2 Integration by Substitution

Transcription

1 86 Chapter 5 Integration 2 Integration by Substitution (a) Find a function P(x) that satisfies these conditions. Use the graphing utility of your calculator to graph this function. (b) Use trace and zoom to determine the level of population 9 months from now. When will the population be 7,590? (c) Suppose the current population were 2,000 (instead of 5,000). Sketch the graph of P(x) with this assumption. Then sketch the graph of P(x) assuming current populations of 4,000 and 6,000. What is the difference between the graphs? 49. A car traveling at 67 ft/sec decelerates at the constant rate of 2 ft/sec 2 when the brakes are applied. (a) Find the velocity v(t) of the car t seconds after the brakes are applied. Then find its distance s(t) from the point where the brakes are applied. (b) Use the graphing utility of your calculator to sketch the graphs of v(t) and s(t) on the same screen (use [0, 5]1 by [0, 200]10). (c) Use trace and zoom to determine when the car comes to a complete stop and how far it travels in that time. How fast is the car traveling when it has traveled 45 feet? Recall (from Section 2.5) that according to the chain rule, the derivative of the function (x 2 x 5) 9 is [(x 2 x 5) 9 ] 9(x 2 x 5) 8 (2x ) Notice that the derivative is a proct and that one of its factors, 2x, is the derivative of an expression, x 2 x 5, which occurs in the other factor. More precisely, the derivative is a proct of the form g(u) where, in this case, g(u) 9u 8 and u x 2 x 5. You can integrate many procts of the form g(u) by applying the chain rule in reverse. Specifically, if G is an antiderivative of g, then since, by the chain rule, To summarize: d g(u) G(u) C d [G(u)] G(u) g(u)

2 Chapter 5 Section 2 Integration by Substitution 87 Integral Form of the Chain Rule If g is a continuous function of u and u(x) is a differentiable function of x, then g(u) g(u) That is, to integrate a proct of the form g(u) in which one of the factors is the derivative of an expression u that appears in the other factor: 1. Find the antiderivative g(u) of the factor g(u) with respect to u. 2. Replace u in the answer by its expression in terms of x. Find Here are two examples. EXAMPLE 2.1 9(x 2 x 5) 8 (2x ). The integrand 9(x 2 x 5) 8 (2x ) is a proct in which one of the factors 2x is the derivative of an expression x 2 x 5 that appears in the other factor. In particular, 9(x 2 x 5) 8 (2x ) g(u) where g(u) 9u 8 and u x 2 x 5 Hence, by the integral form of the chain rule, 9(x 2 x 5) 8 (2x ) g(u) 9u 8 u 9 C (x 2 x 5) 9 C The proct to be integrated in the next example is not exactly of the form g(u). However, it is a constant multiple of such a proct, and you can integrate it by combining the constant multiple rule with the integral form of the chain rule.

3 88 Chapter 5 Integration Refer to Example 2.2. Store 1 x e 2 4 as (1 4)e^(X^ ) in Y1 and e u as 4 (1 4)e^U in Y2 in your graphing utility. Store.42 as X and (.42) 2 4 as U. Evaluate Y1 and Y2 at X and U, respectively. Explain why the results are the same. EXAMPLE 2.2 x 4 2 Find x e. First use the constant multiple rule to rewrite the integral as x e x (4x e x4 2 ) 1 4 4x e x4 2 so that the new integrand 4x e x4 2 is a proct in which one of the factors 4x is the derivative of an expression x 4 2 that appears in the other factor. In particular, 4x e x4 2 g(u) where g(u) e u and u x 4 2 Hence, by the integral form of the chain rule, x e x x e x4 2 1 g(u) 4 1 4e u 1 4 eu C 1 4 ex4 2 C CHANGE OF VARIABLES The integral form of the chain rule may be thought of as a technique for simplifying an integral by changing the variable of integration. In particular, you start with an integral g(u) in which the variable of integration is x and transform it into the simpler integral g(u) in which the variable of integration is u. In this transformation, the expression in the original integral is replaced in the simplified integral by the symbol. You can remember this relationship between by pretending that is a quotient and writing = and

