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24.2 The Average Value of a Function: An Application of the Definite Integral 781 6. The amount of a certain chemical in a mixture varies with time. If g(t) = 5e −t is the number of grams of the chemical at time t, what is the average number of grams of the chemical in the mixture on the time interval [0, 1]? 7. The velocity of an object is given by 3 sin(πt). (a) What is the object’s speed as a function of time? (b) What is the object’s net displacement from t = 0tot=2? (c) How far has the object traveled from t = 0tot=2? (d) What is the object’s average velocity on [0, 2]? (e) What is the object’s average speed on [0, 2]? 8. The graphs of functions f , g, h, and k are given below. Let I denote the average value of f on [0, 4]. Let II denote the average value of g on [0, 4]. Let III denote the average value of h on [0, 4]. Let IV denote the average value of k on [0, 4]. Put I, II, III, and IV, in ascending order, with “<” or “≤” signs between them as appropriate. (2, 5) 04 f (2, 5) 04 g (2, 5) 04 h 2.5 04 k 9. A bicycle speedometer will give the average velocity of a bicyclist over the time period the bicycle is moving. By pressing a button the bicylist can reset the average velocity counter. Suppose a long-distance cyclist has averaged 14 miles per hour for the first two hours of her trip. She resets the average velocity counter. For the next four hours her average velocity is 18 miles per hour. (a) What is the cyclist’s average velocity for the six-hour trip? (b) How far has she traveled? 10. It takes a bicyclist 8 minutes to ride 1 mile uphill and then 2 minutes to ride 1 mile downhill. Explain how we know that the cyclist’s average velocity for the hill is 12 miles per hour. 11. The temperature of a hotplate of radius 5 inches varies with the distance from the center of the plate. For the area within 2 inches of the center the average temperature is 100 degrees. For the area between 2 and 5 inches from the center the average temperature is 80 degrees. What is the average temperature of the plate? 12. Find the average value of | sin(3t)| on [0, 2π]. Explain your reasoning. 782 CHAPTER 24 The Fundamental Theorem of Calculus 13. The graph below shows the birthrate, B(t), and the death rate, D(t), for a population of fish in a lake. t is measured in years, and t = 0 represents 1960. The population of fish in 1960 was 4500. (Assume that births and deaths are the only factors that affect the population—no fishing, no immigration, etc.) 0481216 2428323620 fish/year t D(t) B(t) (a) Write an expression for the total number of births between 1960 and 1996. (b) Write an expression for the average death rate between 1980 and 1990. (c) Write an expression for the fish population in 1996. (d) Approximate the year the population was greatest. (e) Was the 1996 population greater or less than 4500? Explain. PART IX Applications and Computation of the Integral 25 CHAPTER Finding Antiderivatives— An Introduction to Indefinite Integration 25.1 A LIST OF BASIC ANTIDERIVATIVES In order to apply the Fundamental Theorem of Calculus we need to be able to find antideriva- tives. This process can be challenging. Therefore, it is useful to have at our disposal a list of functions we can readily antidifferentiate. We obtain this list by thinking about functions we can readily differentiate and working backward. Definition The symbol f(x)dxstands for the entire family of functions that are antiderivatives of f(x). f(x)dx is called the indefinite integral of f(x).f(x)is the integrand. Recall that if F(x)is an antiderivative of f(x),then F(x)+C is also an antiderivative of f(x)for any constant C. d dx [F(x)+ C]= d dx F(x)+0= f(x) 783 784 CHAPTER 25 Finding Antiderivatives—An Introduction to Indefinite Integration Any two antiderivatives of f differ by an additive constant, so every antiderivative of f can be written in the form F(x)+C. Weknow that d dx sin x = cos x; therefore cos xdx=sin x + C. Equivalently, we can write d dw sin w = cos w; therefore cos wdw=sin w + C. Or d du sin u = cos u; therefore cos udu=sin u + C. The variables x, w, and u are sometimes referred to as “dummy variables,” meaning that the