1 00:00:00,000 --> 00:00:08,960 CHRISTINE BREINER: Welcome back to recitation. 2 00:00:08,960 --> 00:00:11,120 In this video I want to explain to you 3 00:00:11,120 --> 00:00:14,310 where the coefficients we saw in Simpson's rule 4 00:00:14,310 --> 00:00:16,010 actually come from. 5 00:00:16,010 --> 00:00:19,620 So what I'm going to do is take this curve that I have 6 00:00:19,620 --> 00:00:22,970 and show you what the picture of Simpson's rule does. 7 00:00:22,970 --> 00:00:26,570 And then I'm actually going to determine explicitly where 8 00:00:26,570 --> 00:00:28,300 the coefficients come from. 9 00:00:28,300 --> 00:00:30,570 So let's look at this picture first. 10 00:00:30,570 --> 00:00:35,585 The picture I have here, the white curve is going to be my y 11 00:00:35,585 --> 00:00:39,159 equals f of x function. 12 00:00:39,159 --> 00:00:41,450 And so, if you remember, what Simpson's rule is saying, 13 00:00:41,450 --> 00:00:45,270 you have to have two delta x's. 14 00:00:45,270 --> 00:00:48,610 So in this case h is equal to delta x. 15 00:00:48,610 --> 00:00:52,230 So what Simpson's rule is saying is I can find, approximate, 16 00:00:52,230 --> 00:00:54,310 the area under this curve from minus 17 00:00:54,310 --> 00:00:57,680 h to h by a certain expression. 18 00:00:57,680 --> 00:01:01,520 And the expression is, I said delta x is equal to h, 19 00:01:01,520 --> 00:01:09,066 so let me write h over 3 times y_0 plus 4*y_1 plus y_2-- oh, 20 00:01:09,066 --> 00:01:09,566 y_2. 21 00:01:12,212 --> 00:01:15,032 So that's what you got-- 22 00:01:15,032 --> 00:01:16,490 that's what you saw in the lecture, 23 00:01:16,490 --> 00:01:18,802 that this is what the coefficients are. 24 00:01:18,802 --> 00:01:20,510 So I want to explain why this is a 1, why 25 00:01:20,510 --> 00:01:22,915 that's a 4, why that's a 1, and where that 3 comes from. 26 00:01:22,915 --> 00:01:24,540 The picture is not going to explain it, 27 00:01:24,540 --> 00:01:27,740 but the picture will help us understand how to go. 28 00:01:27,740 --> 00:01:34,530 So what Simpson's rule does is it takes those three points, 29 00:01:34,530 --> 00:01:39,170 so the (x_0, y_0) the (x_1, y_1) and the (x_2, y_2) 30 00:01:39,170 --> 00:01:42,370 that are on the curve y equals f of x. 31 00:01:42,370 --> 00:01:45,860 And it finds a parabola through those three points. 32 00:01:45,860 --> 00:01:49,590 So I'm going to do my best to draw what looks like a parabola 33 00:01:49,590 --> 00:01:52,140 through those three points. 34 00:01:52,140 --> 00:01:55,870 Something-- I'll draw it lightly first and then I'll fill it in. 35 00:01:55,870 --> 00:02:01,610 Something along these lines. 36 00:02:01,610 --> 00:02:02,560 Something like that. 37 00:02:02,560 --> 00:02:06,610 That's a, sort of looks like a parabola. 38 00:02:06,610 --> 00:02:09,260 And so the idea is Simpson's rule 39 00:02:09,260 --> 00:02:14,777 is saying I can find the area under the blue curve. 40 00:02:14,777 --> 00:02:16,610 So I can find the area under the blue curve. 41 00:02:19,950 --> 00:02:22,540 So this is a function, we'll call this y equal P of x. 42 00:02:26,010 --> 00:02:29,130 And that's what you were actually told about in lecture. 43 00:02:29,130 --> 00:02:31,575 That you're approximating your function y 44 00:02:31,575 --> 00:02:34,610 equals f of x by a quadratic function that 45 00:02:34,610 --> 00:02:39,020 goes through the values (x_0, y_0), (x_1, y_1), 46 00:02:39,020 --> 00:02:41,160 and (x_2, y_2). 47 00:02:41,160 --> 00:02:45,750 And then you find the area under that parabola. 