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Calculus: Integrating Polynomials by Substitution


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About this Lesson

  • Type: Video Tutorial
  • Length: 15:24
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 166 MB
  • Posted: 11/18/2008

This lesson is part of the following series:

Calculus (279 lessons, $198.00)
Calculus Review (48 lessons, $95.04)
Calculus: Final Exam Test Prep and Review (45 lessons, $64.35)
Calculus: Basics of Integration (14 lessons, $23.76)
Calculus: Integration by Substitution (2 lessons, $3.96)

This lesson will teach you how to do integration of polynomials using substitution. Before digging in on substitution in antidifferentiation, Professor Burger will review notation associated with differentiation and antidifferentation (with respect to x). Next, he will move on to teach you integration by substitution, a technique that is helpful for finding the antiderivative of a composite function. While running through the chain rule backwards, he will highlight several rules and properties of antidifferentiation; for instance, the integral of a product is not necessarily equal to the product of the integrals. He also gives us advice on what expressions to select and replace with a constant when using substitution as a method of integration. To illustrate all of this, you will find the integral of 42x(x^2+4)^20 dx as well as the integral of this expression that contains radicals 2x^2*(5+x^3)^(1/2)dx.

Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hôpital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

About this Author

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The Basics of Integration
Integrating Polynomials by Substitution Page [1 of 3]
Let’s see if we can actually find a method for determining the anti-derivatives, or the integrals, of elaborate functions where we’re trying to untangle the chain rule. So really what I want to think about now is some sort of a fancy method of integration. That’s really what we’re up against here, sort of a fancy, souped-up method of finding anti-derivatives or integrations.
Okay, before we start, I want to sort of re-cap two basic facts with you, which, of course, you already know but will be pivotal sort of here. The first one is what exactly that d,dx means. Remember, that’s sort of a verb, that’s an action, and it means – it’s a command – differentiate with respect to x. And the x here is very important. It tells us what we’re differentiating with respect to…what that variable is. Remember in the related rate problems we did a while back, we actually were differentiating with respect to time; and a lot of times in those max and min problems, we were differentiating with respect to other variables. So differentiating with respect to x.
Now the new notation that we’ve been developing here is sort of this pair of symbols, where there’s a function inside here, and that’s a commandment that says, “Integrate with respect to x.” The same exact idea. x, it’s telling us, is the variable that we have to integrate with respect to, and then we trudge along and integrate.
So both of these ideas I think are reasonably clear. I point them out only because we’re going to really concentrate on this one, especially the notion of integrating with respect to x, being told to us by that dx here. So I want you just to remember that because when I talk about it and it’s like two seconds, I want you to remember, “Oh, yeah, that’s what that means. Integrate with respect to x…x is the variable.”
Okay, armed with that, let’s take a look at how we can run the chain rule backwards. That really is sort of the fancy integration we’re thinking about here is how do you run the chain rule backwards? So to illustrate this actually, I would like to take a look at an example. So let’s look at the example together and see if we can figure out a means for untangling the chain rule. So let’s now try to evaluate this integral: The integral of 42x multiplied by (x2 + 4)20dx. So that’s a real hard integral to figure out, and I just can’t do it directly. Remember, by the way – oh, here’s a great thing, and I hope this is clear – the integral of a product is not the product of the integrals, just as the derivative of a product is not the product of the derivatives. We need to use the product rule. Well here it’s the same kind of thing. You just can’t take the integral of each of these terms and multiply it together. That is wrong.
Instead, let’s take a look at this and notice something interesting. Let’s notice that what makes this problem so darn difficult and so scary is all of this inside stuff here. If that were just one thing, that wouldn’t be too hard of a problem. So why don’t we do the following, why don’t we think about this sort of running the chain rule backwards and think about that whole inside thing as a blop. Notice there is something very pleasant about that choice of blops, and that is the derivative of that blop, which is just 2x, is sort of sitting out here in front. Now I admit, this is not 2x, but it’s just off by 2x by a multiple constant. If I multiply this thing, I guess, by 21, I have exactly 2x. So the key thing to notice here – and this is really important – that the derivative of this piece is actually sitting out here as a multiple.