statements we’ve made are equivalent, regardless of the variable we use. cos xdx= sin x + C, cos wdw=sin w + C, and cos udu=sin u + C all say the same thing. Derivatives Corresponding Antiderivatives d dx x n =nx n−1 x n dx = x n+1 n+1 + C for n =−1 d dx ln x = 1 x 1 x dx =ln |x|+C (An explanation of the need for |x| will follow.) d dx e x =e x e x dx =e x +C d dx b x =(ln b) ·b x b x dx = 1 ln b b x + C d dx sin x = cos x cos xdx=sin x +C d dx cos x =−sin x sin xdx=−cos x +C d dx cos x =−sin x,so d dx [−cos x] = sin x d dx tan x = sec 2 x sec 2 xdx=tan x +C d dx arcsin x = 1 √ 1−x 2 1 √ 1−x 2 dx =arcsin x + C d dx arccos x = −1 √ 1−x 2 −1 √ 1−x 2 dx =arccos x +C d dx arctan x = 1 1+x 2 dx 1 1+x 2 dx =arctan x + C Comments: The “+ C,” in each of these indefinite integrals is necessary; without the “+ C” we have given only one antiderivative as opposed to the entire family of antiderivatives. From a graphical perspective, the “+ C” indicates that all antiderivatives of the function are vertical translates of one another. Why do we need the absolute value bars in 1 x dx =ln |x|+C? Weare looking for an antiderivative of 1 x . If x>0and F(x)=ln x, then F (x) = 1 x . Therefore, ln x is an antiderivative of 1 x for x>0. If x<0,then ln x is undefined. However, if we let F(x)=ln(−x), then F (x) = 1 −x · (−1) = 1 x . Therefore, ln(−x) is an antiderivative of 1 x for x<0. 25.1 A List of Basic Antiderivatives 785 ln |x|= ln x for x>0 ln(−x) for x<0 . Therefore, we can use F(x)=ln |x| as an antiderivative of 1 x for all x = 0. EXERCISE 25.1 On the one hand, − 1 √ 1−x 2 dx = arccos x + C; on the other hand, −1 √ 1−x 2 dx = − 1 √ 1−x 2 dx =−arcsin x + C. Therefore, arccos x and − arcsin x must differ by a constant. − arcsin x = arccos x + C Find C. In addition to the list of specific antiderivatives given in the table on page 784, we can deduce principles for integration of the sum of functions and integration of a constant multiple of a function from the corresponding principles for differentiation. General Differentiation Rules Corresponding General Integration Rules d dx [f(x)+g(x)] = d dx f(x)+ d dx g(x) f(x)+g(x) dx = f(x)dx+ g(x) dx d dx kf(x) =k d dx f(x) kf(x)dx=k f(x)dx The antiderivatives listed in the table on page 784 together with the two general integration rules above, enable us to evaluate a wide variety of definite integrals. Learning how to undo the Chain Rule will allow us to evaluate an even broader range of definite and indefinite integrals. We will do this, using a technique called substitution, in this chapter. We can further expand the type of functions we can integrate by using the Product Rule in reverse and arriving at a technique called integration by parts. This will be taken up in Section 29.1. In the problems below we apply general integration rules in combination with the basic antiderivatives displayed in the table on page 784. Our strategy is to manipulate the integrand algebraically so that it can be expressed as a sum. Then the integral can be pulled apart into the sum of simpler integrals. ◆ EXAMPLE 25.1 Integrate the following. (Try these on your own first; then read the solutions.) (a) 1 + x 2 2 dx (b) (x + 7) √ xdx (c) 2x + 3 x 2 dx SOLUTIONS (a) 1 + x 2 2 dx = x 4 +2x 2 +1 dx = x 5 5 + 2x 3 3 + x + C Note that there’s no need to use three separate constants. 786 CHAPTER 25 Finding Antiderivatives—An Introduction to Indefinite Integration (b) (x + 7) √ xdx= x √ xdx+ 7 √ xdx = x 3/2 dx +7 x 1/2 +C = 2 5 x 5/2 + 7 · 2 3 x 3/2 + C = 2 5 x 5/2 + 14 3 x 3/2 + C (c) 2x + 3 x 2 dx = 2x x 2 + 3 x 2 dx = 2 x dx + 3 x 2 dx =2 1 x dx +3 x −2 dx =2ln|x|+3 x −2+1 −2 +1 + C = 2ln|x|− 3 x + C ◆ PROBLEMS FOR SECTION 25.1 In Problems 1 through 11, compute the integral. 1. 3x 3 + 2x + πdx 2. At n dt, where A and n are constants and n =−1. 3. 3x −1 dx 4. dx 2x 5. 3 sin t − 3 1+t 2 dt 6. 5 dx 7x 7. 2 cos w 3 dw 8. 1 x 2 +1 dx 9. e p 2 dp 10. sec 2 t 5 dt 11. cos t + sec t tan tdt 25.2 Substitution: The Chain Rule in Reverse 787 In Problems 12 through 14, find antiderivatives for the given functions. In other words, for each function f , find a function F such that F = f . Check your answers. 