4 Chapter 5 Section 2 Integration by Substitution 89 These observations suggest the following general integration technique, called integration by substitution, in which the variable u is formally substituted for an appropriate expression in x and the original integral is transformed into a simpler one in which the variable of integration is u. Integration by Substitution Step 1. Introce the letter u to stand for some expression in x that is chosen with the simplification of the integral as the goal. Step 2. Rewrite the integral in terms of u. To rewrite, compute and solve algebraically as if the symbol were a quotient. Step. Evaluate the resulting integral and then replace u by its expression in terms of x in the answer. Note If the integrand is a proct or quotient of two terms and one term is a constant multiple of the derivative of an expression that appears in the other, then this expression is probably a good choice for u. The method of integration by substitution is illustrated in the next example using the integral from Example 2.1. EXAMPLE 2. Find 9(x 2 x 5) 8 (2x ). The integrand is a proct in which one of the factors 2x is the derivative of an expression x 2 x 5 that appears in the other factor. This suggests that you let u x 2 x 5. Then 2x and so (2x ) Substituting u x 2 x 5 and (2x ), you get 9(x 2 x 5) 8 (2x ) 9u 8 u 9 C (x 2 x 5) 9 C

5 90 Chapter 5 Integration Here are some additional examples illustrating the method of integration by substitution. EXAMPLE 2.4 Find x. x 2 1 Observe that Thus the integrand is a quotient in which one term x is a constant multiple of the derivative of an expression x 2 1 that appears in the other factor. This suggests that you let u x 2 1. Then, 2x Substituting u x 2 1 and d (x2 1) 2x 2 (x) 2x or x, you get 2 x x u u ln u C 2 x 2 2 ln x2 1 C Find EXAMPLE 2.5 x 6 2x 2 8x. Observe that d (2x2 8x ) 4x 8 4(x 2) 4 (x 6)

6 Chapter 5 Section 2 Integration by Substitution 91 Thus, the integrand is a quotient in which one term x 6 is a constant multiple of the derivative of an expression 2x 2 8x that appears in the other factor. This suggests that you let u 2x 2 8x. Then, or 4x 8 Substituting u 2x 2 8x and (4x 8) 4(x 2) 4 (x 6) x 6 2x 2 8x 4 (x 6) 4 4 (x 6), you get 1 u 4 u 1/2 4 (2u1/2 ) C 2 2x2 8x C Find EXAMPLE 2.6 (ln x) 2. x Observe that Thus, the integrand 1 is a proct in which one factor is the derivative of an expression ln x that appears x in the other factor. This suggests that you let u = ln x. Then, 1 x (ln x) 2 x d (ln x) 1 x (ln x) 2 1 x or 1 x

7 92 Chapter 5 Integration 1 Substituting u ln x and, you get x (ln x) 2 u 2 1 x u C 1 (ln x) C The next example is designed to show you the versatility of the formal method of substitution. It deals with an integral that does not seem to be of the form g(u) but that nevertheless can be simplified significantly by a clever change of variables. Find EXAMPLE 2.7 x x 1. There is no easy way to integrate this quotient as it stands. But watch what happens if you make the substitution u x 1. Then and x u 1 and so x x 1 u u u u ln u C x 1 ln x 1 C A COMPARISON OF THE TWO TECHNIQUES WHEN SUBSTITUTION FAILS The integral form of the chain rule, which was used in Examples 2.1 and 2.2, is attractive because it involves nothing more than a familiar rule for differentiation applied in reverse. With practice in using this method you should be able to find integrals like those in Examples 2.1 through 2.6 by inspection, without writing down any intermediate steps. Nevertheless, many people prefer to use the method of formal substitution. They like the fact that it involves straightforward manipulation of symbols, and they appreciate the convenience of the notation. This method of substitution is also somewhat more versatile, as you saw in Example 2.7. For many of the integrals you will encounter, both methods work well and you should feel free to use the one with which you are more comfortable. The method of substitution does not always succeed. In the next example, we consider an integral very like the integral in Example 2.2 but just enough different so no substitution will work.

