Calculus II For Dummies, 2nd Edition (2012)

Part II. Indefinite Integrals

In this part . . .

You begin calculating the indefinite integral as an anti-derivative — that is, as the inverse of a derivative. In practice, this calculation is easier for some functions than others. So I show you four important tricks — variable substitution, integration by parts, trig substitution, and integrating with partial fractions — for turning a function you don’t know how to integrate into one that you do.

Chapter 4. Instant Integration: Just Add Water (And C)

In This Chapter

Calculating simple integrals as anti-derivatives

Using 17 integral formulas and 3 integration rules

Integrating more difficult functions using more than one integration tool

First the good news: Because integration is the inverse of differentiation, you already know how to evaluate a lot of basic integrals.

Now the bad news: In practice, integration is often a lot trickier than differentiation. I’m telling you this upfront because a) it’s true; b) I believe in honesty; and c) you should prepare yourself before your first exam. (Buying and reading this book, by the way, are great first steps!)

In this chapter — and also in Chapters 5 through 8 — I focus exclusively on one question: How do you integrate every single function on the planet? Okay, I’m exaggerating, but not by much. I give you a manageable set of integration techniques that you can do with a pencil and paper, and if you know when and how to apply them, you’ll be able to integrate everything but the kitchen sink.

First, I show you how to start integrating by thinking about integration as anti-differentiation — that is, as the inverse of differentiation. I give you a not-too-long list of basic integrals, which mirrors the list of basic derivatives from Chapter 2. I also give you a few rules for breaking down functions into manageable chunks that are easier to integrate.

After that, I show you a few techniques for tweaking functions to make them look like the functions you already know how to integrate. By the end of this chapter, you’ll have the tools to integrate dozens of functions quickly and easily.

Evaluating Basic Integrals

In Calculus I (which I cover in Chapter 2), you find that a few algorithms — such as the Product Rule, Quotient Rule, and Chain Rule — give you the tools to differentiate just about every function your professor could possibly throw at you. In Calculus II, students often greet the news that “there’s no Chain Rule for integration” with celebratory cheers. By the middle of the semester, they usually revise this opinion.

Using the 17 basic anti-derivatives for integrating

In Chapter 2, I give you a list of 17 derivatives to know, cherish, and above all memorize (yes, I said memorize). Reading that list may lead you to believe that I’m one of those harsh über-math dudes who takes pleasure in cruel and unusual curricular activities.

But math is kind of like the Ghost of Christmas Past — the stuff you thought was long ago dead and buried comes back to haunt you. And so it is with derivatives. If you already know them, you’ll find this section easy.

The Fundamental Theorem of Calculus shows that integration is the inverse of differentiation up to a constant C. This key theorem gives you a way to begin integrating. In Table 4-1, I show you how to integrate a variety of common functions by identifying them as the derivatives of functions you already know.

As I discuss in Chapter 3, you need to add the constant of integration C because constants differentiate to 0. For example:

So when you integrate with anti-differentiation, you need to account for the potential presence of this constant:

Three important integration rules

After you know how to integrate using the 17 basic anti-derivatives in Table 4-1, you can expand your repertoire with three additional integration rules: the Sum Rule, the Constant Multiple Rule, and the Power Rule. These three rules mirror those that you know from differentiation.

The Sum Rule for integration

The Sum Rule for integration tells you that integrating long expressions term by term is okay. Here it is formally:

For example:

Note that the Sum Rule also applies to expressions of more than two terms. It also applies regardless of whether the term is positive or negative. (Some books call this variation the Difference Rule, but you get the idea.) Splitting this integral into three parts allows you to integrate each separately by using a different anti-differentiation rule:

Notice that I add only one C at the end. Technically speaking, you should add one variable of integration (say, C₁, C₂, and C₃) for each integral that you evaluate. But, at the end, you can still declare the variable C = C₁ + C₂ + C₃to consolidate all these variables. In most cases when you use the Sum Rule, you can skip this step and just tack a C onto the end of the answer.

