1 00:00:00,000 --> 00:00:02,330 The following content is provided under a Creative 2 00:00:02,330 --> 00:00:03,620 Commons license. 3 00:00:03,620 --> 00:00:05,990 Your support will help MIT OpenCourseWare 4 00:00:05,990 --> 00:00:09,455 continue to offer high quality educational resources for free. 5 00:00:09,455 --> 00:00:12,540 To make a donation, or to view additional materials 6 00:00:12,540 --> 00:00:15,850 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:15,850 --> 00:00:21,864 at ocw.mit.edu. 8 00:00:21,864 --> 00:00:24,280 PROFESSOR: Today we're going to continue with integration. 9 00:00:24,280 --> 00:00:29,670 And we get to do the-- probably the most important thing 10 00:00:29,670 --> 00:00:31,090 of this entire course. 11 00:00:31,090 --> 00:00:33,970 Which is appropriately named. 12 00:00:33,970 --> 00:00:50,250 It's called the fundamental theorem of calculus. 13 00:00:50,250 --> 00:00:54,900 And we'll be abbreviating it FTC and occasionally I'll 14 00:00:54,900 --> 00:00:58,964 put in a 1 here, because there will be two versions of it. 15 00:00:58,964 --> 00:01:00,380 But this is the one that you'll be 16 00:01:00,380 --> 00:01:06,420 using the most in this class. 17 00:01:06,420 --> 00:01:14,540 The fundamental theorem of calculus says the following. 18 00:01:14,540 --> 00:01:26,875 It says that if F' = f, so F'(x) = f(x), 19 00:01:26,875 --> 00:01:30,890 there's a capital F and a little f, 20 00:01:30,890 --> 00:01:45,960 then the integral from a to b of f(x) is equal to F(b) - F(a). 21 00:01:52,160 --> 00:01:52,940 That's it. 22 00:01:52,940 --> 00:01:55,600 That's the whole theorem. 23 00:01:55,600 --> 00:02:00,120 And you may recognize it. 24 00:02:00,120 --> 00:02:03,880 Before, we had the notation that F 25 00:02:03,880 --> 00:02:10,020 was the antiderivative, that is, capital F 26 00:02:10,020 --> 00:02:11,635 was the integral of f(x). 27 00:02:11,635 --> 00:02:12,510 We wrote it this way. 28 00:02:12,510 --> 00:02:14,860 This is this indefinite integral. 29 00:02:14,860 --> 00:02:17,970 And now we're putting in definite values. 30 00:02:17,970 --> 00:02:20,260 And we have a connection between the two 31 00:02:20,260 --> 00:02:22,420 uses of the integral sign. 32 00:02:22,420 --> 00:02:24,530 But with the definite values, we get real numbers 33 00:02:24,530 --> 00:02:26,220 out instead of a function. 34 00:02:26,220 --> 00:02:29,810 Or a function up to a constant. 35 00:02:29,810 --> 00:02:30,990 So this is it. 36 00:02:30,990 --> 00:02:32,120 This is the formula. 37 00:02:32,120 --> 00:02:35,400 And it's usually also written with another notation. 38 00:02:35,400 --> 00:02:40,390 So I want to introduce that notation to you as well. 39 00:02:40,390 --> 00:02:44,950 So there's a new notation here. 40 00:02:44,950 --> 00:02:47,320 Which you'll find very convenient. 41 00:02:47,320 --> 00:02:51,010 Because we don't always have to give a letter f 42 00:02:51,010 --> 00:02:52,710 to the functions involved. 43 00:02:52,710 --> 00:02:54,930 So it's an abbreviation. 44 00:02:54,930 --> 00:02:57,610 For right now there'll be a lot of f's, but anyway. 45 00:02:57,610 --> 00:02:59,660 So here's the abbreviation. 46 00:02:59,660 --> 00:03:04,420 Whenever I have a difference between a function at two 47 00:03:04,420 --> 00:03:09,430 values, I also can write this as F(x) 48 00:03:09,430 --> 00:03:12,440 with an a down here and a b up there. 49 00:03:12,440 --> 00:03:16,550 So that's the notation that we use. 50 00:03:16,550 --> 00:03:19,910 And you can also, for emphasis, and this sometimes 51 00:03:19,910 --> 00:03:23,450 turns out to be important, when there's more than one variable 52 00:03:23,450 --> 00:03:25,750 floating around in the problem. 53 00:03:25,750 --> 00:03:28,200 To specify that the variable is x. 54 00:03:28,200 --> 00:03:32,270 So this is the same thing as x = a. 55 00:03:32,270 --> 00:03:34,100 And x = b. 56 00:03:34,100 --> 00:03:36,040 It indicates where you want to plug in, 57 00:03:36,040 --> 00:03:37,450 what you want to plug in. 58 00:03:37,450 --> 00:03:41,840 And now you take the top value minus the bottom value. 59 00:03:41,840 --> 00:03:43,380 So F(b) - F(a). 60 00:03:43,380 --> 00:03:50,290 So this is just a notation, and in that notation, of course, 61 00:03:50,290 --> 00:03:59,420 the theorem can be written with this set of symbols here. 62 00:03:59,420 --> 00:04:04,160 Equally well. 63 00:04:04,160 --> 00:04:06,250 So let's just give a couple of examples. 64 00:04:06,250 --> 00:04:08,180 The first example is the one that we 65 00:04:08,180 --> 00:04:12,260 did last time very laboriously. 66 00:04:12,260 --> 00:04:19,370 If you take the function F(x), which happens to be x^3 / 3, 67 00:04:19,370 --> 00:04:24,090 then if you differentiate it, you get, well, 68 00:04:24,090 --> 00:04:25,870 the the factor of 3 cancels. 69 00:04:25,870 --> 00:04:29,440 So you get x^2, that's the derivative. 70 00:04:29,440 --> 00:04:32,760 And so by the fundamental theorem, 71 00:04:32,760 --> 00:04:37,290 so this implies by the fundamental theorem, 72 00:04:37,290 --> 00:04:47,140 that the integral from say, a to b of x^3 over - sorry, x^2 dx, 73 00:04:47,140 --> 00:04:50,280 that's the derivative here. 74 00:04:50,280 --> 00:04:55,620 This is the function we're going to use as f(x) here - 75 00:04:55,620 --> 00:05:02,530 is equal to this function here, F(b) - F(a), that's here. 76 00:05:02,530 --> 00:05:04,220 This function here. 77 00:05:04,220 --> 00:05:13,470 So that's F(b) - F(a), and that's equal to b^3 / 3 - 78 00:05:13,470 --> 00:05:19,480 a^3 / 3. 79 00:05:19,480 --> 00:05:23,710 Now, in this new notation, we usually 80 00:05:23,710 --> 00:05:25,060 don't have all of these letters. 81 00:05:25,060 --> 00:05:26,310 All we write is the following. 82 00:05:26,310 --> 00:05:27,970 We write the integral from a to b, 83 00:05:27,970 --> 00:05:29,470 and I'm going to do the case 0 to b, 84 00:05:29,470 --> 00:05:31,803 because that was the one that we actually did last time. 85 00:05:31,803 --> 00:05:35,940 So I'm going to set a = 0 here. 86 00:05:35,940 --> 00:05:39,590 And then, the problem we were faced last time as this. 87 00:05:39,590 --> 00:05:41,940 And as I said we did it very laboriously. 88 00:05:41,940 --> 00:05:47,620 But now you can see that we can do it in ten seconds, 89 00:05:47,620 --> 00:05:48,280 let's say. 90 00:05:48,280 --> 00:05:52,300 Well, the antiderivative of this is x^3 / 3. 91 00:05:52,300 --> 00:05:55,750 I'm going to evaluate it at 0 and at b and subtract. 92 00:05:55,750 --> 00:06:00,540 So that's going to be b^3 / 3 - 0^3 / 3. 93 00:06:00,540 --> 00:06:03,140 Which of course is b^3 / 3. 