Download or Buy eBook Here. Leave a Comment Cancel reply. Go to mobile version. Haveno fear. Pre-Calculus For Dummies is an un-intimidating,hands-on guide that walks you through all the essential topics,from absolute value and quadratic equations to logarithms andexponential functions to trig identities and matrix operations.
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Calculus Workbook For Dummies serves up the concept review and practice problems with an easy-to-follow, practical approach. Small wonder that much of Calculus II focuses on evaluating indefinite integrals.
In Part II, I give you an ordered approach to evaluating indefinite integrals. In practice, this calculation is easier for some functions than others. Now the bad news: In practice, integration is often a lot trickier than differentiation. 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?
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 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. 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. 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 , I show you how to integrate a variety of common functions by identifying them as the derivatives of functions you already know. For example: Part II: Indefinite Integrals 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 , 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. 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. 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.
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. Use the Sum Rule to break the polynomial into its terms and integrate each of these separately.
Use the Constant Multiple Rule to move the coefficient of each term outside its respective integral. 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. 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. Use trig identities to turn all factors into sines and cosines. Cancel factors wherever possible. If necessary, use trig identities to eliminate all fractions.
So you follow these steps to turn it into an expression you can integrate: 1. 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. Turn all three factors into sines and cosines: 2. Cancel sin x in the numerator and denominator: 3.
I show you lots more tricks for integrating trig functions in Chapter 7. 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. 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. Computing integrals For many input functions, integrals are more difficult to compute than derivatives.
You find this method in Chapter 6, where I discuss integration by parts. In comparison, finding derivatives is comparatively simple — you learned most of what you need to know about it in Calculus I. See Chapter 14 for a more in-depth look at elementary functions. 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. Each year, mathematicians find new ways to integrate classes of functions. 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. 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. However, the same function is integrable for all values of x. 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. This fact makes integrating compositions of functions functions within functions a little bit tricky.
The most useful trick for integrating certain common compositions of functions uses variable substitution. The result is a simplified function that you can integrate using the anti-differentiation formulas and the three basic integration rules Sum Rule, Constant Multiple Rule, and Power Rule — all discussed in Chapter 4.
In this chapter, I show you how to use variable substitution. Then I show you how to identify a few common situations where variable substitution is helpful. After you get comfortable with the process, I give you a quick way to integrate by just looking at the problem and writing down the answer.
Finally, I show you how to skip a step when using variable substitution to evaluate definite integrals. But as functions begin to get a little bit more complex, these methods become insufficient.
The sticking point here is the presence of the constant 2 inside the sine function. You have an anti-differentiation rule for integrating the sine of a variable, but how do you integrate the sine of a variable times a constant? The answer is variable substitution, a five-step process that allows you to integrate where no integral has gone before.
Here are the steps: 1. Declare a variable u and set it equal to an algebraic expression that appears in the integral, and then substitute u for this expression in the integral.
Differentiate u to find of the equal sign. Make another substitution to change dx and all other occurrences of x in the integral to an expression that includes du.
Integrate using u as your new variable of integration. Express this answer in terms of x. Finding the integral of nested functions Suppose that you want to integrate the following: The difficulty here lies in the fact that this function is the composition of two functions: the function 2x nested inside a sine function.
If you were differentiating, you could use the Chain Rule. Unfortunately, no Chain Rule exists for integration. Chapter 5: Making a Fast Switch: Variable Substitution Fortunately, this function is a good candidate for variable substitution.
Follow the five steps I give you in the previous section: 1. As it stands, the symbol dx tells you that the variable of integration is still x. To integrate properly, you need to find a way to change dx to an expression containing du. Substitute du for dx into the integral: You can treat the just like any coefficient and use the Constant Multiple Rule to bring it outside the integral: 4.
At this point, you have an expression that you know how to evaluate: 5. Again, variable substitution comes to the rescue: 1. I answer this question later in the chapter. For now, just follow along and get the mechanics of variable substitution. You can substitute this variable into the expression that you want to integrate as follows: Notice that the expression cos x dx still remains and needs to be expressed in terms of u. Substitute du for cos x dx in the integral: 4.
Now you have an expression that you can integrate: 5. See Chapter 2 for a review of the Chain Rule. Substitute for x dx: You can move the fraction outside the integral: 4. Now you have an integral that you know how to evaluate. I take an extra step, putting the square root in exponential form, to make sure that you see how to do this: 5. Recognizing When to Use Substitution In the previous section, I show you the mechanics of variable substitution — that is, how to perform variable substitution.
In this section, I clarify when to use variable substitution. You may be able to use variable substitution in three common situations.
I round out the discussion by showing you how to deal with functions multiplied by functions and functions multiplied by compositions of functions. Integrating nested functions Compositions of functions — that is, one function nested inside another — are of the form f g x.