48 00:02:45,750 --> 00:02:47,990 Between minus h and h. 49 00:02:47,990 --> 00:02:49,890 Now this picture, you might say, Christine, 50 00:02:49,890 --> 00:02:51,807 this looks really like a bad estimate. 51 00:02:51,807 --> 00:02:52,890 This looks kind of stupid. 52 00:02:52,890 --> 00:02:54,931 Maybe you should have picked a different function 53 00:02:54,931 --> 00:02:56,790 to estimate this. 54 00:02:56,790 --> 00:02:59,940 And I did this one because I wanted to be zoomed out far 55 00:02:59,940 --> 00:03:02,650 and to show you, to give you a little intuition about what 56 00:03:02,650 --> 00:03:05,190 happens when we make h smaller. 57 00:03:05,190 --> 00:03:06,810 Notice that if I cut the size of h 58 00:03:06,810 --> 00:03:10,540 in half, so if I started with here was minus h to h, 59 00:03:10,540 --> 00:03:12,540 what would my three points be? 60 00:03:12,540 --> 00:03:16,750 My three points would be this point, y_1, and y_2 61 00:03:16,750 --> 00:03:17,920 would be right here. 62 00:03:17,920 --> 00:03:19,660 Well the quadratic through those points 63 00:03:19,660 --> 00:03:22,870 is much closer to the actual function. 64 00:03:22,870 --> 00:03:26,460 And if I cut that in half again, I'd have points here, 65 00:03:26,460 --> 00:03:27,890 here, and here. 66 00:03:27,890 --> 00:03:29,730 Something like that, and that starts 67 00:03:29,730 --> 00:03:31,660 to look almost exactly like a quadratic. 68 00:03:31,660 --> 00:03:35,222 So if I found the area under this-- 69 00:03:35,222 --> 00:03:36,680 or if I wanted to estimate the area 70 00:03:36,680 --> 00:03:39,070 under this piece of the curve using 71 00:03:39,070 --> 00:03:41,170 a quadratic through those three points, 72 00:03:41,170 --> 00:03:43,280 they would be very close. 73 00:03:43,280 --> 00:03:45,680 So don't be alarmed by Simpson's rule 74 00:03:45,680 --> 00:03:50,371 as an approximation based on this large picture. 75 00:03:50,371 --> 00:03:51,620 So now, what do we have to do? 76 00:03:51,620 --> 00:03:52,703 Remember, what's our goal? 77 00:03:52,703 --> 00:03:55,280 Our goal is to figure out where the coefficients come from. 78 00:03:55,280 --> 00:03:56,970 So what I actually need to do is I need 79 00:03:56,970 --> 00:03:59,870 to evaluate a certain integral. 80 00:03:59,870 --> 00:04:02,220 Or I need to integrate a certain function. 81 00:04:02,220 --> 00:04:04,650 I need to integrate P of x. 82 00:04:04,650 --> 00:04:08,220 So what I'm going to be finding through the rest of this video, 83 00:04:08,220 --> 00:04:13,310 is the integral from minus h to h of P of x dx. 84 00:04:13,310 --> 00:04:16,750 And my goal is to show that this integral is 85 00:04:16,750 --> 00:04:17,840 equal to this expression. 86 00:04:17,840 --> 00:04:20,235 I want to show that these are equal. 87 00:04:20,235 --> 00:04:20,860 That's my goal. 88 00:04:24,270 --> 00:04:26,305 So let's get down to it. 89 00:04:26,305 --> 00:04:26,930 What do I know? 90 00:04:26,930 --> 00:04:29,780 What's the only thing I know right away about P of x? 91 00:04:29,780 --> 00:04:32,166 I know I'm choosing it to be a quadratic function 92 00:04:32,166 --> 00:04:34,165 and I know it goes through three certain points. 93 00:04:34,165 --> 00:04:35,390 Right? 94 00:04:35,390 --> 00:04:39,210 I'll write down what the three points are when we need them. 95 00:04:39,210 --> 00:04:41,330 But first let's just say, in general, 96 00:04:41,330 --> 00:04:45,010 let's take this integral for a general quadratic 97 00:04:45,010 --> 00:04:47,690 and see how much information we actually need. 98 00:04:47,690 --> 00:04:50,940 So let me come over here. 99 00:04:50,940 --> 00:04:55,304 Actually, let me say first, what do I mean my general quadratic? 