94 00:06:03,140 --> 00:06:06,010 And that's the end, that's the answer. 95 00:06:06,010 --> 00:06:08,820 So this is a lot faster than yesterday. 96 00:06:08,820 --> 00:06:10,860 I hope you'll agree. 97 00:06:10,860 --> 00:06:15,060 And we can dispense with those elaborate computations. 98 00:06:15,060 --> 00:06:18,460 Although there's a conceptual reason, a very important one, 99 00:06:18,460 --> 00:06:21,960 for understanding the procedure that we went through. 100 00:06:21,960 --> 00:06:26,829 Because eventually you're going to be using integrals 101 00:06:26,829 --> 00:06:28,370 and these quick ways of doing things, 102 00:06:28,370 --> 00:06:32,280 to solve problems like finding the volumes of pyramids. 103 00:06:32,280 --> 00:06:34,700 In other words, we're going to reverse the process. 104 00:06:34,700 --> 00:06:42,100 And so we need to understand the connection between the two. 105 00:06:42,100 --> 00:06:45,210 I'm going to give a couple more examples. 106 00:06:45,210 --> 00:06:47,280 And then we'll go on. 107 00:06:47,280 --> 00:06:49,700 So the second example would be one 108 00:06:49,700 --> 00:06:52,440 that would be quite difficult to do by this Riemann sum 109 00:06:52,440 --> 00:06:55,310 technique that we described yesterday. 110 00:06:55,310 --> 00:06:57,100 Although it is possible. 111 00:06:57,100 --> 00:06:59,860 It uses much higher mathematics to do it. 112 00:06:59,860 --> 00:07:16,360 And that is the area under one hump of the sine curve, sin x. 113 00:07:16,360 --> 00:07:17,920 Let me just draw a picture of that. 114 00:07:17,920 --> 00:07:20,990 The curve goes like this, and we're talking about this area 115 00:07:20,990 --> 00:07:21,490 here. 116 00:07:21,490 --> 00:07:24,370 It starts out at 0, it goes to pi. 117 00:07:24,370 --> 00:07:28,320 That's one hump. 118 00:07:28,320 --> 00:07:33,710 And so the answer is, it's the integral from 0 to pi of sin 119 00:07:33,710 --> 00:07:37,849 x dx. 120 00:07:37,849 --> 00:07:39,890 And so I need to take the antiderivative of that. 121 00:07:39,890 --> 00:07:42,970 And that's -cos x. 122 00:07:42,970 --> 00:07:46,650 That's the thing whose derivative is sin x. 123 00:07:46,650 --> 00:07:49,820 Evaluating it at 0 and pi. 124 00:07:49,820 --> 00:07:52,070 Now, let's do this one carefully. 125 00:07:52,070 --> 00:07:55,300 Because this is where I see a lot of arithmetic mistakes. 126 00:07:55,300 --> 00:07:57,690 Even though this is the easy part of the problem. 127 00:07:57,690 --> 00:08:02,100 It's hard to pay attention and plug in the right numbers. 128 00:08:02,100 --> 00:08:04,150 And so, let's just pay very close attention. 129 00:08:04,150 --> 00:08:05,530 I'm plugging in pi. 130 00:08:05,530 --> 00:08:08,444 That's -cos pi. 131 00:08:08,444 --> 00:08:09,360 That's the first term. 132 00:08:09,360 --> 00:08:12,240 And then I'm subtracting the value 133 00:08:12,240 --> 00:08:19,980 at the bottom, which is -cos 0. 134 00:08:19,980 --> 00:08:22,350 There are already five opportunities for you 135 00:08:22,350 --> 00:08:24,680 to make a transcription error or an arithmetic 136 00:08:24,680 --> 00:08:26,740 mistake in what I just did. 137 00:08:26,740 --> 00:08:30,090 And I've seen all five of them. 138 00:08:30,090 --> 00:08:34,670 So the next one is that this is -(-1). 139 00:08:34,670 --> 00:08:36,810 Minus negative 1, if you like. 140 00:08:36,810 --> 00:08:41,270 And then this is minus, and here's another -1. 141 00:08:41,270 --> 00:08:44,250 So altogether we have 2. 142 00:08:44,250 --> 00:08:44,920 So that's it. 143 00:08:44,920 --> 00:08:46,810 That's the area. 144 00:08:46,810 --> 00:09:02,960 This area, which is hard to guess, this is area 2. 145 00:09:02,960 --> 00:09:06,020 The third example is maybe superfluous 146 00:09:06,020 --> 00:09:10,280 but I'm going to say it anyway. 147 00:09:10,280 --> 00:09:17,740 We can take the integral, say, from 0 to 1, of x^100. 148 00:09:17,740 --> 00:09:21,260 Any power, now, is within our power. 149 00:09:21,260 --> 00:09:24,050 So let's do it. 150 00:09:24,050 --> 00:09:32,980 So here we have the antiderivative is x^101 / 101, 151 00:09:32,980 --> 00:09:36,550 evaluated at 0 and 1. 152 00:09:36,550 --> 00:09:42,050 And that is just 1 / 101. 153 00:09:42,050 --> 00:09:46,520 That's that. 154 00:09:46,520 --> 00:09:49,770 So that's the fundamental theorem. 155 00:09:49,770 --> 00:09:53,670 Now this, as I say, harnesses a lot 156 00:09:53,670 --> 00:09:58,110 of what we've already learned, all about antiderivatives. 157 00:09:58,110 --> 00:10:05,820 Now, I want to give you an intuitive interpretation. 158 00:10:05,820 --> 00:10:10,120 So let's try that. 159 00:10:10,120 --> 00:10:12,450 We'll talk about a proof of the fundamental theorem 160 00:10:12,450 --> 00:10:14,170 a little bit later. 161 00:10:14,170 --> 00:10:16,210 It's not actually that hard. 162 00:10:16,210 --> 00:10:22,470 But we'll give an intuitive reason, interpretation, 163 00:10:22,470 --> 00:10:28,040 if you like. 164 00:10:28,040 --> 00:10:37,670 Of the fundamental theorem. 165 00:10:37,670 --> 00:10:40,330 So this is going to be one which is not 166 00:10:40,330 --> 00:10:43,990 related to area, but rather to time and distance. 167 00:10:43,990 --> 00:10:55,600 So we'll consider x(t) is your position at time t. 168 00:10:55,600 --> 00:11:04,020 And then x'(t), which is dx/dt, is going to be what we know 169 00:11:04,020 --> 00:11:12,230 as your speed. 170 00:11:12,230 --> 00:11:18,150 And then what the theorem is telling us is the following. 171 00:11:18,150 --> 00:11:25,580 It's telling us the integral from a to b of v(t) dt - 172 00:11:25,580 --> 00:11:31,040 so, reading the relationship - is equal to x (b) - x(a). 173 00:11:35,930 --> 00:11:40,800 And so this is some kind of cumulative sum 174 00:11:40,800 --> 00:11:45,110 of your velocities. 175 00:11:45,110 --> 00:11:48,440 So let's interpret the right-hand side first. 176 00:11:48,440 --> 00:11:57,290 This is the distance traveled. 177 00:11:57,290 --> 00:12:03,030 And it's also what you would read on your odometer. 178 00:12:03,030 --> 00:12:05,550 Right, from the beginning to the end of the trip. 179 00:12:05,550 --> 00:12:07,540 That's what you would read on your odometer. 180 00:12:07,540 --> 00:12:19,500 Whereas this is what you would read on your speedometer. 181 00:12:19,500 --> 00:12:23,210 So this is the interpretation. 182 00:12:23,210 --> 00:12:25,540 Now, I want to just go one step further 183 00:12:25,540 --> 00:12:27,160 into this interpretation, to make 184 00:12:27,160 --> 00:12:32,890 the connection with the Riemann sums that we had yesterday. 185 00:12:32,890 --> 00:12:35,447 Because those are very complicated to understand. 