Integration Techniques and Differential Equations. Differential Equations

Transcription

1 30434_ch06_p qd /7/03 8:5 AM Page 434 Integration Techniques and Differential Equations 6. Integration by Parts 6.2 Integration Using Tables 6.3 Improper Integrals 6.4 Numerical Integration 6.5 Differential Equations 6.6 Further Applications of Differential Equations: Three Models of Growth

2 30434_ch06_p qd /6/03 8:08 AM Page 435 Application Preview Improper Integrals and Eternal Recognition Suppose that after you become rich and famous, you decide to commission a statue of yourself for your hometown. Your town, however, will accept this selfless gesture only if you pay for the perpetual upkeep of the statue by establishing a fund that will generate $2000 annually for every year in the future. Before deciding whether or not to accept this condition, you of course want to know how much it will cost. (Interestingly, we will see that a fund that will generate income forever does not require an infinite amount of money.) On page 279 we found that to realize a yield of $2000 t years from now requires only its present value deposited now in a bank. At an interest rate of, say, 5% compounded continuously, $2000 in t years requires a deposit now of value of $2000 at 5% 2000e Present compounded continuously 0.05t Amount times e 0.05t Therefore, the size of the fund needed to generate $2000 annually forever is found by summing (integrating) this present value over the infinite time interval from zero to infinity ( ). Size of fund to yield an annual $2000 forever 2000e 0.05t dt 0 Integrating out to infinity Present value y 2000 y 2000e 0.05t Area e 0.05t dt Year t On page 466 we will learn how to evaluate such improper integrals, finding that the value of this integral is $40,000, which shows that $40,000 deposited in a bank at 5% compounded continuously will generate $2000 every year forever. That is, $40,000 would pay for the perpetual upkeep of your statue, buying you (or at least your likeness) a kind of immortality. Incidentally, to generate $2000 annually for only the first hundred 00 years would require e 0.05t dt $39,730 (integrating from 0 to 00). This amount is only $270 less than the amount needed to generate the same sum forever. This small additional cost shows that the short term is epensive, but eternity is cheap.

3 30434_ch06_p qd /6/03 8:08 AM Page CHAPTER 6 INTEGRATION TECHNIQUES AND DIFFERENTIAL EQUATIONS 6. INTEGRATION BY PARTS Introduction In this chapter we introduce further techniques for finding integrals: integration by parts, integration by tables, and numerical integration. We also discuss improper integrals (integrals over infinite intervals) and differential equations. You may have felt that integration is harder than differentiation. One reason is that integration is an inverse process, carried out by reversing differentiation, and inverse processes are generally more difficult. Another reason is that, while we have the product and quotient rules for differentiating complicated epressions, there are no product and quotient rules for integration. The method of integration by parts, eplained in this section, is in some sense a product rule for integration in that it comes from interpreting the product rule as an integration formula. Integration by Parts For two differentiable functions u() and v(), hereafter denoted simply u and v, the Product Rule is (uv) uv uv The derivative of a product is the derivative of the first times the second, plus the first times the derivative of the second If we integrate both sides of this equation, integrating the left side undoes the differentiation. uv uv d uv d du dv uv v du u dv Using differential notation, du u d, dv v d Formula above in differential notation Solving this equation for the second integral u dv gives u dv uv v du This formula is the basis for a technique called integration by parts. Integration by Parts For differentiable functions u and v, u dv uv v du

4 30434_ch06_p qd /6/03 8:08 AM Page INTEGRATION BY PARTS 437 We use this formula to solve integrals by a double substitution, substituting u for part of the given integral and dv for the rest, and then epressing the integral in the form uv v du. The point is to choose the u and the dv so that the resulting integral v du is simpler than the original integral u dv. A few eamples will make the method clear. EXAMPLE INTEGRATING BY PARTS Use integration by parts to find e d. Solution e d u dv u dv e d du d d v e d e e d u v v du e e e C From e d Original integral We choose u and dv e d Differentiating u to find du and integrating dv to find v (omitting the C) (the arrows show which part leads to which other part) Replacing the original integral udv by the righthand side of the formula, uv vdu, with u, v e, and du d Finding the new integral e d e to give the final answer (with C) The procedure is not as complicated as it might seem. All the steps may be written together as follows: e d e e d e e C u dv uv v du u dv e d du d v e d e From e d