100 00:04:55,304 --> 00:04:57,220 I mean something of the form big A, capital A, 101 00:04:57,220 --> 00:05:02,990 x squared plus B*x plus C. So I'm not filling in values 102 00:05:02,990 --> 00:05:04,950 for these coefficients yet. 103 00:05:04,950 --> 00:05:08,870 Those coefficients are coming exactly from my three points. 104 00:05:08,870 --> 00:05:10,850 (x_0, y_0), (x_1, y_1), and (x_2, y_2). 105 00:05:13,242 --> 00:05:14,950 So let's just integrate that from minus h 106 00:05:14,950 --> 00:05:18,812 to h first and see what kind of information we need. 107 00:05:18,812 --> 00:05:19,770 So if I come over here. 108 00:05:19,770 --> 00:05:32,570 So what do I get when I actually integrate this? 109 00:05:32,570 --> 00:05:37,180 Well, I get A x to the third over 3, 110 00:05:37,180 --> 00:05:40,724 plus B x squared over 2, plus C*x, 111 00:05:40,724 --> 00:05:42,640 and then I have to evaluate from minus h to h. 112 00:05:46,937 --> 00:05:48,520 Well, if you were thinking about this, 113 00:05:48,520 --> 00:05:51,000 it shouldn't be surprising that when I do this, 114 00:05:51,000 --> 00:05:53,080 there's not going to be a B term. 115 00:05:53,080 --> 00:05:54,840 Why is that? 116 00:05:54,840 --> 00:05:56,790 Well, these two functions are even. 117 00:05:56,790 --> 00:05:59,172 A x squared and C are even functions. 118 00:05:59,172 --> 00:06:00,880 And I'm integrating over something that's 119 00:06:00,880 --> 00:06:03,130 symmetric about the y-axis. 120 00:06:03,130 --> 00:06:05,420 I'm going from minus h to h. 121 00:06:05,420 --> 00:06:07,460 So if you think about A x squared, 122 00:06:07,460 --> 00:06:08,960 and I'm going from minus h to h, I'm 123 00:06:08,960 --> 00:06:12,430 finding the area under a quadratic, from minus h to h, 124 00:06:12,430 --> 00:06:16,200 it's going to be twice the area from 0 to h. 125 00:06:16,200 --> 00:06:19,470 But B*x, that's a line with slope B. 126 00:06:19,470 --> 00:06:24,410 If I wanted to integrate B*x from minus h to h, 127 00:06:24,410 --> 00:06:28,670 that's the area, the signed area under the curve, of a line B*x, 128 00:06:28,670 --> 00:06:29,770 from minus h to h. 129 00:06:29,770 --> 00:06:35,120 If I just draw a quick picture, what does that look like? 130 00:06:35,120 --> 00:06:39,970 It's symmetric with respect to a rotation there. 131 00:06:39,970 --> 00:06:41,210 I'd have this signed area. 132 00:06:41,210 --> 00:06:43,960 This is the curve y equals B*x. 133 00:06:43,960 --> 00:06:47,420 From minus h to h, I get this area is negative 134 00:06:47,420 --> 00:06:49,470 and this are is positive and they're equal. 135 00:06:49,470 --> 00:06:51,940 So, what I'm about to do, you shouldn't be surprised 136 00:06:51,940 --> 00:06:54,100 there won't be a B term. 137 00:06:54,100 --> 00:06:56,370 So what do we actually get when we evaluate this? 138 00:06:56,370 --> 00:07:03,470 We get 2A h cubed, over 3, plus 2*C*h. 139 00:07:06,599 --> 00:07:07,390 That's what we get. 140 00:07:07,390 --> 00:07:09,723 You can actually work it all, put in all the h's and see 141 00:07:09,723 --> 00:07:12,860 that, but I knew I was going to have double 142 00:07:12,860 --> 00:07:15,660 what was here when I evaluated at h, and nothing 143 00:07:15,660 --> 00:07:17,870 from the B term. 144 00:07:17,870 --> 00:07:20,400 So we're getting closer. 145 00:07:20,400 --> 00:07:21,280 We're getting closer. 146 00:07:21,280 --> 00:07:23,113 We might not look like we're getting closer, 147 00:07:23,113 --> 00:07:24,320 but we're getting closer. 