1 00:00:00,916 --> 00:00:06,280 This'll be the second half of my lectures on integration. 2 00:00:06,280 --> 00:00:08,981 So last time we just talked a little bit about 3 00:00:08,981 --> 00:00:13,154 what integration was, the problem it was trying to solve. 4 00:00:13,154 --> 00:00:16,265 we talked about the fundamental theorem of calculus, which 5 00:00:16,265 --> 00:00:19,471 is a tool that lets us actually evaluate definite integrals. 6 00:00:21,012 --> 00:00:23,460 one over a little bit of application of 7 00:00:23,460 --> 00:00:26,588 integrals and then I started talking about techniques 8 00:00:26,588 --> 00:00:27,660 of integration. 9 00:00:27,660 --> 00:00:32,082 So I cover something called integration by parts which was just the product all 10 00:00:32,082 --> 00:00:35,030 backwards and then something called integration by 11 00:00:35,030 --> 00:00:37,911 substitution or the substitution method which was 12 00:00:37,911 --> 00:00:44,508 just the general kind of running backwards Today I'm going to do one more technique 13 00:00:44,508 --> 00:00:47,388 of integration called completing the square for 14 00:00:47,388 --> 00:00:49,400 those of you who are taking 461. 15 00:00:49,400 --> 00:00:51,608 You're going to use this when we try to 16 00:00:51,608 --> 00:00:53,816 transform a normal distribution into 17 00:00:53,816 --> 00:00:56,730 something called the log normal distribution. 18 00:00:56,730 --> 00:00:59,355 And then I'm going to talk about 19 00:00:59,355 --> 00:01:05,345 differentiating definite integrals so a function can 20 00:01:05,345 --> 00:01:11,230 have, could be defined with integral and having the variable in the limit. 21 00:01:11,230 --> 00:01:13,980 So in that case you n actually think of that as a, a function you can 22 00:01:13,980 --> 00:01:14,430 [INAUDIBLE] 23 00:01:14,430 --> 00:01:16,190 the derivative off. 24 00:01:16,190 --> 00:01:20,478 I'll talk about improper integrals so these are either integrals of functions 25 00:01:20,478 --> 00:01:23,850 that take infinite values in either the x or, or y direction. 26 00:01:25,030 --> 00:01:28,115 And the, differentiating improper integrals. 27 00:01:28,115 --> 00:01:34,480 Okay, So, I'm going to get started with my completing the square. 28 00:01:34,480 --> 00:01:36,520 And I'll, I'll introduce this just by trying 29 00:01:36,520 --> 00:01:39,370 to remind everybody of something called the Quadratic Formula. 30 00:01:41,510 --> 00:01:43,600 And so, a quadratic function. 31 00:01:43,600 --> 00:01:48,280 Is just a function, it's a polynomial of degree two, so it's a times x 32 00:01:48,280 --> 00:01:52,770 squared plus b times x plus c, or a b and c are real valued constants. 33 00:01:54,850 --> 00:01:58,550 And we have to add the extra condition that a is not equal to zero. 34 00:01:58,550 --> 00:02:00,110 Because if it were equal to zero this 35 00:02:00,110 --> 00:02:02,090 would just be a linear function, bx plus c. 36 00:02:05,990 --> 00:02:10,939 And suppose I'm interested in finding the value or values 37 00:02:10,939 --> 00:02:15,331 of x that make my quadratic function equal to zero. 38 00:02:15,331 --> 00:02:20,227 I think everybody remembers from high school algebra the 39 00:02:20,227 --> 00:02:25,531 quadratic formula, so it's x is equal to minus b plus or minus 40 00:02:25,531 --> 00:02:31,039 the square root of b squared minus 4 ac all over 2 a, and that's 41 00:02:31,039 --> 00:02:33,997 going to give you the, the values of x 42 00:02:33,997 --> 00:02:38,180 that will make this function be equal to zero. 43 00:02:38,180 --> 00:02:43,822 And the way we actually come up with this formula is the same technique 44 00:02:43,822 --> 00:02:48,385 that we will use in integration for completing the square. 