5 30434_ch06_p qd /6/03 8:08 AM Page CHAPTER 6 INTEGRATION TECHNIQUES AND DIFFERENTIAL EQUATIONS We can check this answer by differentiation. Cancel d d (e e C) e e e e Agrees with the original integrand, so the integration is correct Differentiating e by the Product Rule Remarks on the Integration by Parts Procedure i. The differentials du and dv include the d. ii. We omit the constant C when we integrate dv to get v because one C at the end is enough. iii. The integration by parts formula does not give a final answer, but rather epresses the given integral as uv v du, a product uv (already integrated), and a new integral v du. That is, integration by parts echanges the original integral u dv for another integral v du. The hope is that the second integral will be simpler than the first. In our eample we echanged e d for the simpler e d, which could be integrated immediately by formula 3 (inside back cover). iv. Integration by parts should be used only if the simpler formulas through 7 (inside back cover) fail to solve the integral. How to Choose the u and the dv In Eample, the choice of u echanged the original integral e d for the simpler integral e d. If we had instead cho- dv e d u e sen we would have echanged the original integral for dv d 2 e d (as you may check), which is more difficult than the original (because of the 2 ). Therefore, the first choice was the right choice in that it led to a solution. Generally, one choice for u and dv will be best, and finding it may involve some trial and error. While there is no foolproof rule for finding the best u and dv, the following guidelines often help. Guidelines for Choosing u and dv. Choose dv to be the most complicated part of the integral that can be integrated easily. 2. Choose u so that u is simpler than u.

6 30434_ch06_p qd /6/03 8:08 AM Page INTEGRATION BY PARTS 439 EXAMPLE 2 INTEGRATING BY PARTS Find 2 ln d. Solution None of the easier formulas ( through 7 on the inside back cover) will solve the integral, as you may easily check. Therefore, we try integration by parts. The integrand is a product, 2 times ln. The guidelines say to choose dv to be the most complicated part that can be easily integrated. We can integrate 2 but not ln (we know how to differentiate ln, but not how to integrate it), so we choose dv 2 d, and therefore u ln. 3 3 d 3 3 ln 3 2 d 2 ln d (ln ) 3 3 u ln du d u u v v du dv dv 2 d v 2 d 3 3 Moving the outside and simplifying 3 to ln C 3 3 ln 9 3 C 3 We may check this answer by differentiation. d d 3 3 ln 9 3 C 2 ln Differentiating 3 ln by the Product Rule Agrees with the original 2 ln integrand, so the ln integration is correct 3 From 3 Cancel Practice Problem Use integration by parts to find 3 ln d. Solution on page 444

7 30434_ch06_p qd /6/03 8:08 AM Page CHAPTER 6 INTEGRATION TECHNIQUES AND DIFFERENTIAL EQUATIONS Integration by parts is also useful because it simplifies integrating products of powers of linear functions. EXAMPLE 3 INTEGRATING BY PARTS Use integration by parts to find ( 2)( 4) 8 d. Solution The guidelines recommend that dv be the most complicated part that can be integrated. Both 2 and ( 4) 8 can be integrated. For eample, By the substitution method with ( 4) 8 d ( 4) 9 C 9 u 4 (omitting the details) Since ( 4) 8 is more complicated than ( 2), we take dv ( 4) 8 d. ( 2)( 4) 8 d ( 2) 9 ( 4)9 9 ( 4)9 d u du d dv u 2 dv ( 4) 8 d v ( 4) 8 d 9 ( 4)9 u v v du 9 ( 2)( 4)9 9 ( 4) 9 d Taking out the 9 9 ( 2)( 4)9 9 0 ( 4)0 C Integrating by the substitution method 9 ( 2)( 4)9 90 ( 4)0 C Again we could check this answer by differentiation. (Do you see how the Product Rule, applied to the first part of this answer, will give a piece that will cancel with the derivative of the second part?)