186 00:12:35,447 --> 00:12:37,280 And I want you to understand them viscerally 187 00:12:37,280 --> 00:12:39,090 on several different levels. 188 00:12:39,090 --> 00:12:43,280 Because that's how you'll understand integration better. 189 00:12:43,280 --> 00:12:44,960 The first thing that I want to imagine, 190 00:12:44,960 --> 00:12:46,876 so we're going to do a thought experiment now, 191 00:12:46,876 --> 00:12:50,090 which is that you are extremely obsessive. 192 00:12:50,090 --> 00:12:53,410 And you're driving your car from time a 193 00:12:53,410 --> 00:12:58,010 to time b, place Q to place R, whatever. 194 00:12:58,010 --> 00:13:03,900 And you check your speedometer every second. 195 00:13:03,900 --> 00:13:09,280 OK, so you've read your speedometer in the i-th second, 196 00:13:09,280 --> 00:13:12,620 and you've read that you're going at this speed. 197 00:13:12,620 --> 00:13:16,790 Now, how far do you go in that second? 198 00:13:16,790 --> 00:13:19,530 Well, the answer is you go this speed 199 00:13:19,530 --> 00:13:22,290 times the time interval, which in this case 200 00:13:22,290 --> 00:13:24,950 we're imagining as 1 second. 201 00:13:24,950 --> 00:13:25,940 All right? 202 00:13:25,940 --> 00:13:27,980 So this is how far you went. 203 00:13:27,980 --> 00:13:29,230 But this is the time interval. 204 00:13:29,230 --> 00:13:37,070 And this is the distance traveled 205 00:13:37,070 --> 00:13:46,210 in that-- second number i, in the i-th second. 206 00:13:46,210 --> 00:13:48,210 The distance traveled in the i-th second, that's 207 00:13:48,210 --> 00:13:49,640 a total distance you traveled. 208 00:13:49,640 --> 00:13:53,240 Now, what happens if you go the whole distance? 209 00:13:53,240 --> 00:13:56,860 Well, you travel the sum of all these distances. 210 00:13:56,860 --> 00:14:00,470 So it's some massive sum, where n is some ridiculous number 211 00:14:00,470 --> 00:14:01,730 of seconds. 212 00:14:01,730 --> 00:14:04,250 3600 seconds or something like that. 213 00:14:04,250 --> 00:14:05,070 Whatever it is. 214 00:14:05,070 --> 00:14:09,170 And that's going to turn out to be very similar to what you 215 00:14:09,170 --> 00:14:11,770 would read on your odometer. 216 00:14:11,770 --> 00:14:14,140 Because during that second, you didn't change velocity 217 00:14:14,140 --> 00:14:14,990 very much. 218 00:14:14,990 --> 00:14:17,470 So the approximation that the speed at one 219 00:14:17,470 --> 00:14:21,360 time that you spotted it is very similar to the speed 220 00:14:21,360 --> 00:14:22,790 during the whole second. 221 00:14:22,790 --> 00:14:24,430 It doesn't change that much. 222 00:14:24,430 --> 00:14:26,250 So this is a pretty good approximation 223 00:14:26,250 --> 00:14:29,160 to how far you traveled. 224 00:14:29,160 --> 00:14:33,280 And so the sum is a very realistic approximation 225 00:14:33,280 --> 00:14:34,810 to the entire integral. 226 00:14:34,810 --> 00:14:37,664 Which is denoted this way. 227 00:14:37,664 --> 00:14:39,080 Which, by the fundamental theorem, 228 00:14:39,080 --> 00:14:43,370 is exactly how far you traveled. 229 00:14:43,370 --> 00:14:49,760 So this is x(b) - x(a) Exactly. 230 00:14:49,760 --> 00:14:55,560 The other one is approximate. 231 00:14:55,560 --> 00:15:08,950 OK, again this is called a Riemann sum. 232 00:15:08,950 --> 00:15:17,470 All right, so that's the intro to the fundamental theorem. 233 00:15:17,470 --> 00:15:23,900 And now what I need to do is extend it just a bit. 234 00:15:23,900 --> 00:15:29,170 And the way I'm going to extend it is the following. 235 00:15:29,170 --> 00:15:31,140 I'm going to do it on this example first. 236 00:15:31,140 --> 00:15:35,530 And then we'll do it more formally. 237 00:15:35,530 --> 00:15:39,200 So here's this example where we went someplace. 238 00:15:39,200 --> 00:15:44,370 But now I just want to draw you an additional picture here. 239 00:15:44,370 --> 00:15:49,360 Imagine I start here and I go over to there 240 00:15:49,360 --> 00:15:54,650 and then I come back. 241 00:15:54,650 --> 00:15:56,100 And maybe even I do a round trip. 242 00:15:56,100 --> 00:15:58,090 I come back to the same place. 243 00:15:58,090 --> 00:16:01,070 Well, if I come back to the same place, 244 00:16:01,070 --> 00:16:06,120 then the position is unchanged from the beginning to the end. 245 00:16:06,120 --> 00:16:08,140 In other words, the difference is 0. 246 00:16:08,140 --> 00:16:12,632 And the velocity, technically rather than the speed. 247 00:16:12,632 --> 00:16:14,840 It's the speed to the right and the speed to the left 248 00:16:14,840 --> 00:16:16,640 maybe are the same, but one of them 249 00:16:16,640 --> 00:16:18,330 is going in the positive direction and one of them 250 00:16:18,330 --> 00:16:19,788 is going in the negative direction, 251 00:16:19,788 --> 00:16:22,090 and they cancel each other. 252 00:16:22,090 --> 00:16:25,340 So if you have this kind of situation, 253 00:16:25,340 --> 00:16:26,940 we want that to be reflected. 254 00:16:26,940 --> 00:16:28,550 We like that interpretation and we 255 00:16:28,550 --> 00:16:32,260 want to preserve it even when-- in the case when 256 00:16:32,260 --> 00:16:35,090 the function v is negative. 257 00:16:35,090 --> 00:16:47,280 And so I'm going to now extend our notion of integration. 258 00:16:47,280 --> 00:17:02,320 So we'll extend integration to the case f negative. 259 00:17:02,320 --> 00:17:04,420 Or positive. 260 00:17:04,420 --> 00:17:08,750 In other words, it could be any sign. 261 00:17:08,750 --> 00:17:10,550 Actually, there's no change. 262 00:17:10,550 --> 00:17:12,430 The formulas are all the same. 263 00:17:12,430 --> 00:17:14,891 We just-- If this v is going to be positive, 264 00:17:14,891 --> 00:17:16,140 we write in a positive number. 265 00:17:16,140 --> 00:17:18,640 If it's going to be negative, we write in a negative number. 266 00:17:18,640 --> 00:17:20,270 And we just leave it alone. 267 00:17:20,270 --> 00:17:25,480 And the real-- So here's-- Let me carry out an example 268 00:17:25,480 --> 00:17:29,230 and show you how it works. 269 00:17:29,230 --> 00:17:32,750 I'll carry out the example on this blackboard up here. 270 00:17:32,750 --> 00:17:33,670 Of the sine function. 271 00:17:33,670 --> 00:17:36,020 But we're going to try two humps. 272 00:17:36,020 --> 00:17:38,690 We're going to try the first hump and the one that 273 00:17:38,690 --> 00:17:39,880 goes underneath. 274 00:17:39,880 --> 00:17:40,380 There. 