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Unformatted text preview: Calculus II This document was written and copyrighted by Paul Dawkins. Use of this document and its online version is governed by the Terms and Conditions of Use located at . The online version of this document is available at . At the above web site you will find not only the online version of this document but also pdf versions of each section, chapter and complete set of notes. Preface
Here are my online notes for my Calculus II course that I teach here at Lamar University. Despite the fact that these are my "class notes" they should be accessible to anyone wanting to learn Calculus II or needing a refresher in some of the topics from the class. These notes do assume that the reader has a good working knowledge of Calculus I topics including limits, derivatives and integration by substitution. Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed 1. Because I wanted to make this a fairly complete set of notes for anyone wanting to learn calculus I have included some material that I do not usually have time to cover in class and because this changes from semester to semester it is not noted here. You will need to find one of your fellow class mates to see if there is something in these notes that wasn't covered in class. 2. In general I try to work problems in class that are different from my notes. However, with Calculus II many of the problems are difficult to make up on the spur of the moment and so in this class my class work will follow these notes fairly close as far as worked problems go. With that being said I often don't have time in class to work all of these problems and so you will find that some sections contain problems that weren't worked in class due to time restrictions. 3. Sometimes questions in class will lead down paths that are not covered here. I try to anticipate as many of the questions as possible in writing these up, but the reality is that I can't anticipate all the questions. Sometimes a very good question gets asked in class that leads to insights that I've not included here. You should always talk to someone who was in class on the day you missed and compare these notes to their notes and see what the differences are. 4. This is somewhat related to the previous three items, but is important enough to merit its own item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!! Using these notes as a substitute for class is liable to get you in trouble. As already noted not everything in these notes is covered in class and often material or insights not in these notes is covered in class. 2005 Paul Dawkins 1 Calculus II Integration Techniques
In this chapter we are going to be looking at integration techniques. There are a fair number of them, some easier than others. The point of the chapter is to teach you these new techniques and so this chapter assumes that you've got a fairly good working knowledge of basic substitutions with integrals. In fact, most integrals involving "simple" substitutions will not have any of the substitution work shown. It is going to be assumed that you can verify the substitution portion of the integration yourself. Also, most of the integrals done in this chapter will be indefinite integrals. It is also assumed that once you can do the indefinite integrals you can also do the definite integrals and so to conserve space we concentrate mostly on indefinite integrals. There is one exception to this and that is the Trig Substitution section and in this case there are some subtleties involved with definite integrals that we're going to have to watch out for. Here is a list of topics that are covered in this chapter. Integration by Parts Of all the integration techniques covered in this chapter this is probably the one that students are most likely to run into down the road in other classes. Integrals Involving Trig Functions In this section we look at integrating certain products and quotients of trig functions. Trig Substitutions Here we will look using substitutions involving trig functions an how they can be used to simplify certain integrals. Partial Fractions We will use partial fractions to allow us to do integrals involving rational functions. Integrals Involving Roots We will take a look at a substitution that can, on occasion, be used with integrals involving roots. Integrals Involving Quadratics In this section we are going to look at integrals that involve quadratics. Using Integral Tables Here we look at using Integral Tables as well as relating new integrals back to integrals that we already know how to do. Integration Strategy We give a general set of guidelines for determining how to evaluate an integral. Improper Integrals We will look at integrals with infinite intervals of integration and integrals with discontinuous integrands in this section. 2005 Paul Dawkins 2 Calculus II Comparison Test for Improper Integrals Here we will use the Comparison Test to determine if improper integrals converge or diverge. Approximating Definite Integrals There are many ways to approximate the value of a definite integral. We will look at three of them in this section. Integration by Parts