Substitute for dx in the integral: 4. Evaluate the integral: 5. Substitute du for dx in the integral: 4. As you get more comfortable with the concept, you can use a shortcut to integrate compositions of functions — that is, nested functions of the form f g x. Write down the reciprocal of the coefficient of x.
Multiply by the integral of the outer function, copying the inner function as you would when using the Chain Rule in differentiation.
Add C. So you can integrate this function quickly as follows: 1. Write down the reciprocal of 4: 2. Multiply this reciprocal by the integral of the outer function, copying the inner function: 3.
See Chapter 2 for more on the ins and outs of trig notation. Again, the criteria for variable substitution are met, so make your way through the steps: 1. Write down the reciprocal of 2. Write down the reciprocal of the coefficient 7: 2. Multiply the integral of the outer function, copying down the inner function: 3. As you look over this chart, get a sense of the pattern so that you can spot it when you have an opportunity to integrate quickly.
In the following sections, I show you how to recognize both of these cases and integrate each. As usual, variable substitution helps to fill the gaps left by the absence of a Product Rule and a Chain Rule for integration.
This is a great opportunity to use variable substitution: 1. Perform another substitution: 4. Watch how this plays out in this substitution: 1. The second part of the substitution now becomes clear: Notice how this substitution hinges on the fact that the numerator is the derivative of the inner function in the denominator.
You may think that this is quite a coincidence, but coincidences like these happen all the time on exams! Integration is now quite straightforward: I take an extra step to remove the fraction before I integrate: 5.
For example: Notice that the derivative of x4 — 1 is x3, off by a constant factor. Using Substitution to Evaluate Definite Integrals In the first two sections of this chapter, I cover how and when to evaluate indefinite integrals with variable substitution.
All this information also applies to evaluating definite integrals, but I also have a time-saving trick that you should know. When using variable substitution to evaluate a definite integral, you can save yourself some trouble at the end of the problem. Specifically, you can leave the solution in terms of u by changing the limits of integration.
This is just a notational change to remind you that the limits of integration are values of x. This fact becomes important later in the problem. You can evaluate this equation simply by using variable substitution.
But because this is a definite integral, you still need to express the limits of integration in terms of u rather than x. I review this in Chapter 2, in case you need a refresher. Unfortunately, no formula allows you to integrate the product of any two functions. As a result, a variety of techniques have been developed to handle products of functions on a case-by-case basis.
In this chapter, I show you the most widely applicable technique for integrating products, called integration by parts.
First, I demonstrate how the formula for integration by parts follows the Product Rule. Then I show you how the formula works in practice. After that, I give you a list of the products of functions that are likely to yield to this method. After you understand the principle behind integration by parts, I give you a method — called the DI-agonal method — for performing this calculation efficiently and without errors.
Then I show you examples of how to use this method to integrate the four most common products of functions. Introducing Integration by Parts Integration by parts is a happy consequence of the Product Rule discussed in Chapter 2. In this section, I show you how to tweak the Product Rule to derive the formula for integration by parts. I show you two versions of this formula — a complicated version and a simpler one — and then recommend that you memorize the second.
I show you how to use this formula, and then I give you a heads up as to when integration by parts is likely to work best. Knowing how to derive the formula for integration by parts is less important than knowing when and how to use it, which I focus on in the rest of this chapter. The first step is simple: Just rearrange the two products on the right side of the equation: Next, rearrange the terms of the equation: Now integrate both sides of this equation: Use the Sum Rule to split the integral on the right in two: The first of the two integrals on the right undoes the derivative: This is the formula for integration by parts.
To integrate by parts: 1. Decompose the entire integral including dx into two factors. Let the factor without dx equal u and the factor with dx equal dv. Differentiate u to find du, and integrate dv to find v. Use the formula 5. Evaluate the right side of this equation to solve the integral. Decompose the integral into ln x and x dx. Using these values for u, du, v, and dv, you can use the formula for integration by parts to rewrite the integral as follows: At this point, algebra is useful to simplify the right side of the equation: Part II: Indefinite Integrals 5.
For example, x cos x2 is a job for variable substitution, not integration by parts. To see why, flip to Chapter 5. When you decide to use integration by parts, your next question is how to split up the function and assign the variables u and dv. Fortunately, a helpful mnemonic exists to make this decision: Lovely Integrals Are Terrific, which stands for Logarithmic, Inverse trig, Algebraic, Trig.
If you prefer, you can also use the mnemonic Lousy Integrals Are Terrible. Always choose the first function in this list as the factor to set equal to u, and then set the rest of the product including dx equal to dv.
You can use integration by parts to integrate any of the functions listed in Table Integrating by Parts with the DI-agonal Method The DI-agonal method is basically integration by parts with a chart that helps you organize information. This method is especially useful when you need to integrate by parts more than once to solve a problem. In this section, I show you how to use the DI-agonal method to evaluate a variety of integrals.