275 00:17:40,380 --> 00:17:43,830 So our example here is going to be the integral 276 00:17:43,830 --> 00:17:50,910 from 0 to 2pi of sin x dx. 277 00:17:50,910 --> 00:17:55,710 And now, because the fundamental theorem is so important, and so 278 00:17:55,710 --> 00:17:58,410 useful, and so convenient, we just 279 00:17:58,410 --> 00:18:01,100 assume that it be true in this case as well. 280 00:18:01,100 --> 00:18:07,170 So we insist that this is going to be -cos x, evaluated at 0 281 00:18:07,170 --> 00:18:10,540 and 2pi, with the difference. 282 00:18:10,540 --> 00:18:12,770 Now, when we carry out that difference, 283 00:18:12,770 --> 00:18:19,320 what we get here is -cos 2pi - (-cos 0). 284 00:18:24,670 --> 00:18:33,670 Which is -1 - (-1), which is 0. 285 00:18:33,670 --> 00:18:39,650 And the interpretation of this is the following. 286 00:18:39,650 --> 00:18:43,840 Here's our double hump, here's pi and here's 2pi. 287 00:18:43,840 --> 00:18:47,590 And all that's happening is that the geometric interpretation 288 00:18:47,590 --> 00:18:50,440 that we had before of the area under the curve 289 00:18:50,440 --> 00:18:53,320 has to be taken with a grain of salt. In other words, 290 00:18:53,320 --> 00:18:56,530 I lied to you before when I said that the definite integral was 291 00:18:56,530 --> 00:18:57,610 the area under the curve. 292 00:18:57,610 --> 00:18:59,030 It's not. 293 00:18:59,030 --> 00:19:00,790 The definite integral is the area 294 00:19:00,790 --> 00:19:03,470 under the curve when it's above the curve, 295 00:19:03,470 --> 00:19:08,500 and it counts negatively when it's below the curve. 296 00:19:08,500 --> 00:19:12,690 So yesterday, my geometric interpretation was incomplete. 297 00:19:12,690 --> 00:19:19,180 And really just a plain lie. 298 00:19:19,180 --> 00:19:28,860 So the true geometric interpretation 299 00:19:28,860 --> 00:19:43,760 of the definite integral is plus the area 300 00:19:43,760 --> 00:19:49,890 above the axis, above the x-axis, 301 00:19:49,890 --> 00:20:00,360 minus the area below the x-axis. 302 00:20:00,360 --> 00:20:02,580 As in the picture. 303 00:20:02,580 --> 00:20:04,230 I'm just writing it down in words, 304 00:20:04,230 --> 00:20:08,870 but you should think of it visually also. 305 00:20:08,870 --> 00:20:12,820 So that's the setup here. 306 00:20:12,820 --> 00:20:17,530 And now we have the complete definition of integrals. 307 00:20:17,530 --> 00:20:19,950 And I need to list for you a bunch of their properties 308 00:20:19,950 --> 00:20:21,740 and how we deal with integrals. 309 00:20:21,740 --> 00:20:25,270 So are there any questions before we go on? 310 00:20:25,270 --> 00:20:25,770 Yeah. 311 00:20:25,770 --> 00:20:32,240 STUDENT: [INAUDIBLE] 312 00:20:32,240 --> 00:20:39,000 PROFESSOR: Right. 313 00:20:39,000 --> 00:20:42,733 So the question was, wouldn't the absolute value 314 00:20:42,733 --> 00:20:45,720 of the velocity function be involved? 315 00:20:45,720 --> 00:20:48,300 The answer is yes. 316 00:20:48,300 --> 00:20:51,730 That is, that's one question that you could ask. 317 00:20:51,730 --> 00:20:54,430 One question you could ask is what's 318 00:20:54,430 --> 00:20:57,360 the total distance traveled. 319 00:20:57,360 --> 00:21:00,600 And in that case, you would keep track 320 00:21:00,600 --> 00:21:05,534 of the absolute value of the velocity as you said, 321 00:21:05,534 --> 00:21:06,950 whether it's positive or negative. 322 00:21:06,950 --> 00:21:13,970 And then you would get the total length of this curve here. 323 00:21:13,970 --> 00:21:18,520 That's, however, not what the definite integral measures. 324 00:21:18,520 --> 00:21:21,640 It measures the net distance traveled. 325 00:21:21,640 --> 00:21:24,049 So it's another thing. 326 00:21:24,049 --> 00:21:25,340 In other words, we can do that. 327 00:21:25,340 --> 00:21:27,890 We now have the tools to do both. 328 00:21:27,890 --> 00:21:35,180 We could also-- So if you like, the total distance 329 00:21:35,180 --> 00:21:39,550 is equal to the integral of this. 330 00:21:39,550 --> 00:21:40,730 From a to b. 331 00:21:40,730 --> 00:21:49,200 But the net distance is the one without the absolute value 332 00:21:49,200 --> 00:21:54,620 signs. 333 00:21:54,620 --> 00:21:57,350 So that's correct. 334 00:21:57,350 --> 00:22:03,950 Other questions? 335 00:22:03,950 --> 00:22:04,590 All right. 336 00:22:04,590 --> 00:22:23,950 So now, let's talk about properties of integrals. 337 00:22:23,950 --> 00:22:37,630 So the properties of integrals that I want to mention to you 338 00:22:37,630 --> 00:22:39,030 are these. 339 00:22:39,030 --> 00:22:47,280 The first one doesn't bear too much comment. 340 00:22:47,280 --> 00:22:53,370 If you take the cumulative integral of a sum, 341 00:22:53,370 --> 00:22:58,820 you're just trying to get the sum of the separate integrals 342 00:22:58,820 --> 00:23:01,867 here. 343 00:23:01,867 --> 00:23:03,200 And I won't say much about that. 344 00:23:03,200 --> 00:23:06,660 That's because sums come out, the because the integral 345 00:23:06,660 --> 00:23:07,790 is a sum. 346 00:23:07,790 --> 00:23:15,330 Incidentally, you know this strange symbol here, 347 00:23:15,330 --> 00:23:17,302 there's actually a reason for it historically. 348 00:23:17,302 --> 00:23:18,760 If you go back to old books, you'll 349 00:23:18,760 --> 00:23:21,880 see that it actually looks a little bit more like an S. 350 00:23:21,880 --> 00:23:24,530 This capital sigma is a sum. 351 00:23:24,530 --> 00:23:27,170 S for sum, because everybody in those days 352 00:23:27,170 --> 00:23:28,400 knew Latin and Greek. 353 00:23:28,400 --> 00:23:31,360 And this one is also an S, but gradually it 354 00:23:31,360 --> 00:23:33,390 was such an important S that they made a bigger. 355 00:23:33,390 --> 00:23:35,890 And then they stretched it out and made it a little thinner, 356 00:23:35,890 --> 00:23:40,590 because it didn't fit into one typesetting space. 357 00:23:40,590 --> 00:23:43,550 And so just for typesetting reasons it got stretched. 358 00:23:43,550 --> 00:23:45,390 And got a little bit skinny. 359 00:23:45,390 --> 00:23:48,640 Anyway, so it's really an S. And in fact, in French 360 00:23:48,640 --> 00:23:50,800 they call it sum. 361 00:23:50,800 --> 00:23:53,900 Even though we call it integral. 362 00:23:53,900 --> 00:23:57,690 So it's a sum. 363 00:23:57,690 --> 00:24:00,450 So it's consistent with sums in this way. 364 00:24:00,450 --> 00:24:09,040 And similarly, similarly we can factor constants out of sums. 365 00:24:09,040 --> 00:24:20,820 So if you have an integral like this, the constant factors out. 366 00:24:20,820 --> 00:24:25,140 But definitely don't try to get a function out of this. 367 00:24:25,140 --> 00:24:27,410 That won't happen. 368 00:24:27,410 --> 00:24:30,470 OK, in other words, c has to be a constant. 369 00:24:30,470 --> 00:24:41,320 Doesn't depend on x. 370 00:24:41,320 --> 00:24:44,340 The third property. 