8 Chapter 5 Section 2 Integration by Substitution 9 EXAMPLE 2.8 Evaluate x 4 e x42. The natural substitution is u x 4 2, as in Example 2.2. As before, you find 4x, so x 1. However, the integrand involves x 4, not x, and the extra 4 factor of x satisfies x u 4 2, so when the substitution is made, you have x 4 e x42 xe x42 (x ) u 4 2 e u which is hardly an improvement on the original integral! Try a few other possible substitutions (say, u x 2 or u x ) to convince yourself that nothing works. AN APPLICATION INVOLVING SUBSTITUTION EXAMPLE 2.9 The price p (dollars) of each unit of a particular commodity is estimated to be changing at the rate dp 15x 9 x 2 where x (hundred) units is the consumer demand (the number of units purchased at that price). Suppose 400 units (x 4) are demanded when the price is $0 per unit. (a) Find the demand function p(x). (b) At what price will 00 units be demanded? At what price will no units be demanded? (c) How many units are demanded at a price of $20 per unit? (a) The price per unit demanded p(x) is found by the integrating p(x) with respect to x. To perform this integration, use the substitution u 9 x 2 1, 2x, x 2 to get

9 94 Chapter 5 Integration Since p 0 when x 4, you find that so p(x) 15x 9 x 15 2 u 1 2 1/ u 1/ u1/2 1/2 C 159 x 2 C C C p(x) 159 x (b) When 00 units are demanded, x and the corresponding price is p() $12.24 per unit No units are demanded when x 0 and the corresponding price is p(0) $00 per unit (c) To determine the number of units demanded at a unit price of $20 per unit, you need to solve the equation 159 x x x By squaring both sides of this equation and simplifying, you find that x That is, roughly 409 units will be demanded when the price is $20 per unit. P. R. O. B. L. E. M. S 5.2 P. R. O. B. L. E. M. S 5.2 In Problems 1 through 26, find the indicated integral and check your answer by differentiation. 1. (2x 6) 5 2. e 5x. 4x x 5

10 Chapter 5 Section 2 Integration by Substitution e 1x 6. [(x 1) 5 (x 1) 2 5] 7. xe x2 8. 2xe x t(t 2 1) 5 dt 10. tt 2 8 dt 11. x 2 (x 1) / y4 14. y 5 1 dy y2 (y 5) 2 dy 15. (x 1)(x 2 2x 5) 1/2 16. x5 (x 2 1)e xx x 5x x4 12x 6 x 4 x 2 6 6u 19. 5x4 10x 12 u 20. (u 2 2u 6) 2 4u 2 4u 1 ln 5x x x ln x ln x x(ln x) 2 x 2x ln (x ) ex x 2 1 x 27. Find the function whose tangent has slope xx 2 5 for each value of x and whose graph passes through the point (2, 10). 2x 28. Find the function whose tangent has slope for each value of x and whose 1 x 2 graph passes through the point (0, 5). In Problems 29 through 2, the velocity v(t) x(t) at time t of an object moving along the x axis is given, along with the initial position x(0) of the object. In each case, find: (a) The position x(t) at time t. (b) The position of the object at time t 4. (c) The time when the object is at x. 29. x(t) 2(t 1) 1/2 ; x(0) 4 x 5 e 1x6

11 96 Chapter 5 Integration TREE GROWTH DEPRECIATION POPULATION GROWTH LAND VALUE 0. x(t) (5 2t) ; x(0) 5 t 1. x(t) ; x(0) 0 (t 1) 2 2t 2. x(t) ; x(0) 4 (1 t 2 ) /2 1. A tree has been transplanted and after x years is growing at the rate of 1 (x 1) 2 meters per year. After 2 years it has reached a height of 5 meters. How tall was it when it was transplanted? 4. The resale value of a certain instrial machine decreases at a rate that changes with time. When the machine is t years old, the rate at which its value is changing is 960e t/5 dollars per year. If the machine was bought new for $5,000, how much will it be worth 10 years later? 5. It is projected that t years from now the population of a certain country will be changing at the rate of e 0.02t million per year. If the current population is 50 million, what will be the population 10 years from now? 6. It is estimated that x years from now, the value V(x) of an acre of farmland will be increasing at the rate of V(x) 0.4x 0.2x 4 8,000 dollars per year. The land is currently worth $500 per acre. (a) Find V(x). Use the graphing utility of your calculator to sketch the graph of V(x). (b) Use trace and zoom to determine how much the land will be worth in 10 years. When will the land be worth $1,000 per acre? AIR POLLUTION 7. In a certain suburb of Los Angeles, the level of ozone L(t) at 7:00 A.M. is 0.25 parts per million (ppm). A 12-hour weather forecast predicts that the ozone level t hours later will be changing at the rate of t L(t) 6 16t t 2 parts per million per hour (ppm/hr). (a) Express the ozone level L(t) as a function of t. When does the peak ozone level occur? What is the peak level? (b) Use the graphing utility of your calculator to sketch the graph of L(t) and use trace and zoom to answer the questions in part (a). Then determine at what other time the ozone level will be the same as it is at 11:00 A.M.