8 30434_ch06_p qd /6/03 8:08 AM Page INTEGRATION BY PARTS 44 Practice Problem 2 Use integration by parts to find ( )( ) 3 d. Solution on page 445 Present Value of a Continuous Stream of Income If a business or some other asset generates income continuously at the rate C(t) dollars per year, where t is the number of years from now, then C(t) is called a continuous stream of income.* On page 279 we saw that to find the present value of a sum (the amount now that will later yield the stated sum) under continuous compounding we multiply by e rt, where r is the interest rate and t is the number of years. (We will refer to a rate with continuous compounding as a continuous interest rate. ) Therefore, the present value of the continuous stream C(t) is found by multiplying by e rt and summing (integrating) over the time period. Present Value of a Continuous Stream of Income The present value of the continuous stream of income C(t) dollars per year, where t is the number of years from now, for T years at continuous interest rate r is Present T value 0 C(t)e rt dt EXAMPLE 4 FINDING THE PRESENT VALUE OF A CONTINUOUS STREAM OF INCOME A business generates income at the rate of 2t million dollars per year, where t is the number of years from now. Find the present value of this continuous stream for the net 5 years at the continuous interest rate of 0%. * C(t) must be continuous, meaning that the income is being paid continuously rather than in lump-sum payments. However, even lump-sum payments can be approimated by a continuous stream if the payments are frequent enough or last long enough.

9 30434_ch06_p qd /6/03 8:09 AM Page CHAPTER 6 INTEGRATION TECHNIQUES AND DIFFERENTIAL EQUATIONS Solution Present 5 value 0 2te 0.t dt Multiplying C(t) 2t by e 0.t (since 0% 0.) and integrating from 0 to 5 years This is a definite integral, but we will ignore the limits of integration until after we have found the indefinite integral. None of the formulas through 7 on the inside back cover will find this integral, so we try integration by parts with u 2t and dv e 0.t dt. (Do you see why the guidelines on page 438 suggest this choice?) 2t e 0.t dt (2t)(0e 0.t ) (0e 0.t ) 2 dt u dv u v v du u 2t du 2 dt dv e 0.t dt v e 0.t dt 0e 0.t 20te 0.t 20 e 0.t dt 20te 0.t 20(0)e 0.t C 20te 0.t 200e 0.t C For the definite integral, we evaluate this from 0 to 5: (20te 0.t 200e 0.t ) 5 (20 5e e 0.5 ) (20 0e 0 200e 0 ) 0 Evaluation at t 5 Evaluation at t 0 In millions of 300e dollars (using a calculator) The present value of the stream of income over 5 years is approimately $8 million. This answer means that $8 million at 0% interest compounded continuously would generate the continuous stream C(t) 2t million dollars for 5 years. This method is often used to determine the fair value of a business or some other asset, since it gives the present value of future income.

10 30434_ch06_p qd /6/03 8:09 AM Page INTEGRATION BY PARTS 443 Graphing Calculator Eploration f()d= on [0, 5] by [2, 8] a. Verify the answer to Eample 4 by graphing 2e 0. and finding the area under the curve from 0 to 5. b. Can you eplain why the curve increases less steeply farther to the right? [Hint: Think of the present value of money to be paid in the more distant future.] Why is it necessary to learn integration by parts if integrals like the one in Eample 4 can be evaluated on graphing calculators? One answer is that you should have a variety of ways to approach a problem sometimes geometrically, sometimes analytically, and sometimes numerically (using a calculator). These various approaches mutually support one another; for eample, you can solve a problem one way and check it another way. Furthermore, not all applications involve definite integrals. Eample 4 might have asked for a formula for the present value up to any time t, and such formulas may be found by integrating by hand but not from most graphing calculators. Remember that you should use integration by parts only if the easier formulas ( through 7 on the inside back cover) fail to solve the integral. Practice Problem 3 Which of the following integrals require integration by parts, and which can be found by the substitution formula e u du e u C? (Do not solve the integrals.) a. e b. d e 2 d Solution on page 445

11 30434_ch06_p qd /6/03 8:09 AM Page CHAPTER 6 INTEGRATION TECHNIQUES AND DIFFERENTIAL EQUATIONS 6. Section Summary The integration by parts formula u dv uv v du is simply the integration version of the Product Rule. The guidelines on page 438 lead to the following suggestions for choosing u and dv in some commonly occurring integrals. For Integrals of the Form: n e a d n ln d ( a)( b) n d u n u ln u a Choose: dv e a d dv n d dv ( b) n d Integration by parts can be useful in any situation involving integrals, such as recovering total cost from marginal cost, calculating areas, average values (the definite integral divided by the length of the interval), continuous accumulations, or present values of continuous income streams. Solutions to Practice Problems. 3 ln d (ln ) 4 d u ln dv 3 d du d (ln ) v 3 d (ln) d 4 C 6