45 00:02:48,385 --> 00:02:52,295 So suppose I want to solve my quadratic function, I use 46 00:02:52,295 --> 00:02:56,205 this notation to say I'm setting it equal to zero so its 47 00:02:56,205 --> 00:03:01,560 not defined to be equal to zero, its me I'm choosing a certain level and now 48 00:03:01,560 --> 00:03:06,915 once I've done that these x's aren't vairable anymore, they are sort of 49 00:03:06,915 --> 00:03:12,252 fixed numbers. And I want to find out what they are. 50 00:03:12,252 --> 00:03:16,170 So the first thing I'm going to do is just divide every term by a. 51 00:03:16,170 --> 00:03:18,640 So I multiply both sides of this by one over a, 52 00:03:18,640 --> 00:03:21,630 and I'm allowed to do that because one of my conditions 53 00:03:21,630 --> 00:03:24,690 on my quadratic function was the a is not equal to zero. 54 00:03:24,690 --> 00:03:28,170 So as long as a is not equal to zero, I can divide by a. 55 00:03:28,170 --> 00:03:31,730 And so I get this new form. So it doesn't look too exciting yet. 56 00:03:33,290 --> 00:03:35,080 And then what I'm going to try and do. 57 00:03:35,080 --> 00:03:40,530 So this is sort of the, the meat and potatoes of completing the square. 58 00:03:41,880 --> 00:03:47,090 I want to add zero to this. But I want to choose a particularly 59 00:03:47,090 --> 00:03:54,325 smart choice of zero. So I'm going to, first of all, 60 00:03:54,325 --> 00:04:01,060 write b over a as 2 times b over 2 a. So, that should be still b over a. 61 00:04:01,060 --> 00:04:06,766 And then I want to add something here, and I want to subtract something over here. 62 00:04:06,766 --> 00:04:11,130 And I want to choose the thing that I'm going to add over here. 63 00:04:11,130 --> 00:04:12,306 So that the first 64 00:04:12,306 --> 00:04:17,290 three terms can be written as x plus something squared. 65 00:04:17,290 --> 00:04:24,370 So lets just look really quickly at, what I would want to have so that the 66 00:04:24,370 --> 00:04:31,820 something is going to be b over 2 a and so if I look at x plus b over 2 a squared. 67 00:04:31,820 --> 00:04:34,250 Well on the first term. 68 00:04:34,250 --> 00:04:37,385 So I I use a trick to remind myself called foil, which is first, 69 00:04:37,385 --> 00:04:39,250 outer, inner, last. 70 00:04:39,250 --> 00:04:44,030 So the first two terms are going to be x times x, which gives me x squared. 71 00:04:44,030 --> 00:04:47,620 And then I'm going to have outer, so x times b over 2 a. 72 00:04:47,620 --> 00:04:52,440 And then inner, b over 2 a times x so that's why I get two of these guys. 73 00:04:52,440 --> 00:04:55,280 And then last will just be the last term squared. 74 00:04:57,730 --> 00:05:00,990 So this is what I'm going to choose for my blank here. 75 00:05:00,990 --> 00:05:04,656 And so my intelligent value of 0 is going to be b 76 00:05:04,656 --> 00:05:09,390 squared over 4 a squared minus b squared over 4 a squared. 77 00:05:09,390 --> 00:05:12,700 So I'm going to, oops, stick that into these blanks here. 78 00:05:18,040 --> 00:05:22,360 And now these first three terms that I have here, I can 79 00:05:22,360 --> 00:05:27,250 replace that by x plus b over two a, that quantity squared. 80 00:05:31,870 --> 00:05:34,558 And I've done two things in this line, so I'm also going to 81 00:05:34,558 --> 00:05:37,499 move these last two terms to the other side of the equals sign. 82 00:05:46,360 --> 00:05:48,430 Oh, and, what, I must have done a little bit more here. 83 00:05:48,430 --> 00:05:53,578 Okay, I also multiplied the c over a by 4 a over 84 00:05:53,578 --> 00:05:58,594 4 a so that's where this 4, oops 4 a c over a 85 00:05:58,594 --> 00:06:03,650 squared comes from what, what did I do. 86 00:06:07,260 --> 00:06:07,760 Think. 