371 00:24:44,340 --> 00:24:46,710 What do I want to call the third property here? 372 00:24:46,710 --> 00:24:51,522 I have sort of a preliminary property, yes, here. 373 00:24:51,522 --> 00:24:52,480 Which is the following. 374 00:24:52,480 --> 00:24:53,730 And I'll draw a picture of it. 375 00:24:53,730 --> 00:24:59,410 I suppose you have three points along a line. 376 00:24:59,410 --> 00:25:01,300 So then I'm going to draw a picture of that. 377 00:25:01,300 --> 00:25:03,940 And I'm going to use the interpretation above the curve, 378 00:25:03,940 --> 00:25:05,564 even though that's not the whole thing. 379 00:25:05,564 --> 00:25:07,990 So here's a, here's b and here's c. 380 00:25:07,990 --> 00:25:10,490 And you can see that the area of this piece, 381 00:25:10,490 --> 00:25:14,080 of the first two pieces here, when added together, 382 00:25:14,080 --> 00:25:15,820 gives you the area of the whole. 383 00:25:15,820 --> 00:25:19,400 And that's the rule that I'd like to tell you. 384 00:25:19,400 --> 00:25:23,540 So if you integrate from a to b, and you 385 00:25:23,540 --> 00:25:28,070 add to that the integral from b to c, 386 00:25:28,070 --> 00:25:37,344 you'll get the integral from a to c. 387 00:25:37,344 --> 00:25:39,260 This is going to be just a little preliminary, 388 00:25:39,260 --> 00:25:41,830 because the rule is a little better than this. 389 00:25:41,830 --> 00:25:47,970 But I will explain that in a minute. 390 00:25:47,970 --> 00:25:52,540 The fourth rule is a very simple one. 391 00:25:52,540 --> 00:25:57,910 Which is that the integral from a to a of f(x) dx 392 00:25:57,910 --> 00:26:00,624 is equal to 0. 393 00:26:00,624 --> 00:26:02,790 Now, that you can see very obviously because there's 394 00:26:02,790 --> 00:26:04,140 no area. 395 00:26:04,140 --> 00:26:05,900 No horizontal movement there. 396 00:26:05,900 --> 00:26:08,430 The rectangle is infinitely thin, 397 00:26:08,430 --> 00:26:10,350 and there's nothing there. 398 00:26:10,350 --> 00:26:12,020 So this is the case. 399 00:26:12,020 --> 00:26:17,860 You can also interpret it a F(a) - F(a). 400 00:26:17,860 --> 00:26:21,860 So that's also consistent with our interpretation. 401 00:26:21,860 --> 00:26:24,510 In terms of the fundamental theorem of calculus. 402 00:26:24,510 --> 00:26:28,020 And it's perfectly reasonable that this is the case. 403 00:26:28,020 --> 00:26:33,090 Now, the fifth property is a definition. 404 00:26:33,090 --> 00:26:35,110 It's not really a property. 405 00:26:35,110 --> 00:26:38,080 But it's very important. 406 00:26:38,080 --> 00:26:46,040 The integral from a to b of f(x) dx equal to minus the integral 407 00:26:46,040 --> 00:26:50,960 from b to a, of f( x) dx. 408 00:26:50,960 --> 00:26:57,490 Now, really, the right-hand side here is an undefined quantity 409 00:26:57,490 --> 00:26:58,740 so far. 410 00:26:58,740 --> 00:27:02,380 We never said you could ever do this 411 00:27:02,380 --> 00:27:05,650 where the a is less than the b. 412 00:27:05,650 --> 00:27:09,600 Because this is working backwards here. 413 00:27:09,600 --> 00:27:12,232 But we just have a convention that that's the definition. 414 00:27:12,232 --> 00:27:13,690 Whenever we write down this number, 415 00:27:13,690 --> 00:27:17,120 it's the same as minus what that number is. 416 00:27:17,120 --> 00:27:20,229 And the reason for all of these is again 417 00:27:20,229 --> 00:27:22,520 that we want them to be consistent with the fundamental 418 00:27:22,520 --> 00:27:23,730 theorem of calculus. 419 00:27:23,730 --> 00:27:26,250 Which is the thing that makes all of this work. 420 00:27:26,250 --> 00:27:33,210 So if you notice the left-hand side here is F(b) - F(a), 421 00:27:33,210 --> 00:27:36,410 capital F, the antiderivative of little f. 422 00:27:36,410 --> 00:27:39,420 On the other hand, the other side is minus, 423 00:27:39,420 --> 00:27:42,050 and if we just ignore that, we say these are letters, 424 00:27:42,050 --> 00:27:44,633 if we were a machine, we didn't know which one was bigger than 425 00:27:44,633 --> 00:27:49,980 which, we just plugged them in, we would get here F(a) - F(b), 426 00:27:49,980 --> 00:27:50,690 over here. 427 00:27:50,690 --> 00:27:53,402 And to make these two things equal, what we want 428 00:27:53,402 --> 00:27:54,610 is to put that minus sign in. 429 00:27:54,610 --> 00:27:59,620 Now it's consistent. 430 00:27:59,620 --> 00:28:02,610 So again, these rules are set up so 431 00:28:02,610 --> 00:28:05,420 that everything is consistent. 432 00:28:05,420 --> 00:28:11,190 And now I want to improve on rule 3 here. 433 00:28:11,190 --> 00:28:15,010 And point out to you - so let me just go back to rule 3 434 00:28:15,010 --> 00:28:21,030 for a second - that now that we can evaluate integrals 435 00:28:21,030 --> 00:28:24,920 regardless of the order, we don't have to have a < b, 436 00:28:24,920 --> 00:28:28,070 b < c in order to make sense out of this. 437 00:28:28,070 --> 00:28:31,740 We actually have the possibility of considering integrals 438 00:28:31,740 --> 00:28:34,180 where the a's and the b's and the c's are 439 00:28:34,180 --> 00:28:36,270 in any order you want. 440 00:28:36,270 --> 00:28:38,630 And in fact, with this definition, 441 00:28:38,630 --> 00:28:43,140 with this definition 5, 3 works no matter what the numbers are. 442 00:28:43,140 --> 00:28:44,710 So this is much more convenient. 443 00:28:44,710 --> 00:28:49,570 We don't, this is not necessary. 444 00:28:49,570 --> 00:28:51,070 Not necessary. 445 00:28:51,070 --> 00:28:57,220 It just works using convention 5. 446 00:28:57,220 --> 00:29:04,140 OK, with 5. 447 00:29:04,140 --> 00:29:08,760 Again, before I go on, let me emphasize: 448 00:29:08,760 --> 00:29:11,510 we really want to respect the sign of this velocity. 449 00:29:11,510 --> 00:29:15,110 We really want the net change in the position. 450 00:29:15,110 --> 00:29:18,180 And we don't want this absolute value here. 451 00:29:18,180 --> 00:29:20,740 Because otherwise, all of our formulas are going to mess up. 452 00:29:20,740 --> 00:29:22,260 We won't always be able to check. 453 00:29:22,260 --> 00:29:24,955 Sometimes you have letters rather than 454 00:29:24,955 --> 00:29:26,580 actual numbers here, and you won't know 455 00:29:26,580 --> 00:29:28,160 whether a is bigger than b. 456 00:29:28,160 --> 00:29:31,370 So you'll want to know that these formulas work and are 457 00:29:31,370 --> 00:29:36,410 consistent in all situations. 458 00:29:36,410 --> 00:29:39,550 OK, I'm going to trade these again. 459 00:29:39,550 --> 00:29:47,250 In order to preserve the ordering 1 through 5. 460 00:29:47,250 --> 00:29:54,200 And now I have a sixth property that I want to talk about. 