Looking at the DI-agonal chart The DI-agonal method avoids using u and dv, which are easily confused especially if you write the letters u and v as sloppily as I do! Instead, a column for differentiation is used in place of u, and a column for integration replaces dv. You may also notice that the D and the I are placed diagonally in the chart — yes, the name DI-agonal method works on two levels so to speak. Using the DI-agonal method Earlier in this chapter, I provide a list of functions that you can integrate by parts.
The DI-agonal method works for all these functions. I also give you Chapter 6: Integration by Parts the mnemonic Lovely Integrals Are Terrific which stands for Logarithmic, Inverse trig, Algebraic, Trig to help you remember how to assign values of u and dv — that is, what to differentiate and what to integrate.
To use the DI-agonal method: 1. Write the value to differentiate in the box below the D and the value to integrate omitting the dx in the box below the I. Differentiate down the D column and integrate down the I column. Add the products of all full rows as terms. I explain this step in further detail in the examples that follow.
Add the integral of the product of the two lowest diagonally adjacent boxes. I also explain this step in greater detail in the examples. I show you how to use the DI-agonal method to integrate products that include logarithmic, inverse trig, algebraic, and trig functions.
L is for logarithm You can use the DI-agonal method to evaluate the product of a log function and an algebraic function. For example, suppose that you want to evaluate the following integral: Whenever you integrate a product that includes a log function, the log function always goes in the D column.
Write the log function in the box below the D and the rest of the function value omitting the dx in the box below the I. Differentiate ln x and place the answer in the D column. Notice that in this step, the minus sign already in the box attaches to. Integrate x2 and place the answer in the I column.
Add the integral of the two lowest diagonally adjacent boxes that are circled. Whenever you integrate a product that includes an inverse trig function, this function always goes in the D column. For example, suppose that you want to integrate the following: Part II: Indefinite Integrals 1. Write the inverse trig function in the box below the D and the rest of the function value omitting the dx in the box below the I.
Differentiate arccos x and place the answer in the D column, and then integrate 1 and place the answer in the I column. Substituting 1 — x2 for u and simplifying gives you this answer: Therefore,. For example, suppose that you want to integrate the following: This example is a product of functions, so integration by parts is an option.
But it does include the algebraic factor x3, so place this factor in the D column and the rest in the I column. If evaluating this expression looks like fun and if you think you can do it quickly on an exam without dropping a minus sign along the way!
If not, I show you a better way. Read on. To integrate an algebraic function multiplied by a sine, a cosine, or an exponential function, place the algebraic factor in the D column and the other factor in the I column.
Differentiate the algebraic factor down to zero, and then integrate the other factor the same number of times. You can then copy the answer directly from the chart. In the D column, continue alternating plus and minus signs and differentiate until you reach 0. And in the I column, continue integrating. The very pleasant surprise is that you can now copy the answer from the chart. Actually, no — but , which explains where that final C comes from.
But the chart keeps track of everything. For example, suppose that you want to evaluate the following integral: When integrating either a sine or cosine function multiplied by an exponential function, make your DI-agonal chart with five rows rather than four.
Then place the trig function in the D column and the exponential in the I column. Oddly enough, however, this similarity makes solving the integral possible. It allows you to integrate functions that contain radicals of polynomials such as and other similar difficult functions. Trig substitution may remind you of variable substitution, which I discuss in Chapter 5. With both types of substitution, you break the function that you want to integrate into pieces and express each piece in terms of a new variable.
With trig substitution, however, you express these pieces as trig functions. So before you can do trig substitution, you need to be able to integrate a wider variety of products and powers of trig functions. The first few parts of this chapter give you the skills that you need.
After that, I show you how to use trig substitution to express complicated-looking radical functions in terms of trig functions. Integrating the Six Trig Functions You already know how to integrate sin x and cos x from Chapter 4, but for completeness, here are the integrals of all six trig functions: Part II: Indefinite Integrals Please commit these to memory — you need them!
For practice, you can also try differentiating each result to show why each of these integrals is correct. Integrating Powers of Sines and Cosines Later in this chapter, when I show you trig substitution, you need to know how to integrate powers of sines and cosines in a variety of combinations. In this section, I show you what you need to know. Odd powers of sines and cosines You can integrate any function of the form sinm x cosn x when m is odd, for any real value of n.
Peel off a sin x and place it next to the dx: 2. You can expand the function out, turning it into a polynomial. Similarly, you integrate any function of the form sinm x cosn x when n is odd, for any real value of m. These steps are practically the same as those in the previous example.
Peel off a cos x and place it next to the dx: 2. Use the half-angle identity for cosine to rewrite the integral in terms of cos 2x: 2. Use the Constant Multiple Rule to move the denominator outside the integral: 3. Distribute the function and use the Sum Rule to split it into several integrals: 4.
Use the two half-angle identities to rewrite the integral in terms of cos 2x: 2. Engineering students would find this calculus learning book very helpful to get a complete understanding of this subject.
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