12 30434_ch06_p qd /6/03 8:09 AM Page ( )( ) 3 d u dv ( )3 d du d v ( ) ( ) ( ) 4 3 d 4 ( )4 ( ) 4 d 4 4 ( )( ) 4 ( ) 5 C ( )( ) 4 ( ) 5 C a. Requires integration by parts b. Can be solved by the substitution u 2 6. INTEGRATION BY PARTS Eercises Integration by parts often involves finding integrals like the following when integrating dv to find v. Find the following integrals without using integration by parts (using formulas through 7 on the inside back cover). Be ready to find similar integrals during the integration by parts procedure.. e 2. 2 d 5 d 3. ( 2) d d ( 3) 4 d 8. ( ) d e 0.5t dt ( 5) 6 d Use integration by parts to find each integral. 9. e 0. 2 d e 3 d ln d 4 ln d 3. ( 2)e d [Hint: Take u 2.] 4. ( )e d [Hint: Take u.] 5. ln d 6. 3 ln d 7. ( 3)( 4) 5 d 8. ( 2)( 5) 5 d 9. te 0.5t dt ln t t dt s(2s ) 4 ds d e d 29. e a d (a 0) te 0.2t dt ln t t dt d e 3 ln( ) d d

13 30434_ch06_p qd /6/03 8:09 AM Page CHAPTER 6 INTEGRATION TECHNIQUES AND DIFFERENTIAL EQUATIONS 30. ( b)e a d (a 0) n ln a d 3. (a 0, n ) 32. ( a) n ln( a) d (n ) 33. ln d [Hint: Take u ln, dv d.] 34. ln 2 d [Hint: Take u ln 2, dv d.] e 2 d [Hint: Take u 2, dv e 2 and use a substitution to find v from dv.] ( 2 ) 6 d [Hint: Take u 2, Find in two different ways. 47. ( 2) 5 d a. Use integration by parts. b. Use the substitution u 2 (so is replaced by u 2) and then multiply out the integrand. 48. ( 4) 6 d a. Use integration by parts. b. Use the substitution u 4 (so is replaced by u 4) and then multiply out the integrand. Derive each formula by using integration by parts on the left-hand side. (Assume n 0.) dv ( 2 ) 6 d find v from dv.] and use a substitution to 49. n e d n e n n e d Find each integral by integration by parts or a substitution, as appropriate. (ln ) 37. a. 3 e 2 d b. d c. ln 2 d d. 38. a. ln b. d 7 c. ln 3 d d. Evaluate each definite integral using integration by parts. (Leave answers in eact form.) e d ln d z(z 2) 4 dz 44. z(z 4) 6 dz 0 0 ln 4 e 45. te 46. t dt ln d e 4 d 2 e 3 d e 4 d 3 e d e ln d 50. (ln ) n d (ln ) n n (ln ) n d 5. Use the formula in Eercise 49 to find the integral 2 e d. [Hint: Apply the formula twice.] 52. Use the formula in Eercise 50 to find the integral (ln ) 2 d. [Hint: Apply the formula twice.] 53. a. Find the integral d by integration by parts (using u and dv d), obtaining d ( 2 ) d which gives d d b. Subtract the integral from both sides of this last equation, obtaining 0. Eplain this apparent contradiction. 54. We omit the constant of integration when we integrate dv to get v. Including the constant C

14 30434_ch06_p qd /6/03 8:09 AM Page INTEGRATION BY PARTS 447 in this step simply replaces v by v C, giving the formula u dv u(v C) (v C) du Multiplying out the parentheses and epanding the last integral into two gives u dv uv Cu v du C du Show that the second and fourth terms on the right cancel, giving the old integration by parts formula u dv uv v du. This shows that including the constant in the dv to v step gives the same formula. One constant of integration at the end is enough. APPLIED EXERCISES 55. BUSINESS: Revenue If a company s marginal revenue function is MR() e /4, find the revenue function. [Hint: Evaluate the constant C so that revenue is 0 at 0.] 56. BUSINESS: Cost A company s marginal cost function is MC() e /2 and fied costs are 200. Find the cost function. [Hint: Evaluate the constant C so that the cost is 200 at 0.] 57. BUSINESS: Present Value of a Continuous Stream of Income An electronics company generates a continuous stream of income of 4t million dollars per year, where t is the number of years that the company has been in operation. Find the present value of this stream of income over the first 0 years at a continuous interest rate of 0%. For Eercises 58 59: a. Solve without using a graphing calculator. b. Verify your answer to part (a) using a graphing calculator. 58. BUSINESS: Present Value of a Continuous Stream of Income An oil well generates a continuous stream of income of 60t thousand dollars per year, where t is the number of years that the rig has been in operation. Find the present value of this stream of income over the first 20 years at a continuous interest rate of 5%. 59. BIOMEDICAL: Drug Dosage A drug taken orally is absorbed into the bloodstream at the rate of te 0.5t milligrams per hour, where t is the number of hours since the drug was taken. Find the total amount of the drug absorbed during the first 5 hours. mg 60. ENVIRONMENTAL SCIENCE: Pollution Contamination is leaking from an underground waste-disposal tank at the rate of t ln t thousand gallons per month, where t is the number of months since the leak began. Find the total leakage from the end of month to the end of month GENERAL: Area Find the area under the curve y ln and above the -ais from to y amount absorbed 5 Hours y ln 62. POLITICAL SCIENCE: Fund Raising A politician can raise campaign funds at the rate 2 absorption rate t (continues)