87 00:06:10,570 --> 00:06:15,080 And so then I'm going to take the square root of both sides, so I get x 88 00:06:15,080 --> 00:06:19,590 plus b over 2 a is equal to and, we're going to end up with two 89 00:06:19,590 --> 00:06:23,444 solutions now because, providing that this value here 90 00:06:23,444 --> 00:06:25,986 is greater than zero I'm going to end 91 00:06:25,986 --> 00:06:28,528 up with a positive value of the square 92 00:06:28,528 --> 00:06:31,660 root and a negative value of the square root. 93 00:06:35,090 --> 00:06:38,782 And then the rest of it is just isolating the x and making it look 94 00:06:38,782 --> 00:06:42,800 pretty so I'll just move the b over two a to the other side and then. 95 00:06:44,620 --> 00:06:47,740 A square root of a fraction, I can think of that as a square root 96 00:06:47,740 --> 00:06:50,980 of the top divided by the square root of the bottom so I'll just go 97 00:06:50,980 --> 00:06:54,520 ahead and do the square root of the bottom because 4 a squared has a 98 00:06:54,520 --> 00:06:56,440 nice square root and it also happens to 99 00:06:56,440 --> 00:06:59,860 be the same denominator as a half-up frontier. 100 00:06:59,860 --> 00:07:05,906 And so that gives me the Quadratic Formula, so X is equal 101 00:07:05,906 --> 00:07:10,760 to minus b plus or minus the square root of b squared minus 4 a c all over 2 a. 102 00:07:14,220 --> 00:07:17,650 So now let's look at I've sort of taken the important 103 00:07:17,650 --> 00:07:21,842 part out of the integral you're going to have to, to solve. 104 00:07:21,842 --> 00:07:24,254 In 461 when I ask you to find the expected 105 00:07:24,254 --> 00:07:27,478 value and the variance of a log normal random variable. 106 00:07:27,478 --> 00:07:33,394 So, I want to compute the integral from zero to infinity of e to the, so 107 00:07:33,394 --> 00:07:39,616 it's a minus fraction here, and then I have log of y minus Mu, that whole 108 00:07:39,616 --> 00:07:43,800 quantity squared Divided by two sigma squared. 109 00:07:45,110 --> 00:07:48,580 And here, mu and sigma are going to be real valued constants. 110 00:07:49,682 --> 00:07:51,190 you can think of them as just the 111 00:07:51,190 --> 00:07:54,460 mean and the standard deviation of a statistical distribution. 112 00:07:57,150 --> 00:08:03,130 So, I'm going to use my integration by substitution trick from, from last time. 113 00:08:04,730 --> 00:08:07,538 And this thing, the log of y minus mu, that 114 00:08:07,538 --> 00:08:12,790 looks like something I wouldn't be particularly happy to have. 115 00:08:12,790 --> 00:08:17,530 So I'd like to get rid of that and just replace it by a single value u. 116 00:08:23,070 --> 00:08:26,710 So if I do that then I can solve for y in terms 117 00:08:26,710 --> 00:08:30,290 of u, and I get e to the u times e to the mu. 118 00:08:34,560 --> 00:08:40,034 And then if I take the derivative of y, I just end up with 119 00:08:40,034 --> 00:08:45,389 y du, which is, I'll put that as my value of y, but in 120 00:08:45,389 --> 00:08:50,860 this particular case it's going to cancel out nicely. 121 00:08:58,460 --> 00:09:00,180 So now I make my substitution. 122 00:09:01,220 --> 00:09:02,872 So lat time I told you for definite 123 00:09:02,872 --> 00:09:05,390 integrals, I can also change the limits here. 124 00:09:05,390 --> 00:09:07,997 So this is going to be a case where probably 125 00:09:07,997 --> 00:09:10,950 be nicer if you the limits just into u. 126 00:09:10,950 --> 00:09:14,037 Rather than remember that this is u equals something, and 127 00:09:14,037 --> 00:09:16,120 then try and switch back to y at the end. 128 00:09:17,370 --> 00:09:19,260 And so it turns out if the mu doesn't 129 00:09:19,260 --> 00:09:22,330 really matter, since I'm going from zero to infinity. 