Let's start off with this section with a couple of integrals that we should already be able to do to get us started. First let's take a look at the following. x x e dx = e + c So, that was simple enough. Now, let's take a look at, xe
u = x2 x2 dx To do this integral we'll use the following substitution. du = 2 x dx
x2 x dx = 1 du 2 1 u 1 u 1 x2 e du = 2 e + c = 2 e + c 2 Again, simple enough to do provided you remember how to do substitutions. By the way make sure that you can do these kinds of substitutions quickly and easily. From this point on we are going to be doing these kinds of substitutions in our head. If you have to stop and write these out with every problem you will find that it will take to significantly longer to do these problems. xe dx = Now, let's look at the integral that we really want to do. 6x xe dx If we just had an x by itself or e6 x by itself we could do the integral easily enough. But, we don't have them by themselves, they are instead multiplied together. There is no substitution that we can use on this integral that will allow us to do the integral. So, at this point we don't have the knowledge to do this integral. To do this integral we will need to use integration by parts so let's derive the integration by parts formula. We'll start with the product rule. ( f g ) = f g + f g Now, integrate both sides of this. ( f g ) dx = f g + f g dx
The left side is easy enough to integrate and we'll split up the right side of the integral. fg = f g dx + f g dx 2005 Paul Dawkins 3 Calculus II Note that technically we should have had a constant of integration show up on the left side after doing the integration. We can drop it at this point since other constants of integration will be showing up down the road and they would just end up absorbing this one. Finally, rewrite the formula as follows and we arrive that the integration by parts formula. f g dx = fg - f g dx This is not the easiest formula to use however. So, let's do a couple of substitutions. u = f ( x) v = g ( x)
du = f ( x ) dx dv = g ( x ) dx Both of these are just the standard Calc I substitutions that hopefully you are used to by now. Don't get excited by the fact that we are using two substitutions here. They will work the same way. Using these substitutions gives us the formula that most people think of as the integration by parts formula. u dv = uv - v du To use this formula we will need to identify u and dv, compute du and v and then use the formula. Note as well that computing v is very easy. All we need to do is integrate dv. v = dv So, let's take a look at the integral we wrote down above.
Example 1 Evaluate the following integral. 6x xe dx Solution So, on some level, the problem here is the x that is in front of the exponential. If that wasn't there we could do the integral. Notice as well that anything that we choose for u will be differentiated and so that seems like choosing u=x will be a good choice since upon differentiating the x will drop out. Once u is chosen we know that dv will be everything else that remains. So, here are the choices for u and dv as well as du and v. u=x dv = e6 x dx
du = dx
1 v = e6 x dx = e6 x 6 The integral is then, 2005 Paul Dawkins 4 Calculus II x 1 dx = e6 x - e6 x dx 6 6 x 1 = e6 x - e6 x + c 6 36 Once we have done the last integral in the problem we will add in the constant of integration to get our final answer. xe 6x Next let's a look at integration by parts for definite integrals. In this case the formula is, b b a u dv = uv a - v du
b a b Note that the uv a in the first term is just the standard integral evaluation notation that you should be familiar with at this point. All we do is evaluate at b then subtract off the evaluation at a.
Example 2 Evaluate the following integral. 2 -1 xe6 x dx Solution This is the same integral that we looked at in the first example so we'll use the same u and dv to get, 2 2 x 6x 1 2 6x xe dx = e - e6 x dx -1 6 6 -1 -1 x 1 = e6 x - e6 x 6 36 -1 11 7 = e12 + e-6 36 36 2 2 -1 Since we need to be able to do the indefinite integral in order to do the definite integral and doing the definite integral amounts to nothing more than evaluating the indefinite integral at a couple of points we will concentrate on doing indefinite integrals in the rest of this section. In fact, through out most of this chapter this will be the case. We will be doing far more indefinite integrals than definite integrals. Let's take a look at some more examples.
Example 3 Evaluate the following integral. t ( 3t + 5 ) cos dt 4 Solution There are two ways to proceed with this example. For many, the first thing that they try is multiplying the cosine through the parenthesis, splitting up the integral and then doing integration by parts on the first integral. 2005 Paul Dawkins 5 Calculus II While that is a perfectly acceptable way of doing the problem it's more work than we really need to do. Instead of splitting the integral up let's instead use the following choices for u and dv. t u = 3t + 5 dv = cos dt 4 t du = 3 dt v = 4sin 4 The integral is then, t t t ( 3t + 5 ) cos dt = 4 ( 3t + 5 ) sin - 12 sin dt 4 4 4 t t = 4 ( 3t + 5 ) sin + 48cos + c 4 4 Notice that we pulled any constants out of the integral when we used the integration by parts formula. We will usually do this in order to simplify the integral a little.