So given that as a potential way of untangling the chain rule, let’s actually blop this whole thing together, and we usually think of this as blop, but right now let me be a little bit more formal and let me call it u. So why don’t we let u – this is just going to give me the blop here and I’m going to try to put it all together – equal that entire orange inside.
Okay, now how could I re-write this thing? Well, I could re-write this thing and make it a little easier as follows. I can now say so now we have…well, what would it look like? Integral 42x. Now in place of the orange, I’m putting in a u, so I see u20. Notice how nice and tidy that makes things. And then I have a dx. So, good news. The good news is that now it’s a much simpler object because I don’t have that inside anymore. Bad news. The bad news is now I’ve got a new variable in there, u, and remember what we just talked about was that this symbol means integrate with respect to x. So I can’t have an integral with a lot of different variables here. I just can’t integrate this and say like because I’m supposed to integrate with respect to x. So the good news is we’ve made it simpler; the bad news is we’ve made it harder. A problem.
So what we have to do in order to do this problem is convert everything in terms of u. Now you might say, “Okay, I’ll make that a u and that a u,” but that’s not right because u equals this. So you just can’t let everything be u because then you’d be changing the problem. u is equal to that quantity, so I can’t put a u in there. And here I can’t put a u in because this is sort of a change in x, and I don’t know how u is changing. So here’s a way to resolve this potential problem. Not hard. What I’m going to do here is come back to this relationship that tells me what u is, and I’m going to now take the derivative, so I take the derivative of u with respect to x. Now watch me. – not a big deal.
Now, if you remember from early on in our discussions about this symbol, , I told you it’s literally just the symbol which represents the derivative of u with respect to x. It’s not really a fraction, it’s just a symbol. But in some examples, for example in the chain rule discussion, we saw that if we fantasize that it’s a fraction, it makes the chain rule a little easier to sort of feel out.
Now what I’d like to do is actually fantasize once again – this is fantasy math, now – that this really is a fraction. If it really is a fraction, then I can multiply both sides by the dx, and the dx would cancel on the bottom there. So let’s have a little fantasy – a little math fantasy – let’s multiply both sides by dx. I then notice that, in fact, these things cancel, and what am I left with? Well, I’m left with just du on the left and 2x(dx) on the right. Well, that’s pretty cool because, you see, look, I want to put in a du in here, but I see exactly that du actually can be written in terms of dx. In fact, if you notice I could divide through by this two; and if I divide through by this two, I would see then a ½(du), and that would equal x(dx). And notice that x(dx) is the very thing we have here. So in that funny-shaped bubble I actually see x multiplied by dx, but now I see what that equals in terms of use. It equals just ½(du). So in place of that bubble, I can replace this bubble because they’re the same thing. Well when I make that substitution, I get that this integral equals integral y(42) out in front, the bubble, which is going to be a ½(du), and then I have this term, u20. I can simplify this a little bit by cancelling, and I see the integral 21u20du. Well, that integral is actually really, really easy. The variable is u, but since all of the u’s are here and I have a du, that means I can just integrate this like I would normally take the derivative of if I had used it. I could just integrate this since everything is u. And if I integrate this, what do I do? I add one to the exponent and divide by that number. So I add one to get 21, and then I divide by 21. So therefore I see , so this would just be a coefficient of one, and this would equal u21 plus a constant.
Let’s check the answer. Take the derivative of this with respect to u. The derivative of u21 is 21u20. That’s correct. Well, that sounds like our answer, but it’s not because we, or in particular you, brought u into the problem. The original problem had no u’s in it. The original problem was posed in terms of x’s. So how do we get this back into x’s? Well, it’s very simple. We know what u equals – u equals all of that stuff. So whenever I see a u, I just plug in (x2 + 4). And when I do that, what do I see? Well, I’m going to do that right over here. What is get is I get (x2 + 4) 21 plus a constant, and that, I claim, is the answer to this.