The Constant Multiple Rule for integration

The Constant Multiple Rule tells you that you can move a constant outside of a derivative before you integrate. Here it is expressed in symbols:

For example:

As you can see, this rule mirrors the Constant Multiple Rule for differentiation. With the constant out of the way, integrating is now easy using an anti-differentiation rule:

= 3 sec x + C

The Power Rule for integration

The Power Rule for integration allows you to integrate any real power of x (except –1). Here’s the Power Rule expressed formally:

For example:

The Power Rule works fine for negative powers of x, which are powers of x in the denominator. For example:

= –x^–1 + C

The Power Rule also works for rational powers of x, which are roots of x. For example:

The only real-number power that the Power Rule doesn’t work for is –1. Fortunately, you have an anti-differentiation rule to handle this case:

= ln |x| + C

What happened to the other rules?

Integration contains formulas that mirror the Sum Rule, the Constant Multiple Rule, and the Power Rule for differentiation. But it lacks formulas that look like the Product Rule, Quotient Rule, and Chain Rule. This fact may sound like good news, but the lack of formulas makes integration a lot trickier in practice than differentiation is.

In fact, Chapters 5 through 8 focus on a bunch of methods that mathematicians have devised for getting around this difficulty. Chapter 5 focuses on variable substitution, which is a limited form of the Chain Rule. And in Chapter 6, I show you integration by parts, which is an adaptation of the Product Rule.

Evaluating More Difficult Integrals

The anti-differentiation rules for integrating, which I explain earlier in this chapter, greatly limit how many integrals you can compute easily. In many cases, however, you can tweak a function to make it easier to integrate.

In this section, I show you how to integrate certain fractions and roots using the Power Rule. I also show you how to use the trig identities in Chapter 2 to stretch your capacity to integrate trig functions.

Integrating polynomials

You can integrate any polynomial in three steps using the rules from this section:

1. Use the Sum Rule to break the polynomial into its terms and integrate each of these separately.

2. Use the Constant Multiple Rule to move the coefficient of each term outside its respective integral.

3. Use the Power Rule to evaluate each integral. (You only need to add a single C to the end of the resulting expression.)

For example, suppose that you want to evaluate the following integral:

1. Break the expression into four separate integrals:

2. Move each of the four coefficients outside its respective integral:

3. Integrate each term separately using the Power Rule:

You can integrate any polynomial using this method. Many integration methods I introduce later in this book rely on this fact. So practice integrating polynomials until you feel so comfortable that you could do it in your sleep.

Integrating rational expressions

In many cases, you can untangle hairy rational expressions and integrate them using the anti-differentiation rules plus the other three rules in this chapter.

For example, here’s an integral that looks like it may be difficult:

You can split the function into several fractions, but without the Product Rule or Quotient Rule, you’re then stuck. Instead, expand the numerator and put the denominator in exponential form:

Next, split the expression into five terms:

Then use the Sum Rule to separate the integral into five separate integrals and the Constant Multiple Rule to move the coefficient outside the integral in each case:

Now you can integrate each term separately using the Power Rule:

Using identities to integrate trig functions

At first glance, some products or quotients of trig functions may seem impossible to integrate using the formulas I give you earlier in this chapter. But you’ll be surprised how much headway you can often make when you integrate an unfamiliar trig function by first tweaking it using the Basic Five trig identities that I list in Chapter 2.

The unseen power of these identities lies in the fact that they allow you to express any combination of trig functions into a combination of sines and cosines. Generally speaking, the trick is to simplify an unfamiliar trig function and turn it into something that you know how to integrate.

When you’re faced with an unfamiliar product or quotient of trig functions, follow these steps:

1. Use trig identities to turn all factors into sines and cosines.

2. Cancel factors wherever possible.

3. If necessary, use trig identities to eliminate all fractions.

For example:

In its current form, you can’t integrate this expression by using the rules from this chapter. So you follow these steps to turn it into an expression you can integrate:

1. Use the identities cot x = and sec x = :

2. Cancel both sin x and cos x in the numerator and denominator:

In this example, even without Step 3, you have a function that you can integrate.