130 00:09:23,476 --> 00:09:28,720 if I, if I have log of zero, that's going to be minus infinity. 131 00:09:28,720 --> 00:09:32,620 And if I have log of infinity, that's going to be plus infinity. 132 00:09:32,620 --> 00:09:35,600 So the limits on u are going to be minus infinity to infinity. 133 00:09:35,600 --> 00:09:38,960 And then I go ahead and substitute in everything 134 00:09:38,960 --> 00:09:43,310 that I worked out in the, the, second bullet-point here. 135 00:09:45,750 --> 00:09:46,940 And then replace 136 00:09:50,410 --> 00:09:52,066 My y by e to the u e to the Mu and 137 00:09:52,066 --> 00:09:56,098 now one of the, the nice things that's going to happen remember Mu is 138 00:09:56,098 --> 00:09:59,698 a constant so I can, that's just scaling my integral so one 139 00:09:59,698 --> 00:10:04,060 of my properties allows me to pull that outside of the integral sign. 140 00:10:07,850 --> 00:10:13,746 And so now I end up with e to the Mu, e to the Mu integral 141 00:10:13,746 --> 00:10:19,240 and I have e to some power of u times e to the u. 142 00:10:19,240 --> 00:10:21,160 So really whats going to end up happening here. 143 00:10:22,260 --> 00:10:26,079 If I have e to a power times e to a power I 144 00:10:26,079 --> 00:10:29,330 can also think of that as e to the sum of those two powers. 145 00:10:32,580 --> 00:10:35,810 So, I'm going to pause for a second and say take a look at the intergrand. 146 00:10:35,810 --> 00:10:37,460 So, inter-grand is just this thing. 147 00:10:37,460 --> 00:10:45,297 It's the function that I'm integrating. 148 00:10:45,297 --> 00:10:51,407 And so if I give them both a common denominator then I can go ahead and do the 149 00:10:51,407 --> 00:10:55,073 sum and now I'm starting to see something sort 150 00:10:55,073 --> 00:10:57,987 of like what I had, when I was looking 151 00:10:57,987 --> 00:11:03,063 at the completing the square when i was, so the same problem that 152 00:11:03,063 --> 00:11:08,212 I needed the solve when I wanted to solve a quadratic equation. 153 00:11:08,212 --> 00:11:13,732 And so here, it turns out that the, 154 00:11:13,732 --> 00:11:18,516 the intelligent value to add is 155 00:11:18,516 --> 00:11:23,484 sigma to the fourth and then I can 156 00:11:23,484 --> 00:11:28,918 write this in this form here. And 157 00:11:28,918 --> 00:11:34,750 this thing here is just going 158 00:11:34,750 --> 00:11:40,585 to be u minus sigma squared. 159 00:11:40,585 --> 00:11:44,550 So I have u minus. Sorry, not e minus sigma squared. 160 00:11:44,550 --> 00:11:47,060 I have to say that twice. So it's u minus sigma squared. 161 00:11:48,190 --> 00:11:48,582 And that 162 00:11:48,582 --> 00:11:51,158 whole quantity is squared as well so that if you, if you 163 00:11:51,158 --> 00:11:54,800 expanded this out you just get the thing that's in parentheses here. 164 00:11:59,290 --> 00:12:04,815 And so now I have, so here when I had Two 165 00:12:04,815 --> 00:12:08,715 functions raise to an exponent multiplied together so e to one 166 00:12:08,715 --> 00:12:11,565 power times e to another power I was able to add 167 00:12:11,565 --> 00:12:16,530 those powers together and just write it as a single exponential. 168 00:12:16,530 --> 00:12:18,660 And now I'm just going to do the same thing but backwards. 169 00:12:19,730 --> 00:12:24,460 So I'm integrating with respect to you. So that means this Sigma 170 00:12:24,460 --> 00:12:28,945 to the fourth over two sigma squared, that's just going to be a constant number. 