Calculus: Making u-Substitutions

U-Substitutions explained very well!

06/02/2009

~ brittanie

Hahaha, I love the use of sound effects and props to emphasize a calculus rule like the chain rule. Professor Burger knows how to keep an audience engaged. However, sometimes his calculus jokes can be a little nerdy. But he is the nerd of all nerds and so likable. He's just like Ellen Degeneres!

The calculus problems he uses to explain u-substitutions are very helpful. I was in need of some serious calculus help.

It's great to have calculus on the web!

• View all learner reviews »

This was very helpful!

11/09/2009

~ Roulet

This was the best explanation for U substitutions that I found on the internet. The full lesson was even better than the preview indicated - I like his sense of humor and use of props and sound effects (for a calculus lesson - go figure!)

• Comment on this lesson!
• View all lesson comments »

Techniques of Integration
Integration Using Tables
Making U-Substitutions Page [1 of 2]
One really fundamental, but very useful technique for integrating functions that are a little bit more elaborate than just some of the classical vanilla, look at it, do it kind of things, is the udu-substitution. And I just wanted to sort of remind you of this technique and to point out its utility and how we can actually use it in practice.
Suppose we wanted to actually integrate the following: integral of the quantity (3x2 – 1)100 x dx. So when you look at that integral, there’s no way to immediately just see what the anti-derivative is, because it’s sort of a complicated thing. I have some thing here raised to the hundredth power; I’ve got some stuff out here and so forth. But, actually, the key to an issue like this is to notice something. And, in fact, with all integration type questions and all techniques of integration, we have to constantly remember to look at the integral and look for patterns. All of mathematics and, in fact, all of life is just a search for an understanding of patterns. And once we recognize the pattern, we really understand the idea.
Now, when I look at this, what I see is that there’s an inside thing, a big blop, and that I’m raising it to the one-hundredth power. And then I’ve got that inside thing to deal with. So this really conjures up the thinking about the chain rule. In fact, the chain rule is that mechanism that allows us to take derivatives of things with insides and outside. We chain the inside and the outside together. So the question really is can I somehow use the chain rule in reverse? And that’s exactly what u-substitution is. The u-substitution is the integral version of the chain rule. The chain rule is the differentiation thing; the integration analog will be u-substitution and here’s how it works.
The thing to notice is that if I just look at the inside and take the derivative of it, that’s going to be 6x. And that sort of looks like what’s hanging out here. Now, I admit there’s no 6, but a constant multiple doesn’t scare me, because I can just pull out constants as I wish. So the fact is the inside, as I see it, is just basically x with a constant. And that’s what I see here. So if I make a u-substitution – so let’s just let u be a new variable, which will be the entire blop. So basically I think of u as being the inside blop. So in this case, its’ going to be 3x2 – 1. If I make that substitution, let’s now rewrite this integral. So if I rewrite this integral, it becomes what? Well, it becomes the integral – and in replace of this really complicated-looking thing, I can replace it by just happily u100. And that’s seems like an enormous savings. Unfortunately, there’s more stuff here, so how do I deal with that? Well, I’ll differentiate this with respect to x. So = 6x. And so, for thinking purposes, we can view this as – if I think about cross-multiplying or multiplying both sides by the differential dx, then I see that du = 6x dx. Now, notice what we have in the original integral. In the original integral we have x dx in that bubble. And here I'm really close to getting that. I have 6x dx. So if I divide both sides by 6, what I would see is . So that little bubble of x dx can now be replaced by its equivalent mate, 6du. And this thing we already replaced by u100. And so now we put in the bubble. The bubble is . And now I have a new integral, which just has u’s in it. There are no x’s at all. Notice that everything was converted. Everything was converted to u’s, not just this piece, but, in fact, everything. And now that is a multiple, a constant multiple, so that can be migrated out beside, and so I’m just looking at the integral of u100du. But that’s actually an easy integral, that’s just . So that equals . But I'm not done, because the original question was just posed in terms of x’s, and now I have an answer that has u’s in it. So I've got to remember that u is just a shorthand way of saying 3x2 – 1. So if I insert that right now, what I see is the integral will equal . That is the value of this integral, using u-substitution. So that’s the power of u-substitution when you see an inside thing, where the derivative of the inside thing is sort of sitting on the outside. Undo the chain rule and use u-substitution. Let me try to really just whip out one last fast one so you can see this in action.
Let’s look at the integral . How would you do this? Well, I look at this and I see this is way too complicated to do. However, I notice that if I look at the inside of this piece here and take the derivative of it, I see –4x. And that’s almost exactly like the 1x I have up here. So, in fact, let’s make a substitution. Let’s let u = 1 – 2x2. In that case, what I would see is = -4x. Now, what would the substitution look like if I made this? Well, if I make this substitution, I see the integral – and now, downstairs I can replace all that stuff by . So in place of all this stuff, I just put in . So that’s pretty great. The only problem is I’m still left with this stuff. How do I change that windshield- shaped thing into stuff with u’s? Well, I look at this derivative and I say I’m pretty close. If I multiply by the differential dx, I see du = -4x dx. And here I see 1x dx. So if I divide by sides by –4, I see . And so, in place of the windshield-shaped thing, I can actually put in . And that’s the windshield-shaped-looking thing. So everything has been converted. And now I have a much easier integral. That is just a constant multiple. I pull that out and what’s ? That’s just u-1/2du. And so what’s the integral of that? Well, you compute the integral of that and we have the integral of u-1/2du. So I add 1 and if I add 1, I see u1/2 and if I divide by ½, that’s multiplying by 2. So what I see here is that this equals . So what is that? That’s just . But what’s u? You introduced u into the problem. The original question was posed in terms of x’s. So let’s replace that by what it equals. That was the shorthand way of saying 1 – 2x2. And then . And what’s ½ power? Well, that’s . And you can check by taking the derivative of this function, the answer, and seeing that the derivative is indeed this. So, in fact, you can check and see that this really is the anti-derivative of this.
And so this u-substitution, the idea of undoing the chain rule, is a really powerful integration technique that we’ll use again and again. I thought it would be fun just to have us think about it together. I see you at the next lecture.