1 00:00:00,690 --> 00:00:04,957 [MUSIC]. 2 00:00:04,957 --> 00:00:10,963 Even integrating something as "easy" as a polynomial can be rather surprising. 3 00:00:10,963 --> 00:00:19,103 Let's integrate from x equals 0 to 1 just the polynomial x plus 10 times x minus 1 4 00:00:19,103 --> 00:00:27,238 to the 10th power dx. So there's some integration problem, just 5 00:00:27,238 --> 00:00:33,000 an integration problem for a polynomial. Your first inclination might be oh, I 6 00:00:33,000 --> 00:00:38,350 know what to do, just have no fear. I'll just expand everything in sight. 7 00:00:38,350 --> 00:00:43,130 So, we've got x plus 10, times x minus 1, to the 10th. 8 00:00:43,130 --> 00:00:46,395 Which you could expand by, you know the binomial theorem if you want. 9 00:00:46,395 --> 00:00:52,451 We're going to get x to the 11th minus 55 x to the 9th. 10 00:00:52,451 --> 00:01:00,011 Plus 330 x to the 8th minus 990 x to the 7th plus 1,848 times x to the 6th mi, I 11 00:01:00,011 --> 00:01:08,085 mean this is ridiculous. Arr, yeah, so I mean yeah, you, you could 12 00:01:08,085 --> 00:01:11,060 do that. You could just expand everything out. 13 00:01:11,060 --> 00:01:14,630 But there's a better way right. We can make use of U-substitution. 14 00:01:14,630 --> 00:01:17,580 Yeah, we're not going to do that, let's do something else. 15 00:01:17,580 --> 00:01:26,250 So instead let's set u equal x minus 1, and I could solve this for x. 16 00:01:26,250 --> 00:01:34,880 That means that x is u plus 1. All right, if I just add 1 to both sides. 17 00:01:34,880 --> 00:01:37,970 And I could make this substitution in this problem. 18 00:01:37,970 --> 00:01:46,310 So this integral from x goes from 0 to 1, is, what's x plus 10 if x is u plus 1. 19 00:01:46,310 --> 00:01:53,020 I could write u plus 1 plus 10. And whats x minus one, thats just u to 20 00:01:53,020 --> 00:01:57,732 the tenth power, and whats dx well if u is x du is dx. 21 00:01:57,732 --> 00:02:02,002 So its the same as trying to do this integration problem. 22 00:02:02,002 --> 00:02:06,812 In some text books this technique of startign with the use of substitution but 23 00:02:06,812 --> 00:02:11,194 then solving for x. This is sometimes called back 24 00:02:11,194 --> 00:02:15,100 substitution. regardless of the name, let's proceed. 25 00:02:15,100 --> 00:02:23,940 So this is the integral, x goes from 0 to 1 of u plus 11 times u to the 10th du. 26 00:02:23,940 --> 00:02:29,025 I could expand this, and that's the same as the 27 00:02:29,025 --> 00:02:35,074 Integral x goes from zero to one of u to the 11th. 28 00:02:35,074 --> 00:02:40,361 Plus 11 u to the 10th du. But this is just asking me to anti 29 00:02:40,361 --> 00:02:44,305 differentiate a polynomial which I can definitely do. 30 00:02:44,305 --> 00:02:51,440 Alright. This is u to the 12 over 12. 31 00:02:51,440 --> 00:02:57,140 Plus, what's an anti-derivative of 11 u to the 10th, it's just u to the 11th. 32 00:02:57,140 --> 00:02:59,340 But now I've got a little bit of an issue, right? 33 00:02:59,340 --> 00:03:05,330 I've got x going from 0 to 1. So here is u going from what to what? 34 00:03:05,330 --> 00:03:08,210 Yeah, so how do we deal with those end points? 35 00:03:08,210 --> 00:03:15,180 Well, when x is 0, u is minus 1. So I'll put a minus 1 there. 36 00:03:15,180 --> 00:03:19,035 And when x is 1, u is 0. So I'll put a zero there. 37 00:03:19,035 --> 00:03:22,467 And now, this is just as easy as plugging in zero, plugging in minus 1, and taking 38 00:03:22,467 --> 00:03:26,090 the difference. So what do I get when I plug in zero? 39 00:03:26,090 --> 00:03:29,000 I just get zero. And what do I get when I plug in minus 1? 40 00:03:29,000 --> 00:03:35,396 Well, I get minus 1 to the 12th over 12. Plus minus 1 to the 11th. 41 00:03:35,396 --> 00:03:42,638 Alright minus 1 to the 12 is just 1. So this is 1 12th. 42 00:03:42,638 --> 00:03:48,700 And minus 1 to the 11th well that's negative 1. 43 00:03:48,700 --> 00:03:55,900 So I've got 0 minus what's this a 12 minus one that's negative 11th 12th. 44 00:03:55,900 --> 00:03:59,430 But at 0 minus negative 11, 12ths. So the integral is 11, 12ths. 45 00:03:59,430 --> 00:04:03,950 So the calculation turned out to be pretty nice. 46 00:04:03,950 --> 00:04:09,530 I mean 11 12ths, isn't a particularly difficult fraction to come up with. 47 00:04:09,530 --> 00:04:13,188 The crazy thing to think about is that we could have approached this just by 48 00:04:13,188 --> 00:04:17,999 expanding everything in sight. Yeah, I really could have expanded this 49 00:04:17,999 --> 00:04:22,459 thing out, and then just started anti-differentiating you know. 50 00:04:22,459 --> 00:04:26,869 And finding this really was you know, I don't know x to the 12th over 12, minus I 51 00:04:26,869 --> 00:04:31,090 think it's 11 halves, x to the tenth plus a hundered and ten thirds x to the ninth 52 00:04:31,090 --> 00:04:35,903 right. And it would keep on goign and id just 53 00:04:35,903 --> 00:04:39,935 evaluate that, exit one exit zero and take the differenc i mean i really could 54 00:04:39,935 --> 00:04:45,450 have done this by expanding. And somehow that would of ended up giving 55 00:04:45,450 --> 00:04:48,783 me the same answer right, of eleven twelths. 56 00:04:48,783 --> 00:04:52,815 Except that had I gone down that path, I would have been certain to have walked 57 00:04:52,815 --> 00:04:58,676 right into an arithmetic error. The world of mathematical solutions is 58 00:04:58,676 --> 00:05:02,120 not divided into correct solutions and incorrect solutions. 59 00:05:02,120 --> 00:05:06,016 It's really more subtle than that. Mathematical solutions come in a ton of 60 00:05:06,016 --> 00:05:08,522 varieties. Right, there's easy solutions and hard 61 00:05:08,522 --> 00:05:11,230 solutions. There's solutions that are defending you 62 00:05:11,230 --> 00:05:15,164 against arithmetic errors. And then there's solutions that are just 63 00:05:15,164 --> 00:05:18,330 walking you right into an arithmetic mess. 64 00:05:18,330 --> 00:05:21,988 The fun of u substitution isn't just being able to do integrals, it's being 65 00:05:21,988 --> 00:05:25,941 able to do integrals in an efficient way that protects you from those arithmetic 66 00:05:25,941 --> 00:05:28,090 errors.