12. (a) f(x)=e 3x (b) f(x)= 3 e x 13. (a) f(x)= −1 2x (b) f(x)= 4 1+x 2 14. (a) f(x)=sin 2x (b) f(x)=cos(x/3) 15. (a) Differentiate f(x)=5tan(x 2 ) + arctan 3x. (b) Find 10x sec 2 (x 2 ) + 3 1+9x 2 dx 16. (a) Differentiate y = −π cos 3x 3 . (b) Find A sin Bx dx, where A and B are constants. 17. (a) Suppose the velocity of an object is given by v(t) = 1 1+t 2 miles per hour. Find its net change in position from t = 0tot=1, t measured in hours. What is the total distance traveled on [0, 1]? (b) Suppose that the velocity of an object is given by v(t) =5t(t − 1) meters per second. What is the net change in position between t = 0 and t = 2, t measured in seconds? What is the total distance traveled on [0, 2]? 18. Find the following indefinite integrals. (a) 2+x x dx (b) 3 x 2 dx (c) 3 1+x 2 dx (d) t 3 4 + 4 √ t dt 19. Evaluate the following integrals. (a) (x + π)x 2 dx (b) kx √ x dx (c) 3t 2 +t 6t 3 dt (d) 2 − 1 x √ xdx (e) (x + 1) √ 5xdx 25.2 SUBSTITUTION: THE CHAIN RULE IN REVERSE In Section 25.1, we used our knowledge of the derivatives of basic building block functions (like sin x, x n , and ln x) to create a list of “basic” or “familiar” integrals to keep at our fingertips. We used two general rules of differentiation, the rules for differentiating the sum of functions and the constant multiple of a function, to obtain two general rules of integration. In this section we will use the Chain Rule in reverse to arrive at an integration technique referred to as substitution. This will enable us to greatly expand the type of functions we can antidifferentiate. From functions h(x) and u(x) we can build the composite function h(u(x)) in which the output of u is the input of h. The Chain Rule tells us how to differentiate the composite of differentiable functions: d dx h(u(x)) = h (u(x)) · u (x). The derivative of h(u(x)) is the product of the derivative of h evaluated at u(x) and the derivative of u. 788 CHAPTER 25 Finding Antiderivatives—An Introduction to Indefinite Integration In the examples below we use the Chain Rule to differentiate composite functions and then give the corresponding antiderivative problem. After presenting these problems (all variations on a theme) we will generalize. ◆ EXAMPLE 25.2 Theme and Variations. In parts (a)–(c) we differentiate and then give the corresponding antiderivative problem. We apply the results of part (c) to parts (d)–(f). (a) d dx sin(5x) =cos(5x) ·5 cos(5x) ·5 dx =sin(5x) +C (b) d dx sin(x 2 ) = cos(x 2 ) · 2x cos(x 2 ) · 2xdx=sin(x 2 ) + C (c) d dx sin(u(x)) = cos(u(x)) · u (x) cos(u(x)) · u (x) dx = sin(u(x)) + C or, equivalently, d dx sin(u(x)) = cos(u(x)) · du dx cos(u(x)) · du dx dx =sin(u(x)) + C (d) cos(e x ) · e x dx has the underlying structure cos(u(x)) · u (x) dx. cos(e x ) · e x dx =sin(e x ) +C cos(u(x)) · u (x) dx = sin(u(x)) + C, where u(x) = e x and u (x) = e x (e) cos(x + 5)dxhas the underlying structure cos(u(x)) · u (x) dx. cos(x + 5) · 1 dx =sin(x + 5) + C cos(u(x)) · u (x) dx = sin(u(x)) + C, where u(x) = x + 5 and u (x) = 1 (f) cos( √ x) 2 √ x dx has the underlying structure cos(u(x)) · u (x) dx. cos √ x · 1 2 √ x dx =sin( √ x) + C cos(u(x)) · u (x) dx = sin(u(x)) + C, where u(x) = √ x and u (x) = 1 2 √ x ◆ There is a whole world of integrals accessible to us using the ideas of Example 25.2. Before we generalize, we distinguish between what we are able to do and what we are unable to do in the next example. ◆ EXAMPLE 25.3 Which of the following antiderivatives can be computed by using the Chain Rule in reverse? Compute those we can do in this way. (a) cos 2xdx (b) x cos(x 2 )dx (c) 2 cos(x 2 )dx SOLUTIONS (a) cos(2x) dx Can we put this integral into the form cos(u(x)) · u (x) dx? If so, u(x) = 2x. Therefore, u (x) = 2. We’d like to see u (x) in the integrand; we’re missing a 2. 