15 30434_ch06_p qd /6/03 8:09 AM Page CHAPTER 6 INTEGRATION TECHNIQUES AND DIFFERENTIAL EQUATIONS of 50te 0.t thousand dollars per week during the first t weeks of a campaign. Find the average amount raised during the first 5 weeks. 63. BUSINESS: Product Recognition A company begins advertising a new product and finds that after t weeks the product is gaining customer recognition at the rate of t 2 ln t thousand customers per week (for t ). Find the total gain in recognition from the end of week to the end of week GENERAL: Population The population of a town is increasing at the rate of 400te 0.02t people per year, where t is the number of years from now. Find the total gain in population during the net 5 years. Repeated Integration by Parts Sometimes an integral requires two or more integrations by parts. As an eample, we apply integration by parts to the integral 2 e d. e 2 d 2 e 2 e d 2 e d 2 e u dv u u 2 du 2 d The new integral e d is solved by a second integration by parts. Continuing with the previous solution, we choose new u and dv: 2 e 2 e d 2 e 2 e uv u dv e d 2 e 2(e e ) C 2 e 2e 2e C v v dv e d v e d e v du u du d dv e d v e After reading the preceding eplanation, find each integral by repeated integration by parts e d e 2 d du 67. ( ) 2 e d (ln ) 2 d For each definite integral: a. Evaluate it by integration by parts. (Give answer in its eact form.) b. Verify your answer to part (a) using a graphing calculator e d 72. Repeated Integration by Parts Using a Table The solution to a repeated integration by parts problem can be organized in a table. As an eample, we solve 2 e 3 d. We begin by choosing u 2 (ln ) 2 d 3 e d 5 (ln ) 2 d dv v d e 3 d We then make a table consisting of the following three columns: Alternating Signs u 2 and Its Derivatives 2 e v e 3 and Its Antiderivatives 3 e3 9 e3 27 e3 Using the formula for e a d Stop when you get to 0 Finally, the solution is found by adding the signed products of the diagonals shown in the table: 2 e 3 d 3 2 e e e3 C After reading the preceding eample, find each integral by repeated integration by parts using a table e d e 2 d

16 30434_ch06_p qd /6/03 8:09 AM Page INTEGRATION USING TABLES e 2 d 3 e d 77. ( ) e 3 d ( ) 2 ( 2) 5 d 6.2 INTEGRATION USING TABLES Introduction There are many techniques of integration, and only some of the most useful ones will be discussed in this book. Many of the advanced techniques lead to integration formulas, which can then be collected into a table of integrals. In this section we will see how to find integrals by choosing an appropriate formula from such a table.* On the inside back cover is a short table of integrals that we shall use. The formulas are grouped according to the type of integrand (for eample, Forms Involving 2 a 2 ). Look at the table now (formulas 9 through 23) to see how it is organized. Using Integral Tables Given a particular integral, we first look for a formula that fits it eactly. EXAMPLE INTEGRATING USING AN INTEGRAL TABLE Find Solution The denominator 2 4 is of the form 2 a 2 (with a 2), so we look in the table of integrals under Forms Involving 2 a 2. Formula 5, with a d. 2 a d 2 2a ln a a C becomes our answer: 2 4 d 4 ln 2 2 C Formula 5 Formula 5 with a 2 substituted on both sides * Another way is to use a more advanced graphing calculator (see pages ) or a computer software package such as Maple, Mathcad, or Mathematica.