= –cos x + C

Here’s another example:

Again, this integral looks like a dead end before you apply the Basic Five trig identities to it:

1. Turn all three factors into sines and cosines:

2. Cancel sin x in the numerator and denominator:

3. Use the identity cos x = to eliminate the fraction:

= tan x + C

Again, you turn an unfamiliar function into one of the ten trig functions that you know how to integrate. I show you lots more tricks for integrating trig functions in Chapter 7.

Understanding Integrability

By now, you’ve probably figured out that, in practice, integration is usually harder than differentiation. The lack of any set rules for integrating products, quotients, and compositions of functions makes integration something of an art rather than a science.

So you may think that a large number of functions are differentiable, with a smaller subset of these being integrable. It turns out that this conclusion is false. In fact, the set of integrable functions is larger, with a smaller subset of these being differentiable. To understand this fact, you need to be clear on what the words integrable and differentiable really mean.

In this section, I shine some light on two common mistakes that students make when trying to understand what integrability is all about. After that, I discuss what it means for a function to be integrable, and I show you why many functions that are integrable aren’t differentiable.

Taking a look at two red herrings of integrability

In trying to understand what makes a function integrable, you first need to understand two related issues: difficulties in computing integrals and representing integrals as functions. These issues are valid, but they’re red herrings — that is, they don’t really affect whether a function is integrable.

Computing integrals

For many input functions, integrals are more difficult to compute than derivatives. For example, suppose that you want to differentiate and integrate the following function:

y = 3x⁵e^2x

You can differentiate this function easily by using the Product Rule (I take an additional step to simplify the answer):

= 3(5x⁴e^2x + 2e^2xx⁵)

= 3x⁴e^2x(2x + 5)

Because no such rule exists for integration, in this example you’re forced to seek another method. (You find this method in Chapter 6, where I discuss integration by parts.)

Finding solutions to integrals can be tricky business. In comparison, finding derivatives is comparatively simple — you learned most of what you need to know about it in Calculus I.

Representing integrals as functions

Beyond difficulties in computation, the integrals of certain functions simply can’t be represented by using the functions that you’re used to.

More precisely, some integrals can’t be represented as elementary functions — that is, as combinations of the functions you know from Pre-Calculus. (See Chapter 14 for a more in-depth look at elementary functions.)

For example, consider the following function:

You can find the derivative of the function easily using the Chain Rule:

However, the integral of the same function, , can’t be expressed as a function — at least, not any function that you’re used to.

Instead, you can express this integral either exactly — as an infinite series — or approximately — as a function that approximates the integral to a given level of precision. (See Part IV for more on infinite series.) Alternatively, you can just leave it as an integral, which also expresses it just fine for some purposes.

Getting an idea of what integrable really means

When mathematicians discuss whether a function is integrable, they aren’t talking about the difficulty of computing that integral — or even whether a method has been discovered. Each year, mathematicians find new ways to integrate classes of functions. However, this fact doesn’t mean that previously nonintegrable functions are now integrable.

Similarly, a function’s integrability also doesn’t hinge on whether its integral can be easily represented as another function, without resorting to infinite series.

In fact, when mathematicians say that a function is integrable, they mean only that the integral is well defined — that is, that the integral makes mathematical sense.

In practical terms, integrability hinges on continuity: If a function is continuous on a given interval, it’s integrable on that interval. Additionally, if a function has only a finite number of discontinuities on an interval, it’s also integrable on that interval.

You probably remember from Calculus I that many functions — such as those with discontinuities, sharp turns, and vertical slopes — are nondifferentiable. Discontinuous functions are also nonintegrable. However, functions with sharp turns and vertical slopes are integrable.

For example, the function y = |x| contains a sharp point at x = 0, so the function is nondifferentiable at this point. However, the same function is integrable for all values of x. This is just one of infinitely many examples of a function that’s integrable but not differentiable in the entire set of real numbers.

So, surprisingly, the set of differentiable functions is actually a subset of the set of integrable functions. In practice, however, computing the integral of most functions is more difficult than computing the derivative.