Example 4 Evaluate the following integral. 2 w sin (10w) dw Solution For this example we'll use the following choices for u and dv. u = w2 dv = sin (10w ) dw du = 2w dw The integral is then,
2 w sin (10w) dw = - v=- 1 cos (10 w ) 10 w2 1 cos (10 w ) + w cos (10 w ) dw 10 5 In this example, unlike the previous examples, the new integral will also require integration by parts. For this second integral we will use the following choices. u=w dv = cos (10w ) dw du = dw So, the integral becomes,
2 w sin (10w) dw = - v= 1 sin (10w ) 10 w2 1 w 1 cos (10w ) + sin (10 w ) - sin (10 w ) dw 10 5 10 10 w2 1 w 1 cos (10w ) + sin (10 w ) + cos (10w ) + c 10 5 10 100 w2 w 1 cos (10w ) + sin (10w ) + cos (10w ) + c 10 50 500 =- =- 2005 Paul Dawkins 6 Calculus II Be careful with the coefficient on the integral for the second application of integration by 1 parts. Since the integral is multiplied by we need to make sure that the results of 5 1 actually doing the integral is also multiplied by . Forgetting to do this is one of the 5 more common mistakes with integration by parts problems. As this last example has shown us, we will sometimes need more than one application of integration by parts to complete a problem. This is something that will happen so don't get excited about it when it happens. In this next example we need to acknowledge an important point about integration techniques. Some integrals can be done in using several different techniques. That is the case with the integral in the next example.
Example 5 Evaluate the following integral x x + 1 dx (a) Using Integration by Parts. (b) Using a standard Calculus I substitution.
Solution (a) First notice that there are no trig functions or exponentials in this integral. While a good many integration by parts integrals will involve trig functions and/or exponentials not all of them will so don't get too locked into the idea of expecting them to show up. In this case we'll use the following choices for u and dv. u=x dv = x + 1 dx du = dx The integral is then,
3 3 2 2 x ( x + 1) 2 - ( x + 1) 2 dx 3 3 3 5 2 4 = x ( x + 1) 2 - ( x + 1) 2 + c 3 15 (b) Now let's do the integral with a substitution. We can use the following substitution. u = x +1 x = u -1 du = dx v= 3 2 x + 1) 2 ( 3 x x + 1 dx = Notice that we'll actually use the substitution twice, once for the quantity under the square root and once for the x in front of the square root. The integral is then, 2005 Paul Dawkins 7 Calculus II x x + 1 dx = ( u - 1) u du = u 2 - u 2 du 2 5 2 3 = u2 - u2 + c 5 3 5 3 2 2 = ( x + 1) 2 - ( x + 1) 2 + c 5 3
3 1 So, we used two different integration techniques in this example and we got two different answers. The obvious question then should be : Did we do something wrong? Actually, we didn't do anything wrong. We need to remember the following fact from Calculus I.
If f ( x ) = g ( x ) then f ( x ) = g ( x ) + c In other words, if two functions have the same derivative then they will differ by no more than a constant. So, how does this apply to the above problem? First define the following, f ( x) = g ( x) = x x +1 Then we can compute f(x) and g(x) by integrating as follows, f ( x ) = f ( x ) dx g ( x ) = g ( x ) dx We'll use integration by parts for the first integral and the substitution for the second integral. Then according to the fact f(x) and g(x) should differ by no more than a constant. Let's verify this and see if this is the case. We can verify that they differ my no more than a constant if we take a look at the difference of the two.
3 5 5 3 4 2 2 2 x ( x + 1) 2 - ( x + 1) 2 - ( x + 1) 2 - ( x + 1) 2 15 3 3 5 3 4 2 2 2 = ( x + 1) 2 x - ( x + 1) - ( x + 1) + 15 5 3 3 = ( x + 1) 2 ( 0 ) =0 So, in this case it turns out the two functions are exactly the same function since the difference is zero. Note that this won't always happen. Sometimes the difference will yield a nonzero constant. For an example of this check out the Constant of Integration section in my Calculus I notes. 3 2005 Paul Dawkins 8 Calculus II So just what have we learned? First, there will, on occasion, be more than one method for evaluating an integral. Secondly, we saw that different methods will often lead to different answers. Last, even though the answers are different it can be shown that they differ by no more than a constant. When we are faced with an integral the first thing that we'll need to decide is if there is more than one way to do the integral. If there is more than one way we'll then need to determine which method we should use. The general rule of thumb that I use in my classes is that you should use the method that you find easiest. This may not be the method that others find easiest, but that doesn't make it the wrong method. One of the more common mistakes with integration by parts is for people to get too locked into perceived patterns. For instance, all of the previous examples used the basic pattern of taking u to be the polynomial that sat in front of another function and then letting dv be the other function. This will not always happen so we need to be careful and not get locked into any patterns that we think we see. Let's take a look at some integrals that don't fit into the above pattern.