Let’s check it really fast and see. Remember, with integrals, it’s always easy to know if your answer is correct. All you have to do is carefully differentiate and see if you get the original function. The answer is always given to you. The answer is always given to you. Let’s check it really fast. So let’s check. So I take the derivative of this, blop21 – 21blop21. I peeled off the 21, take the derivative of that and I see 2 times x, and notice that 21 times 2 is 42. I’ve got an x, and I’ve got a (x2 + 4) 20. The derivative of a constant is zero. That’s the derivative, and that’s exactly what the original question was.
So now we see how to untangle a problem that may have actually arose from the chain rule. The idea is to isolate the blop – to isolate the inside ¬– and then replace that complicated piece and notice what remains is in a different variable – now it’s in terms of x’s – so to get rid of that, I differentiate this thing here. And if I take the derivative of that and then fantasize that I bring the dx over to this side, which actually mathematically we can do for this kind of problem, then I can actually take this thing. Notice that that’s sort of what’s left over. Substitute it in, and when I substitute, I get now an easy problem. I took a hard problem in x, I converted it to an easy problem in u by substitution, I did that problem and then took the u and brought it back in in terms of x by plugging back in. Not surprisingly, this technique is called the integration by substitution. So I substitute the inside.
All right, let’s take a look at one more example before I close this section right here. We’re going to do a lot of practice on these so you get a sense of how these go. How about this one: integral of 3x2 dx? Well, that problem looks really, really hard. There’s a square root, there’s stuff here, it looks really awful. But notice there’s an inside here. And further note that if we just look at that inside and we take the derivative of it, what do we see? I see 3x2, and that’s exactly what’s on the outside. So this looks like it’s been a derivative that was taken by using the chain rule. So let’s make a substitution. Let’s let u equal the inside. If I make that substitution, what happens when I plug that in for here? Well, I can plug that in, you see, right into here, and I see the integral becomes now integral 3x2 – not of the complicated stuff now, but just square root of u – dx. And now the problem is that I’ve got all of this stuff with x’s in it and stuff with u’s in it. That is not good. So what do I do? Well, I come back to this formula, and I take the derivative of it with respect to x, and I see that equals…that’s zero, and this is 3x2. When I cross-multiply by the dx, it comes over here, and I see du = 3x2dx. And notice that is exactly this bubble right here. So that bubble is 3x(dx). So I can replace it by the much simpler value, du. And when I insert that in, look what I get. When I insert it, I get the integral – well, all of that bubble just becomes du – and I’m just left with . And what’s ? Well, that equals u½. And how do you integrate that? I add one to the exponent and get 3/2, and then I divide by 3/2, which when you think about it, if you visualize the 3/2 in the denominator, when I invert multiply, it becomes (2/3 + c), and again, you can check that by taking the derivative. Bring the 3/2 in front, they kill each other off with the 2/3, and I have (u3/2 - 1), which is . So that’s the answer, but that’s not the answer to the original question, which was posed in terms of x’s. You brought u into the problem, so you have to take it out. But you know what u equals, so you just plug that in for u. And if we do that, what do we see? Well, what we see is that the answer would be 2/3 times u, which is (5 + x3)3/2 + c, and you can actually check that that is the anti-derivative of this. How? By taking the derivative of this and getting that. Notice if you take the derivative of this, you have to use the chain rule – not surprising because this method of substitution is untangling the chain rule. If you take the derivative of this, you bring the 3/2 in front, that will kill off the 2/3, and I’m left with blop½ – well, that’s the square root of all of the blops – then you have to multiply this by the derivative of the inside, which is 3x2, and that’s exactly what’s sitting here.
So you can really see that this is untangling the chain rule, this method of substitution. Okay, we’ll take a look at a whole bunch of examples to really see first of all how this works in practice, this business of letting u equal that and taking the derivative, and more importantly – more interestingly – the notion of what do you pick for your choice of u? We’ll have a little u marathon in a bit. I’ll see you at our next example. Bye.

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