148 00:07:24,320 --> 00:07:28,210 So let's simplify this expression a little bit. 149 00:07:28,210 --> 00:07:31,680 Actually, what I ultimately need to do is I need to figure out C 150 00:07:31,680 --> 00:07:33,870 and I need to figure out something over here. 151 00:07:33,870 --> 00:07:36,960 I need to actually figure out A h squared. 152 00:07:36,960 --> 00:07:40,080 And let me explain why I need to figure out A h squared. 153 00:07:40,080 --> 00:07:42,790 In the end, if I come over back to what 154 00:07:42,790 --> 00:07:45,620 I want to show-- let me even box it, so we see it-- 155 00:07:45,620 --> 00:07:49,340 I want to show that this integral, which 156 00:07:49,340 --> 00:07:51,360 I've started to calculate, is equal to h over 157 00:07:51,360 --> 00:07:53,100 3 times this quantity. 158 00:07:53,100 --> 00:07:55,610 So I want to keep one around, but I 159 00:07:55,610 --> 00:07:57,600 want the other, any other powers of h 160 00:07:57,600 --> 00:07:59,440 to not be there when I'm done. 161 00:07:59,440 --> 00:07:59,940 Right? 162 00:07:59,940 --> 00:08:01,290 I need one power of h there. 163 00:08:01,290 --> 00:08:03,570 In fact I need an h over 3 there. 164 00:08:03,570 --> 00:08:05,540 So I think what I'll do is I'll start off 165 00:08:05,540 --> 00:08:08,470 and I'll pull out an h over 3, and then 166 00:08:08,470 --> 00:08:11,320 I'll try and figure out if I can get the rest of my expression 167 00:08:11,320 --> 00:08:13,220 to look like what's in the parentheses. 168 00:08:13,220 --> 00:08:16,140 That's really, ultimately, what I'd like to do. 169 00:08:16,140 --> 00:08:19,130 So let's come back over here. 170 00:08:19,130 --> 00:08:21,355 I'm going to pull out an h over 3 from both of these 171 00:08:21,355 --> 00:08:23,146 and we're going to see what we end up with. 172 00:08:23,146 --> 00:08:28,359 OK so if I pull out an h over 3 here, what do I end up with? 173 00:08:28,359 --> 00:08:28,900 This is easy. 174 00:08:28,900 --> 00:08:30,340 That's 2A h squared. 175 00:08:33,660 --> 00:08:35,340 And this one's a little bit trickier. 176 00:08:35,340 --> 00:08:39,470 But I have to multiply by 3 over 3 to get a 3 there. 177 00:08:39,470 --> 00:08:40,570 So I end up with 6C. 178 00:08:44,751 --> 00:08:45,250 OK. 179 00:08:45,250 --> 00:08:47,041 Let's make sure I didn't make any mistakes. 180 00:08:47,041 --> 00:08:50,660 If I multiply through here I get 2A h cubed over 3. 181 00:08:50,660 --> 00:08:56,790 If I multiply through here I get 2h*C. Looking good so far. 182 00:08:56,790 --> 00:09:01,295 Now I have to figure out A and C, or A h squared and C. Well, 183 00:09:01,295 --> 00:09:02,970 C is pretty easy to find. 184 00:09:02,970 --> 00:09:04,550 Let's think about why that is. 185 00:09:04,550 --> 00:09:06,200 I have this polynomial. 186 00:09:06,200 --> 00:09:08,840 The polynomial was equal to-- if we come over here and we look 187 00:09:08,840 --> 00:09:12,730 again, it's A x squared plus B*x plus C. 188 00:09:12,730 --> 00:09:15,350 And the polynomial has to go through those three points I 189 00:09:15,350 --> 00:09:16,340 had. 190 00:09:16,340 --> 00:09:22,120 So when x is 0, P of 0 is C. So let's come back 191 00:09:22,120 --> 00:09:23,190 and look at our picture. 192 00:09:23,190 --> 00:09:28,240 When x is 0, what's the output on the white curve? 193 00:09:28,240 --> 00:09:29,640 It's y_1. 194 00:09:29,640 --> 00:09:32,290 So C is exactly y_1. 195 00:09:32,290 --> 00:09:33,910 How am I going to find A h squared? 196 00:09:33,910 --> 00:09:36,940 Well, what I need to do is use these other two points. 197 00:09:36,940 --> 00:09:41,280 I'm going to evaluate the function P of x at minus h. 198 00:09:41,280 --> 00:09:43,510 And its output has to be y_0. 