461 00:29:54,200 --> 00:30:02,010 This one is called estimation. 462 00:30:02,010 --> 00:30:05,470 And it says the following. 463 00:30:05,470 --> 00:30:18,270 If f(x) <= g(x), then the integral from a to b of f(x) dx 464 00:30:18,270 --> 00:30:23,180 is less than or equal to the integral from a to b of g(x) 465 00:30:23,180 --> 00:30:28,300 dx. 466 00:30:28,300 --> 00:30:36,860 Now, this one says that if I'm going more slowly than you, 467 00:30:36,860 --> 00:30:40,840 then you go farther than I do. 468 00:30:40,840 --> 00:30:41,760 OK. 469 00:30:41,760 --> 00:30:43,710 That's all it's saying. 470 00:30:43,710 --> 00:30:47,870 For this one, you'd better have a < b. 471 00:30:47,870 --> 00:30:49,100 You need it. 472 00:30:49,100 --> 00:30:55,360 Because we flip the signs when we flip the order of a and b. 473 00:30:55,360 --> 00:30:59,410 So this one, it's essential that the lower limit be smaller 474 00:30:59,410 --> 00:31:04,600 than the upper limit. 475 00:31:04,600 --> 00:31:06,930 But let me just emphasize, because we're dealing 476 00:31:06,930 --> 00:31:08,370 with the generalities of this. 477 00:31:08,370 --> 00:31:10,350 Actually if one of these is negative 478 00:31:10,350 --> 00:31:14,650 and the other one is negative, then it also works. 479 00:31:14,650 --> 00:31:17,510 This one ends up being, if f is more negative than g, 480 00:31:17,510 --> 00:31:22,380 then this added up thing is more negative than that one. 481 00:31:22,380 --> 00:31:25,270 Again, under the assumption that a is less than b. 482 00:31:25,270 --> 00:31:34,760 So as I wrote it it's in full generality. 483 00:31:34,760 --> 00:31:37,490 Let's illustrate this one. 484 00:31:37,490 --> 00:31:47,030 And then we have one more property to learn after that. 485 00:31:47,030 --> 00:31:58,970 So let me give you an example of estimation. 486 00:31:58,970 --> 00:32:01,970 The example is the same as one that I already gave you. 487 00:32:01,970 --> 00:32:05,080 But this time, because we have the tool of integration, 488 00:32:05,080 --> 00:32:11,290 we can just follow our noses and it works. 489 00:32:11,290 --> 00:32:14,400 I start with the inequality, so I'm 490 00:32:14,400 --> 00:32:16,090 trying to illustrate estimation, so I 491 00:32:16,090 --> 00:32:17,673 want to start with an inequality which 492 00:32:17,673 --> 00:32:19,290 is what the hypothesis is here. 493 00:32:19,290 --> 00:32:21,480 And I'm going to integrate the inequality 494 00:32:21,480 --> 00:32:22,900 to get this conclusion. 495 00:32:22,900 --> 00:32:25,860 And see what conclusion it is. 496 00:32:25,860 --> 00:32:30,160 The inequality that I want to take is that e^x >= 1, 497 00:32:30,160 --> 00:32:32,800 for x >= 0. 498 00:32:32,800 --> 00:32:37,620 That's going to be our starting place. 499 00:32:37,620 --> 00:32:39,070 And now I'm going to integrate it. 500 00:32:39,070 --> 00:32:40,850 That is, I'm going to use estimation 501 00:32:40,850 --> 00:32:42,420 to see what that gives. 502 00:32:42,420 --> 00:32:45,100 Well, I'm going to integrate, say, from 0 to b. 503 00:32:45,100 --> 00:32:50,010 I can't integrate below 0 because it's only true above 0. 504 00:32:50,010 --> 00:32:54,390 This is e^x dx greater than or equal to the integral from 0 505 00:32:54,390 --> 00:33:01,990 to b of 1 dx. 506 00:33:01,990 --> 00:33:05,980 Alright, let's work out what each of these is. 507 00:33:05,980 --> 00:33:14,370 The first one, e^x dx, is, the antiderivative is e^x, 508 00:33:14,370 --> 00:33:16,120 evaluated at 0 and b. 509 00:33:16,120 --> 00:33:18,930 So that's e^b - e^0. 510 00:33:18,930 --> 00:33:23,510 Which is e^b - 1. 511 00:33:23,510 --> 00:33:28,560 The other one, you're supposed to be 512 00:33:28,560 --> 00:33:32,270 able to get by the rectangle law. 513 00:33:32,270 --> 00:33:35,840 This is one rectangle of base b and height 1. 514 00:33:35,840 --> 00:33:37,350 So the answer is b. 515 00:33:37,350 --> 00:33:44,100 Or you can do it by antiderivatives, but it's b. 516 00:33:44,100 --> 00:33:49,010 That means that our inequality says if I just combine these 517 00:33:49,010 --> 00:33:55,640 two things together, that e^b - 1 >= b. 518 00:33:55,640 --> 00:34:02,040 And that's the same thing as e^b >= 1 + b. 519 00:34:02,040 --> 00:34:05,010 Again, this only works for b >= 0. 520 00:34:05,010 --> 00:34:10,520 Notice that if b were negative, this would be a well 521 00:34:10,520 --> 00:34:13,520 defined quantity. 522 00:34:13,520 --> 00:34:18,240 But this estimation would be false. 523 00:34:18,240 --> 00:34:22,010 We need that the b > 0 in order for this to make sense. 524 00:34:22,010 --> 00:34:24,590 So this was used. 525 00:34:24,590 --> 00:34:28,640 And that's a good thing, because this inequality is suspect. 526 00:34:28,640 --> 00:34:32,060 Actually, it turns out to be true when b is negative. 527 00:34:32,060 --> 00:34:38,830 But we certainly didn't prove it. 528 00:34:38,830 --> 00:34:42,260 I'm going to just repeat this process. 529 00:34:42,260 --> 00:34:46,400 So let's repeat it. 530 00:34:46,400 --> 00:34:49,890 Starting from the inequality, the conclusion, 531 00:34:49,890 --> 00:34:51,390 which is sitting right here. 532 00:34:51,390 --> 00:34:59,970 But I'll write it in a form e^x >= 1 + x, for x >= 0. 533 00:34:59,970 --> 00:35:02,160 And now, if I integrate this one, 534 00:35:02,160 --> 00:35:06,245 I get the integral from 0 to b, e^x dx is greater than or equal 535 00:35:06,245 --> 00:35:12,360 to the integral from 0 to b, (1 + x) dx, 536 00:35:12,360 --> 00:35:16,020 and I remind you that we've already calculated this one. 537 00:35:16,020 --> 00:35:19,000 This is e^b - 1. 538 00:35:19,000 --> 00:35:21,710 And the other one is not hard to calculate. 539 00:35:21,710 --> 00:35:25,930 The antiderivative is x + x^2 / 2. 540 00:35:25,930 --> 00:35:28,690 We're evaluating that at 0 and b. 541 00:35:28,690 --> 00:35:34,830 So that comes out to be b + b^2 / 2. 542 00:35:34,830 --> 00:35:43,550 And so our conclusion is that the left side, which is e^b - 543 00:35:43,550 --> 00:35:48,300 1 >= b + b^2 / 2. 544 00:35:48,300 --> 00:35:51,930 And this is for b >= 0. 545 00:35:51,930 --> 00:36:00,510 And that's the same thing as e^b >= 1 + b + b^2 / 2. 546 00:36:00,510 --> 00:36:04,050 This one actually is false for b negative, 547 00:36:04,050 --> 00:36:09,280 so that's something that you have 548 00:36:09,280 --> 00:36:15,560 to be careful with the b positive's here. 549 00:36:15,560 --> 00:36:17,350 So you can keep on going with this, 550 00:36:17,350 --> 00:36:20,360 and you didn't have to think. 