25.2 Substitution: The Chain Rule in Reverse 789 cos(2x) dx = cos(2x) · 2 · 1 2 dx Multiply the integrand by 2 2 = 1. = 1 2 cos(2x) · 2 dx kf(x)·dx = k f(x)dx = 1 2 cos(u(x)) · u (x) dx, where u(x) = 2x = 1 2 sin(u(x)) + C = 1 2 sin(2x) + C (b) x cos(x 2 )dx Can we put this integral into the form cos(u(x)) · u (x) dx? If so, u(x) = x 2 . Therefore, u (x) = 2x.We’dlike to see u (x) in the integrand; we’ve got the x but are missing a 2. x cos(x 2 )dx= cos(x 2 ) · xdx = 1 2 cos(x 2 ) · 2xdx = 1 2 cos(u(x)) · u (x) dx, where u(x) = x 2 = 1 2 sin(u(x)) + C = 1 2 sin(x 2 ) + C (c) 2 cos(x 2 )dx Can we put this integral into the form cos(u(x)) · u (x) dx? If so, u(x) = x 2 . Therefore, u (x) = 2x.We’dlike to see u (x) in the integrand; we’re missing an x. 2 cos(x 2 )dx= cos(x 2 ) · 2 dx Now we’re stuck. We can multiply by x x (for x = 0), but we can’t pull the 1 x out of the integral. kf(x)dx =k f(x)dx, only if k is constant. A missing constant in u (x) need not worry us, but a missing variable makes all the difference in the world; we can’t find an antiderivative using the Chain Rule in reverse. 1 ◆ Just as the Chain Rule allows us to generalize each shortcut to differentiation we know, e.g., d dx ln x = 1 x generalizing to d dx ln(u(x)) = 1 u(x) · du dx , so too does it allow us to generalize each antiderivative in our table of antiderivatives. 1 In fact, we can’t find an antiderivative at all unless we’re willing to express it as an infinite polynomial. 790 CHAPTER 25 Finding Antiderivatives—An Introduction to Indefinite Integration x n dx = x n+1 n+1 + C generalizes to u(x) n u (x) dx = [ u(x) ] n+1 n+1 + C, n =−1 1 x dx = ln |x|+C generalizes to 1 u(x) · u (x) dx = ln |u(x)|+C e x dx = e x + C generalizes to e u(x) · u (x) dx = e u(x) + C sin xdx=−cos x + C generalizes to sin(u(x)) · u (x) dx =−cos(u(x)) + C cos xdx=sin x + C generalizes to cos(u(x)) · u (x) dx = sin(u(x)) + C 1 1+x 2 dx = arctan x + C generalizes to 1 1+ [ u(x) ] 2 · u (x) dx = arctan(u(x)) + C and so on. This generalized table of integrals incorporates the antidifferentiation analogue of the Chain Rule. This table is unnecessary, however, if we streamline our notation using the method of substitution. Essentially we will replace u(x) by u and u (x) dx by du. The Mechanics of Substitution When faced with an unfamiliar integral, we can use the technique of substitution to attempt to transform it (altering its form but not its substance) into a familiar integral. In other words, some integrals that look intimidating are really sheep in wolves’ clothing; structurally they are familiar integrals, but they are in disguise. Substitution is a long-standing method we use to uncover the underlying structure of a problem. 2 For instance, in Example 25.3(b) we looked at x cos(x 2 )dx.Inthis form it is certainly not one of the integrals with which we are familiar. Substitution allows us to see that essentially this integral is cos udu in disguise. We make the substitution u = x 2 and transform the integral x cos(x 2 )dxto an integral in u. cos(x 2 ) becomes cos u. We must write xdxin terms of u. Since u = x 2 du dx = 2x. We’ve differentiated both sides with respect to x. We can write du = 2xdx. In other words, du dx dx = 2xdxand we equate du dx dx with du. Going from du dx = 2x to du = 2xdx should strike you as a little underhanded (and conceivably illegal). We’re treating du dx as if it were a fraction, but it is not. We have actually not defined du and dx independently. However, this abuse of notation turns out to be all right; we will justify it momentarily. First, let’s finish the example so you appreciate the handiness of it. Writing du = 2xdxallows us to express our original integral entirely in terms of u. 2 Similarly, when working with equations like x 4 − 5x 2 + 4 = 0 the substitution u = x 2 helped expose the underlying quadratic structure of the equation: u 2 − 5u + 4 = 0, so (u − 4)(u − 1) = 0. We solve for u and then return to the original variable. . integration technique referred to as substitution. This will enable us to greatly expand the type of functions we can antidifferentiate. From functions h(x) and u(x) we can build the composite function. f(x) 783 784 CHAPTER 25 Finding Antiderivatives An Introduction to Indefinite Integration Any two antiderivatives of f differ by an additive constant, so every antiderivative of f can be written in the form. allow us to evaluate an even broader range of definite and indefinite integrals. We will do this, using a technique called substitution, in this chapter. We can further expand the type of functions
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