199 00:09:43,510 --> 00:09:47,430 And I'm going to evaluate the function P of x at h 200 00:09:47,430 --> 00:09:50,054 and its output has to be y_2. 201 00:09:50,054 --> 00:09:51,970 So we're going to come over to the other side, 202 00:09:51,970 --> 00:09:53,780 but that's really what we're doing next. 203 00:09:53,780 --> 00:09:59,900 So let's come over here and let's evaluate P of h 204 00:09:59,900 --> 00:10:03,590 and P of minus h. 205 00:10:03,590 --> 00:10:12,290 So P of h is A h squared plus B*h plus C. 206 00:10:12,290 --> 00:10:23,240 And P of minus h is A h squared minus B*h plus C. OK. 207 00:10:23,240 --> 00:10:23,740 What else? 208 00:10:23,740 --> 00:10:27,340 Again we said this one has to be y_2, the output, 209 00:10:27,340 --> 00:10:29,300 and this one has to be y_0. 210 00:10:29,300 --> 00:10:31,420 Because the output at h has to agree 211 00:10:31,420 --> 00:10:33,530 with the output of the function f of x. 212 00:10:33,530 --> 00:10:35,530 And its output at h was y_2. 213 00:10:35,530 --> 00:10:39,080 The output at minus h of P has to be 214 00:10:39,080 --> 00:10:41,490 the same as the output at minus h of f. 215 00:10:41,490 --> 00:10:43,644 And that was y_0. 216 00:10:43,644 --> 00:10:45,310 Now this might not look fun yet but what 217 00:10:45,310 --> 00:10:48,350 if I add these two things together. 218 00:10:48,350 --> 00:10:49,440 What happens? 219 00:10:49,440 --> 00:10:54,110 I get 2A h squared. 220 00:10:54,110 --> 00:10:56,270 These drop out. 221 00:10:56,270 --> 00:11:06,980 And then I get plus 2C is equal to y_0 plus y_2. 222 00:11:06,980 --> 00:11:08,060 I'm getting closer. 223 00:11:08,060 --> 00:11:09,610 I'm getting much closer. 224 00:11:09,610 --> 00:11:11,300 Because now notice what I have. 225 00:11:11,300 --> 00:11:14,412 I have 2A h squared, I can isolate that. 226 00:11:14,412 --> 00:11:16,370 And that's something that I want to figure out. 227 00:11:16,370 --> 00:11:17,786 I want to figure out 2A h squared. 228 00:11:17,786 --> 00:11:20,620 So let's figure out what that is. 229 00:11:20,620 --> 00:11:29,140 2A h squared is equal to y_0 plus y_2 minus, well 230 00:11:29,140 --> 00:11:30,640 what did we say C was? 231 00:11:30,640 --> 00:11:34,210 C was the output at x equals 0. 232 00:11:34,210 --> 00:11:35,240 Which is y1. 233 00:11:35,240 --> 00:11:40,840 So it's minus 2C, which is minus 2y_1 So now we're very close, 234 00:11:40,840 --> 00:11:42,740 we're very close to getting what we want. 235 00:11:42,740 --> 00:11:45,716 And that's good, because I'm almost out of room. 236 00:11:45,716 --> 00:11:47,090 So let's take that expression, we 237 00:11:47,090 --> 00:11:49,450 were working on this expression right here, 238 00:11:49,450 --> 00:11:51,930 and let's start to fill in what we get. 239 00:11:51,930 --> 00:11:56,260 We get h over 3 times 2A h squared, 240 00:11:56,260 --> 00:12:02,030 which is y_0 plus y_2 minus 2y_1, 241 00:12:02,030 --> 00:12:03,980 and then I have to add 6C. 242 00:12:03,980 --> 00:12:05,010 Now what's C? 243 00:12:05,010 --> 00:12:06,970 C is y_1. 244 00:12:06,970 --> 00:12:10,390 So I'm going to add 6y_1. 245 00:12:10,390 --> 00:12:18,940 And if I simplify that, I get y_0-- bingo-- plus 4y_1 246 00:12:18,940 --> 00:12:21,770 plus y_2. 247 00:12:21,770 --> 00:12:23,290 And that's what we wanted. 248 00:12:23,290 --> 00:12:26,090 So let me take us through kind of where all this came 249 00:12:26,090 --> 00:12:29,920 from again, what the big pieces were and we'll see now we 250 00:12:29,920 --> 00:12:33,330 understand how we do Simpson's rule, and these small chunks 251 00:12:33,330 --> 00:12:34,680 of Simpson's rule. 252 00:12:34,680 --> 00:12:36,388 So let's come back to the very beginning. 