Example 6 Evaluate the following integral. ln x dx Solution So, unlike any of the other integral we've done to this point there is only a single function in the integral and no polynomial sitting in front of the logarithm. The first choice of many people here is to try and fit this into the pattern from above and make the following choices for u and dv. u =1 dv = ln x dx This leads to a real problem however since that means v must be, v = ln x dx In other words, we would need to know the answer ahead of time in order to actually do the problem. So, this choice simply won't work. Therefore, if the logarithm doesn't belong in the dv it must belong instead in the u. So, let's use the following choices instead u = ln x dv = dx 1 du = dx v=x x The integral is then, 2005 Paul Dawkins 9 Calculus II ln x dx = x ln x - x x dx = x ln x - dx = x ln x - x + c 1 Example 7 Evaluate the following integral. x 5 x 3 + 1 dx Solution So, if we again try to use the pattern from the first few examples for this integral our choices for u and dv would probably be the following. u = x5 dv = x 3 + 1 dx However, as with the previous example this won't work since we can't easily compute v. v = x 3 + 1 dx This is not an easy integral to do. However, notice that if we had an x2 in the integral along with the root we could very easily do the integral with a substitution. Also notice that we do have a lot of x's floating around in the original integral. So instead of putting all the x's (outside of the root) in the u let's split them up as follows.
u = x3 du = 3 x 2 dx dv = x 2 x3 + 1 dx v=
3 2 3 x + 1) 2 ( 9 The integral is then,
5 3 x x + 1 dx = 3 3 2 3 3 2 x ( x + 1) 2 - x 2 ( x3 + 1) 2 dx 9 3 3 5 2 4 = x 3 ( x 3 + 1) 2 - ( x 3 + 1) 2 + c 9 45 So, in the previous two examples we saw cases that didn't quite fit into any perceived pattern that we might have gotten from the first couple of examples. This is always something that we need to be on the lookout for with integration by parts. Let's take a look at another example that also illustrates another integration technique that sometimes arises out of integration by parts problems.
Example 8 Evaluate the following integral. e cos d Solution Okay, to this point we've always picked u in such a way that upon differentiating it would make that portion go away or at the very least put it the integral into a form that would make it easier to deal with. In this case no matter which part we make u it will never go away in the differentiation process. 2005 Paul Dawkins 10 Calculus II It doesn't much matter which we choose to be u so we'll choose in the following way. Note however that we could choose the other way as well and we'll get the same result. u = cos dv = e d
du = - sin d v = e The integral is then, e cos d = e cos + e sin d So, it looks like we'll do integration by parts again. Here are our choices this time. u = sin dv = e d
du = cos d v = e The integral is now, e cos d = e cos + e sin - e cos d Now, at this point it looks like we're just running in circles. However, notice that we now have the same integral on both sides and one the right side its got a minus sign in front of it. This means that we can add the integral to both sides to get, 2 e cos d = e cos + e sin All we need to do now is divide by 2 and we're done. The integral is, 1 e cos d = 2 ( e cos + e sin ) + c Notice that after dividing by the two we add in the constant of integration at that point. This idea of using integration by parts until you get the same integral on both sides of the equal sign and then simply solving for the integral is kind of nice to remember. It doesn't show up all that often, but when it does it may be the only way to actually do the integral. We've got one more example to do. As we will see some problems could require us to do integration by parts numerous times and there is a short hand method that will allow us to do multiple applications of integration by parts quickly and easily.
Example 9 Evaluate the following integral. x e x 4 2 dx Solution We start off by choosing u and dv as we always would. However, instead of computing du and v we put these into the following table. We then diffe...
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