551 00:36:20,360 --> 00:36:23,330 And you'll produce a very interesting polynomial, 552 00:36:23,330 --> 00:36:25,340 which is a good approximation to e^b. 553 00:36:30,720 --> 00:36:34,520 So that's it for the basic properties. 554 00:36:34,520 --> 00:36:38,980 Now there's one tricky property that I need to tell you about. 555 00:36:38,980 --> 00:36:47,920 It's not that tricky, but it's a little tricky. 556 00:36:47,920 --> 00:37:07,390 And this is change of variables. 557 00:37:07,390 --> 00:37:09,600 Change of variables in integration, 558 00:37:09,600 --> 00:37:11,220 we've actually already done. 559 00:37:11,220 --> 00:37:14,190 We called that, the last time we talked about it, 560 00:37:14,190 --> 00:37:23,430 we called it substitution. 561 00:37:23,430 --> 00:37:26,210 And the idea here, if you may remember, 562 00:37:26,210 --> 00:37:31,550 was that if you're faced with an integral like this, 563 00:37:31,550 --> 00:37:38,340 you can change it to, if you put in u = u(x) and you have a du, 564 00:37:38,340 --> 00:37:42,700 which is equal to u'(x) du-- dx, sorry. 565 00:37:42,700 --> 00:37:45,900 Then you can change the integral as follows. 566 00:37:45,900 --> 00:37:51,630 This is the same as g(u(x)) u'(x) dx. 567 00:37:51,630 --> 00:37:58,920 This was the general procedure for substitution. 568 00:37:58,920 --> 00:38:05,500 What's new today is that we're going to put in the limits. 569 00:38:05,500 --> 00:38:10,350 If you have a limit here, u_1, and a limit here, u_2, 570 00:38:10,350 --> 00:38:12,790 you want to know what the relationship is 571 00:38:12,790 --> 00:38:15,690 between the limits here and the limits when you change 572 00:38:15,690 --> 00:38:18,780 variables to the new variables. 573 00:38:18,780 --> 00:38:21,290 And it's the simplest possible thing. 574 00:38:21,290 --> 00:38:25,760 Namely the two limits over here are in the same relationship 575 00:38:25,760 --> 00:38:29,130 as u(x) is to this symbol u here. 576 00:38:29,130 --> 00:38:35,830 In other words, u_1 = u(x_1), and u_2 = u(x_2). 577 00:38:35,830 --> 00:38:39,310 That's what works. 578 00:38:39,310 --> 00:38:42,620 Now there's only one danger here, 579 00:38:42,620 --> 00:38:48,230 there's one subtlety which is, this only works 580 00:38:48,230 --> 00:39:02,010 if u' does not change sign. 581 00:39:02,010 --> 00:39:04,680 I've been worrying a little bit about going backwards 582 00:39:04,680 --> 00:39:06,400 and forwards, and I allowed myself 583 00:39:06,400 --> 00:39:08,525 to reverse and do all kinds of stuff, right, 584 00:39:08,525 --> 00:39:09,400 with these integrals. 585 00:39:09,400 --> 00:39:11,560 So we're sort of free to do it. 586 00:39:11,560 --> 00:39:14,560 Well, this is one case where you want to avoid it, OK? 587 00:39:14,560 --> 00:39:15,850 Just don't do it. 588 00:39:15,850 --> 00:39:17,990 It is possible, actually, to make sense out of it, 589 00:39:17,990 --> 00:39:21,810 but it's also possible to get yourself infinitely confused. 590 00:39:21,810 --> 00:39:24,350 So just make sure that-- Now, it's 591 00:39:24,350 --> 00:39:27,580 OK if u' is always negative, or always going one way, 592 00:39:27,580 --> 00:39:30,034 so OK if u' is always positive, you're always 593 00:39:30,034 --> 00:39:31,700 going the other way, but if you mix them 594 00:39:31,700 --> 00:39:39,960 up you'll get yourself mixed up. 595 00:39:39,960 --> 00:39:46,210 Let me give you an example. 596 00:39:46,210 --> 00:39:54,230 The example will be maybe close to what we did last time. 597 00:39:54,230 --> 00:39:57,900 When we first did substitution, I mean. 598 00:39:57,900 --> 00:40:02,160 So the integral from 1 to 2, this time I'll put in definite 599 00:40:02,160 --> 00:40:09,690 limits, of x^2 plus-- sorry, maybe I call this x^3. x^3 + 2, 600 00:40:09,690 --> 00:40:17,950 let's say, I don't know, to the 5th power, x^2 dx. 601 00:40:17,950 --> 00:40:20,310 So this is an example of an integral 602 00:40:20,310 --> 00:40:25,900 that we would have tried to handle by substitution before. 603 00:40:25,900 --> 00:40:36,590 And the substitution we would have used is u = x^3 + 2. 604 00:40:36,590 --> 00:40:38,560 And that's exactly what we're going to do here. 605 00:40:38,560 --> 00:40:44,710 But we're just going to also take into account the limits. 606 00:40:44,710 --> 00:40:47,730 The first step, as in any substitution or change 607 00:40:47,730 --> 00:40:54,070 of variables, is this. 608 00:40:54,070 --> 00:40:57,056 And so we can fill in the things that we 609 00:40:57,056 --> 00:40:58,180 would have done previously. 610 00:40:58,180 --> 00:41:01,790 Which is that this is the integral and this is u^5. 611 00:41:01,790 --> 00:41:09,300 And then because this is 3x^2, we see that this is 3. 612 00:41:09,300 --> 00:41:13,130 Sorry, let's write it the other way. 613 00:41:13,130 --> 00:41:17,480 1/3 du = x^2 dx. 614 00:41:17,480 --> 00:41:20,370 So that's what I'm going to plug in for this factor here. 615 00:41:20,370 --> 00:41:26,480 So here's 1/3 du, which replaces that. 616 00:41:26,480 --> 00:41:29,430 But now there's the extra feature. 617 00:41:29,430 --> 00:41:31,660 The extra feature is the limits. 618 00:41:31,660 --> 00:41:35,160 So here, really in disguise, because, and now 619 00:41:35,160 --> 00:41:38,280 this is incredibly important. 620 00:41:38,280 --> 00:41:44,070 This is one of the reasons why we use this notation dx and du. 621 00:41:44,070 --> 00:41:47,120 We want to remind ourselves which variable 622 00:41:47,120 --> 00:41:49,170 is involved in the integration. 623 00:41:49,170 --> 00:41:52,470 And especially if you're the one naming the variables, 624 00:41:52,470 --> 00:41:54,760 you may get mixed up in this respect. 625 00:41:54,760 --> 00:41:59,200 So you must know which variable is varying between 1 and 2. 626 00:41:59,200 --> 00:42:01,510 And the answer is, it's x is the one that's 627 00:42:01,510 --> 00:42:04,150 varying between 1 and 2. 628 00:42:04,150 --> 00:42:06,870 So in disguise, even though I didn't write it, 629 00:42:06,870 --> 00:42:10,260 it was contained in this little symbol here. 630 00:42:10,260 --> 00:42:11,840 This reminded us which variable. 631 00:42:11,840 --> 00:42:14,140 You'll find this amazingly important when you 632 00:42:14,140 --> 00:42:16,480 get to multivariable calculus. 633 00:42:16,480 --> 00:42:18,810 When there are many variables floating around. 634 00:42:18,810 --> 00:42:21,390 So this is an incredibly important distinction to make. 635 00:42:21,390 --> 00:42:23,300 So now, over here we have a limit. 636 00:42:23,300 --> 00:42:26,270 But of course it's supposed to be with respect to u, now. 637 00:42:26,270 --> 00:42:29,490 So we need to calculate what those corresponding limits are. 638 00:42:29,490 --> 00:42:33,270 And indeed it's just, I plug in here u_1 is going to be equal 639 00:42:33,270 --> 00:42:37,430 to what I plug in for x = 1, that's going to be 1^3 + 2, 640 00:42:37,430 --> 00:42:38,760 which is 3. 