171 00:12:28,945 --> 00:12:35,749 So I'm going to rewrite this as e to the minus quantity u minus sigma squared, 172 00:12:35,749 --> 00:12:43,200 squared divided by two sigma squared times e to the sigma squared over two. 173 00:12:43,200 --> 00:12:45,267 But now this is just another number that I could 174 00:12:45,267 --> 00:12:47,460 bring outside of my integral sign if I wanted to. 175 00:12:50,610 --> 00:12:56,210 So let's go back now that, now that I have this little intermediate result. 176 00:12:56,210 --> 00:12:58,440 And look at the function I'm trying to integrate again. 177 00:13:00,840 --> 00:13:06,073 So I can replace the inter-grand now with 178 00:13:06,073 --> 00:13:11,050 this function that I've this transformation of it that I've, so all 179 00:13:11,050 --> 00:13:15,930 I've done is taken what used to be there and replaced it with this value. 180 00:13:17,510 --> 00:13:21,210 And now I can also bring this, this part of the product out in 181 00:13:21,210 --> 00:13:26,020 front of the integral sign because it, it doesn't depend on you any more as, 182 00:13:26,020 --> 00:13:28,314 as far as I'm concerned its just a, a 183 00:13:28,314 --> 00:13:31,290 scalar quantity so I can write it like this. 184 00:13:31,290 --> 00:13:35,890 So now I have e to the Mu times sigma squared over 2. 185 00:13:35,890 --> 00:13:38,390 Integral from minus infinity to infinity. 186 00:13:40,850 --> 00:13:43,880 And then, sooner or later, you're going to learn to recognize this form. 187 00:13:43,880 --> 00:13:46,730 Here, I just had to make a hint. 188 00:13:46,730 --> 00:13:49,910 Sort of one of the problems with mathematics, it's all sort of intertwined. 189 00:13:49,910 --> 00:13:52,440 So when you try to put it in order, so that I can give them 190 00:13:52,440 --> 00:13:54,695 in sequential lectures, eventually you want to 191 00:13:54,695 --> 00:13:57,860 use something that you haven't talked about yet. 192 00:13:57,860 --> 00:13:59,940 And so I stuck that in a little gray box here. 193 00:14:01,720 --> 00:14:05,861 But it turns out that this integral, so if I have e 194 00:14:05,861 --> 00:14:11,850 to the u minus something, so it doesn't matter what number is there. 195 00:14:11,850 --> 00:14:15,036 In this particular case it happens to be sigma squared, which is 196 00:14:15,036 --> 00:14:18,870 the same value down here, but that's just because I was lucky. 197 00:14:18,870 --> 00:14:21,536 So if I have e to the minus, and then my 198 00:14:21,536 --> 00:14:28,650 variable u minus something, that quantity squared, divided by two sigma squared. 199 00:14:28,650 --> 00:14:31,770 This is something called the normal density function. 200 00:14:32,954 --> 00:14:36,149 if I integrate that from minus infinity to infinity, 201 00:14:36,149 --> 00:14:39,010 we get the square root of two pi sigma. 202 00:14:39,010 --> 00:14:41,386 And then so this is something in, in fourth week, 203 00:14:41,386 --> 00:14:44,580 we'll, we'll talk about how to actually evaluate this integral. 204 00:14:46,530 --> 00:14:50,534 But if I look back up here that's exactly the integral I have on 205 00:14:50,534 --> 00:14:54,570 the right hand side so I can just replace that with 2 pi sigma. 206 00:14:57,200 --> 00:15:02,279 So I end up with 2 pi sigma e to the u plus sigma squared over 2. 207 00:15:04,010 --> 00:15:07,628 So, that probably doesn't seem like the most useful example, until you get to 208 00:15:07,628 --> 00:15:11,270 the homework in 461, where you have to do this again for the first time. 209 00:15:11,270 --> 00:15:13,760 And then I think you'll find these two slides very, very useful.