253 00:12:40,580 --> 00:12:43,570 OK, in the very beginning, what we had was a function 254 00:12:43,570 --> 00:12:45,120 f of x, that was the white curve. 255 00:12:45,120 --> 00:12:47,270 y equals f of x was the white curve. 256 00:12:47,270 --> 00:12:49,590 And then I found a quadratic that 257 00:12:49,590 --> 00:12:54,597 went through minus h, y0, (0, y_1), and (h, y_2). 258 00:12:54,597 --> 00:12:55,930 Went through those three points. 259 00:12:55,930 --> 00:12:59,314 And I called that quadratic P of x. 260 00:12:59,314 --> 00:13:00,730 And then what I was doing, we know 261 00:13:00,730 --> 00:13:03,720 Simpson's rule says find the area under the curve of P 262 00:13:03,720 --> 00:13:06,190 of x between minus h and h. 263 00:13:06,190 --> 00:13:08,390 So what I did was I came over here 264 00:13:08,390 --> 00:13:12,640 and I said OK, P of x, integral of P of x from minus h to h. 265 00:13:12,640 --> 00:13:16,200 I wrote P of x in a very general form. 266 00:13:16,200 --> 00:13:17,900 And then I actually found an integral. 267 00:13:17,900 --> 00:13:21,275 I came through and after writing it out, 268 00:13:21,275 --> 00:13:23,400 calculating the integral, I got to this expression. 269 00:13:23,400 --> 00:13:28,180 This is actually the integral of P of x from minus h to h. 270 00:13:28,180 --> 00:13:30,900 So it's the area under the curve of P of x from minus h to h 271 00:13:30,900 --> 00:13:32,270 is that. 272 00:13:32,270 --> 00:13:33,680 And now I had to figure this out, 273 00:13:33,680 --> 00:13:38,250 how to write this in terms of the outputs of f of x. 274 00:13:38,250 --> 00:13:40,340 Which were y_0, y_1, and y_2. 275 00:13:40,340 --> 00:13:42,550 Those were the ones we were interested in. 276 00:13:42,550 --> 00:13:45,270 So I ultimately knew I wanted an h over three in front. 277 00:13:45,270 --> 00:13:48,160 I knew I wanted my integral in my quadratic to be 278 00:13:48,160 --> 00:13:52,940 h over 3 times something, so I pulled out an h over 3, 279 00:13:52,940 --> 00:13:55,144 and then I looked at what this expression was. 280 00:13:55,144 --> 00:13:56,560 And if I could get this expression 281 00:13:56,560 --> 00:13:58,880 to look like the inside of what I wanted, 282 00:13:58,880 --> 00:14:03,460 which was y_0 plus 4y_1 plus y_2, I was in business. 283 00:14:03,460 --> 00:14:06,300 And so then I used outputs that I 284 00:14:06,300 --> 00:14:10,190 knew for P of x to find 2A h squared 285 00:14:10,190 --> 00:14:14,720 and to find C. I know P of h is the output of the f of x 286 00:14:14,720 --> 00:14:17,560 function at the far right, which was y_2. 287 00:14:17,560 --> 00:14:19,710 And I knew P of minus h was the output of the f 288 00:14:19,710 --> 00:14:21,070 of x function at the far left. 289 00:14:21,070 --> 00:14:22,500 That's y_0. 290 00:14:22,500 --> 00:14:25,600 I actually then evaluate P at h and minus h, 291 00:14:25,600 --> 00:14:29,220 and when I add those together, I get my 2A h squared 292 00:14:29,220 --> 00:14:31,600 in terms of y_0, y_1, and y_2. 293 00:14:31,600 --> 00:14:34,320 Let me do this in terms of y_0, y_1, and y_2. 294 00:14:34,320 --> 00:14:36,972 Because I also knew C was y_1. 295 00:14:36,972 --> 00:14:38,180 We talked about that as well. 296 00:14:38,180 --> 00:14:39,820 C had to be y_1. 297 00:14:39,820 --> 00:14:43,000 So I can then do the substitution I need right here 298 00:14:43,000 --> 00:14:46,640 and show where the coefficients in Simpson's rule come from. 299 00:14:46,640 --> 00:14:48,390 So hopefully that was informative for you. 300 00:14:48,390 --> 00:14:49,990 And I think I'll stop there.