641 00:42:38,760 --> 00:42:46,390 And then u_2 is 2^3 + 2, which is equal to 10, right? 642 00:42:46,390 --> 00:42:47,960 8 + 2 = 10. 643 00:42:47,960 --> 00:42:57,700 So this is the integral from 3 to 10, of u^5 1/3 du. 644 00:42:57,700 --> 00:43:00,380 And now I can finish the problem. 645 00:43:00,380 --> 00:43:06,390 This is 1/18 u^6, from 3 to 10. 646 00:43:06,390 --> 00:43:10,000 And this is where the most common mistake occurs 647 00:43:10,000 --> 00:43:12,360 in substitutions of this type. 648 00:43:12,360 --> 00:43:14,850 Which is that if you ignore this, 649 00:43:14,850 --> 00:43:17,020 and you plug in these 1 and 2 here, 650 00:43:17,020 --> 00:43:20,090 you think, oh I should just be putting it at 1 and 2. 651 00:43:20,090 --> 00:43:22,940 But actually, it should be, the u-value 652 00:43:22,940 --> 00:43:26,400 that we're interested in, and the lower limit is u = 3 653 00:43:26,400 --> 00:43:29,660 and u = 10 is the upper limit. 654 00:43:29,660 --> 00:43:31,405 So those are suppressed here. 655 00:43:31,405 --> 00:43:35,620 But those are the ones that we want. 656 00:43:35,620 --> 00:43:37,360 And so, here we go. 657 00:43:37,360 --> 00:43:41,770 It's 1/18 times some ridiculous number which I won't calculate. 658 00:43:41,770 --> 00:43:44,260 10^6 - - 3^6. 659 00:43:47,820 --> 00:43:48,630 Yes, question. 660 00:43:48,630 --> 00:44:07,250 STUDENT: [INAUDIBLE] 661 00:44:07,250 --> 00:44:10,380 PROFESSOR: So, if you want to do things 662 00:44:10,380 --> 00:44:16,290 with where you're worrying about the sign change, 663 00:44:16,290 --> 00:44:20,820 the right strategy is, what you suggested works. 664 00:44:20,820 --> 00:44:23,090 And in fact I'm going to do an example right now 665 00:44:23,090 --> 00:44:24,220 on this subject. 666 00:44:24,220 --> 00:44:30,160 But, the right strategy is to break it up into pieces. 667 00:44:30,160 --> 00:44:34,950 Where u' has one sign or the other, OK? 668 00:44:34,950 --> 00:44:37,430 Let me show you an example. 669 00:44:37,430 --> 00:44:40,700 Where things go wrong. 670 00:44:40,700 --> 00:44:47,900 And I'll tell you how to handle it, roughly. 671 00:44:47,900 --> 00:44:55,790 So here's our warning. 672 00:44:55,790 --> 00:45:00,280 Suppose you're integrating from -1 to 1, x^2 dx. 673 00:45:00,280 --> 00:45:02,810 Here's an example. 674 00:45:02,810 --> 00:45:09,654 And you have the temptation to plug in u = x^2. 675 00:45:09,654 --> 00:45:11,570 Now, of course, we know how to integrate this. 676 00:45:11,570 --> 00:45:16,390 But let's just pretend we were stubborn and wanted 677 00:45:16,390 --> 00:45:19,260 to use substitution. 678 00:45:19,260 --> 00:45:26,610 Then we have du = 2x dx. 679 00:45:26,610 --> 00:45:30,170 And now if I try to make the correspondence, 680 00:45:30,170 --> 00:45:37,130 notice that the limits are u_1 = (-1)^2, 681 00:45:37,130 --> 00:45:38,860 that's the bottom limit. 682 00:45:38,860 --> 00:45:40,830 And u_2 is the upper limit. 683 00:45:40,830 --> 00:45:43,450 That's 1^2, that's also equal to 1. 684 00:45:43,450 --> 00:45:45,250 Both limits are 1. 685 00:45:45,250 --> 00:45:47,750 So this is going from 1 to 1. 686 00:45:47,750 --> 00:45:53,350 And no matter what it is, we know it's going to be 0. 687 00:45:53,350 --> 00:45:55,610 But we know this is not 0. 688 00:45:55,610 --> 00:45:58,620 This is the integral of a positive quantity. 689 00:45:58,620 --> 00:46:03,139 And the area under a curve is going to be a positive area. 690 00:46:03,139 --> 00:46:04,430 So this is a positive quantity. 691 00:46:04,430 --> 00:46:07,300 It can't be 0. 692 00:46:07,300 --> 00:46:12,090 If you actually plug it in, it looks equally strange. 693 00:46:12,090 --> 00:46:16,160 You put in here this u and then, so that would be for the u^2. 694 00:46:16,160 --> 00:46:22,110 And then to plug in for dx, you would write dx = 1/(2x) du. 695 00:46:22,110 --> 00:46:27,760 And then you might write that as this. 696 00:46:27,760 --> 00:46:31,500 And so what I should put in here is this quantity here. 697 00:46:31,500 --> 00:46:33,360 Which is a perfectly OK integral. 698 00:46:33,360 --> 00:46:37,870 And it has a value, I mean, it's what it is. 699 00:46:37,870 --> 00:46:39,140 It's 0. 700 00:46:39,140 --> 00:46:45,270 So of course this is not true. 701 00:46:45,270 --> 00:46:51,540 And the reason is that u was equal to x^2, 702 00:46:51,540 --> 00:46:57,340 and u'(x) was equal to 2x, which was positive for x positive, 703 00:46:57,340 --> 00:47:00,070 and negative for x negative. 704 00:47:00,070 --> 00:47:03,470 And this was the sign change which causes us trouble. 705 00:47:03,470 --> 00:47:08,160 If we break it off into its two halves, then it'll be OK 706 00:47:08,160 --> 00:47:09,710 and you'll be able to use this. 707 00:47:09,710 --> 00:47:12,120 Now, there was a mistake. 708 00:47:12,120 --> 00:47:15,540 And this was essentially what you were saying. 709 00:47:15,540 --> 00:47:19,050 That is, it's possible to see this happening as you're doing 710 00:47:19,050 --> 00:47:21,680 it if you're very careful. 711 00:47:21,680 --> 00:47:23,610 There's a mistake in this process, 712 00:47:23,610 --> 00:47:26,290 and the mistake is in the transition. 713 00:47:26,290 --> 00:47:28,230 This is a mistake here. 714 00:47:28,230 --> 00:47:33,340 Maybe I haven't used any red yet today, 715 00:47:33,340 --> 00:47:34,940 so I get to use some red here. 716 00:47:34,940 --> 00:47:36,750 Oh boy. 717 00:47:36,750 --> 00:47:38,310 This is not true, here. 718 00:47:38,310 --> 00:47:39,200 This step here. 719 00:47:39,200 --> 00:47:40,470 So why isn't it true? 720 00:47:40,470 --> 00:47:43,500 It's not true for the standard reason. 721 00:47:43,500 --> 00:47:50,670 Which is that really, x = plus or minus square root of u. 722 00:47:50,670 --> 00:47:53,930 And if you stick to one side or the other, 723 00:47:53,930 --> 00:47:55,660 you'll have a coherent formula for it. 724 00:47:55,660 --> 00:47:58,243 One of them will be the plus and one of them will be the minus 725 00:47:58,243 --> 00:48:01,980 and it will work out when you separate it into its pieces. 726 00:48:01,980 --> 00:48:02,880 So you could do that. 727 00:48:02,880 --> 00:48:04,100 But this is a can of worms. 728 00:48:04,100 --> 00:48:06,250 So I avoid this. 729 00:48:06,250 --> 00:48:10,112 And just do it in a place where the inverse is well defined. 730 00:48:10,112 --> 00:48:11,570 And where the function either moves 731 00:48:11,570 --> 00:48:13,880 steadily up or steadily down.