00:29:10,430 --> 00:29:16,095 from 'a' to 'x1', 'f of x' 'dx', which in turn is a limit 600 00:29:16,095 --> 00:29:20,120 of a network of rectangles, namely the limit as 'n' 601 00:29:20,120 --> 00:29:24,420 approaches infinity, summation 'k' goes from 1 to 'n', 'f of 602 00:29:24,420 --> 00:29:27,040 'c sub k'' times 'delta x'. 603 00:29:27,040 --> 00:29:32,150 This then is perhaps the most beautiful lecture in calculus, 604 00:29:32,150 --> 00:29:33,280 not necessarily the way I've given it, 605 00:29:33,280 --> 00:29:34,620 but in terms of what? 606 00:29:34,620 --> 00:29:39,250 Relating two apparently diverse branches of calculus, 607 00:29:39,250 --> 00:29:41,330 integral and differential calculus. 608 00:29:41,330 --> 00:29:43,210 This is a difficult subject. 609 00:29:43,210 --> 00:29:45,980 I have taken great pains to try to write this clearly in 610 00:29:45,980 --> 00:29:47,250 the supplementary notes. 611 00:29:47,250 --> 00:29:50,850 I have tried to pick typical exercises that bring home 612 00:29:50,850 --> 00:29:52,550 these highlights. 613 00:29:52,550 --> 00:29:57,470 And in addition, our next two lectures will emphasize the 614 00:29:57,470 --> 00:30:02,150 relationship between derivatives and integrals as 615 00:30:02,150 --> 00:30:04,540 we study volume and arc length. 616 00:30:04,540 --> 00:30:06,670 At any rate, until next time, goodbye. 617 00:30:09,710 --> 00:30:12,240 MALE SPEAKER: Funding for the publication of this video was 618 00:30:12,240 --> 00:30:16,960 provided by the Gabriella and Paul Rosenbaum Foundation. 619 00:30:16,960 --> 00:30:21,130 Help OCW continue to provide free and open access to MIT 620 00:30:21,130 --> 00:30:25,330 courses by making a donation at ocw.mit.edu/donate.