12 Chapter 5 Section 2 Integration by Substitution 97 PRODUCTION DEPRECIATION MARGINAL COST RETAIL PRICES DEMAND SUPPLY 8. Bejax Corporation has set up a proction line to manufacture a new type of cellular telephone. The rate of proction of the telephones is dp units/month dt 1,500 2 t 2t 5 How many telephones are proced ring the third month? [That is, find P() P(2).] 9. The resale value of a certain instrial machine decreases at a rate that depends on its age. When the machine is t years old, the rate at which its value is changing is 960e t/5 dollars per year. (a) Express the value of the machine in terms of its age and initial value. (b) If the machine was originally worth $5,200, how much will it be worth when it is 10 years old? 40. At a certain factory, the marginal cost is (q 4) 2 dollars per unit when the level of output is q units. (a) Express the total proction cost in terms of the overhead (the cost of procing 0 units) and the number of units proced. (b) What is the cost of procing 14 units if the overhead is $46? 41. In a certain section of the country, the price of chicken is currently $ per kilogram. It is estimated that x weeks from now the price will be increasing at the rate of x 1 cents per week. How much will chicken cost 8 weeks from now? 42. The price p (dollars) of each pair of Yike sports sneakers is estimated to be changing at the rate p(x) where x (hundred) units is the consumer demand. Suppose 500 pairs of sneakers (x 5) are demanded when the price is $75 per pair. (a) Find the demand function p(x). (b) At what price will 400 pairs of sneakers be demanded? At what price will no pairs be demanded? (c) How many pairs will be demanded at a price of $90 per pair? (d) Find the revenue function R(x) xp(x) and the marginal revenue R(x). For what quantity demanded x is revenue maximized? 4. The owner of the Dog Gone hot dog restaurant chain estimates that the dollar price of her newest proct, Weenie Babies, is changing at the rate p(x) 150x (144 x 2 ) /2 0x ( x) 2

13 98 Chapter 5 Integration when x (thousand) Weenies are supplied for purchase. The current price is $2.25 per Weenie. (a) Find the supply function p(x). (b) At what price will 4,000 additional Weenies (x 4) be supplied? (c) How many more Weenies will be supplied at a price of $ per Weenie? MARGINAL PROFIT 44. A company determines that the marginal revenue from the proction of x units is R(x) 7 x 4x 2 hundred dollars per unit, and the corresponding marginal cost is C(x) 5 2x hundred dollars per unit. By how much does the profit change when the level of proction is raised from 5 to 9 units? 11 x MARGINAL PROFIT 45. Repeat Problem 44 for marginal revenue R(x) and for the marginal cost C(x) 2 x x x Introction to Differential Equations A differential equation is an equation that involves a derivative or differential. For example, dy x2 5 or dp dt kp are all differential equations. Differential equations are among the most useful tools for modeling continuous phenomena, including important situations that occur in business and economics and the social and life sciences. In this section, we introce techniques for solving basic differential equations and examine a few practical applications. and The simplest type of differential equation has the form dy g(x) dy 2 dy 2y ex in which the derivative of the quantity y is given explicitly as a function of the independent variable x. Such an equation can be solved by simply finding the indefinite integral of g(x). A complete characterization of all possible solutions of the equation is called a general solution, and a solution that satisfies specified side conditions is called a particular solution. This terminology is illustrated in the following examples. EXAMPLE.1 Find the general solution of the differential equation dy x2 x and the particular solution that satisfies y 2 when x 1.