17 30434_ch06_p qd /6/03 8:09 AM Page CHAPTER 6 INTEGRATION TECHNIQUES AND DIFFERENTIAL EQUATIONS Note that the epression 2 a 2 does not require that the last number be a perfect square. For eample, 2 3 can be written 2 a 2 with a 3. Integral tables are useful in many applications, such as integrating a rate to find the total accumulation. Sales EXAMPLE y Sales ( ) rate 2 total sales S() Weeks 9 FINDING TOTAL SALES FROM THE SALES RATE A company s sales rate is sales per week after weeks. Find 9 a formula for the total sales after weeks. Solution To find the total sales S() we integrate the rate of sales. S() 9 d In the table on the inside back cover under Forms Involving a b we find a b Formula 3 This formula with a and b 9 gives the integral 2 36 Formula 3 with S() d C a and b C d 2a 4b 3a 2 a b C Simplifying To evaluate the constant C we use the fact that total sales at time 0 must be zero: S(0) 0. (2) 9 C 0 36 C 0 Simplifying Therefore, C 36. Substituting this into S() gives the formula for the total sales in the first weeks. S() ( 2 3 2) 9 C at 0 set equal to zero S() ( 2 2) 3 9 C with C 36

18 30434_ch06_p qd /6/03 8:09 AM Page INTEGRATION USING TABLES 45 Modern methods of biotechnology are being used to develop many new products, including powerful antibiotics, disease-resistant crops, and bacteria that literally eat oil spills. These gene splicing techniques require evaluating definite integrals such as the following. EXAMPLE 3 GENETIC ENGINEERING Under certain circumstances, the number of generations of bacteria needed to increase the frequency of a gene from 0.2 to 0.5 is Find n (rounded to the nearest integer). Solution fraction. Formula 2 (inside back cover) integrates a similar-looking 2 (a b) d b a b ln a b C Formula 2 To make (a b) into ( ), we take a and b, so the left-hand side of the formula becomes From formula 2 2 ( ) d or 2 ( ) d with a, b Ecept for replacing by q, this is the same as our integral. Therefore, the indefinite integral is found by formula 2 with a and b (which we epress in the variable q). Formula 2 with a and b q 2 ( q) dq q ln q q C For the definite integral from 0.2 to 0.5, we evaluate and subtract. 0.5 ln ln Evaluation at q 0.5 n q 2 ( q) dq 0.2 Evaluation at q 0.2 (2 ln ) 5 ln ln ln 0.25 Using a calculator

19 30434_ch06_p qd /6/03 8:09 AM Page CHAPTER 6 INTEGRATION TECHNIQUES AND DIFFERENTIAL EQUATIONS We multiply this by the 2.5 in front of the original integral. (2.5)(4.39) 0.98 Therefore, generations are needed to raise the gene frequency from 0.2 to 0.5. Graphing Calculator Eploration on [0, ] by [0, 00] (shaded from 0.2 to 0.5) a. Verify the answer to Eample 3 by finding the definite integral of y 2.5 from 0.2 to ( ) b. From the graph of the function shown on the left, which would require more generations: increasing the gene frequency from 0. to 0.2 or from 0.5 to 0.6? c. Can you think of a reason for this? [Hint: Genes reproduce from other similar genes.] Sometimes a substitution is needed to transform a formula to fit a given integral. In such cases both the and the d must be transformed. A few eamples will make the method clear. EXAMPLE 4 USING A TABLE WITH A SUBSTITUTION Find 4 d. Solution The table of integrals on the inside back cover has no formula involving 4. However, 4 ( 2 ) 2, so a formula involving 2, along with a substitution, might work. Formula 8 looks promising: 2 a 2 d ln 2 a 2 C 2 d ln 2 C The means: use either the upper sign or the lower sign on both sides Formula 8 with a and the upper sign

20 30434_ch06_p qd /6/03 8:09 AM Page INTEGRATION USING TABLES 453 With the substitution this becomes z 2 d 2z dz Differential of z 2 z 4 2z dz ln z2 z 4 C Dividing by 2 and replacing z by gives the integral that we wanted: 4 d 2 ln 2 4 C Above formula with z 2 and d 2z dz Dropping the absolute value bars since 2 4 is positive Given a particular integral, how do we choose a formula? How To Choose a Formula Find a formula that matches the most complicated part of the integral, making appropriate substitutions to change the formula into the given integral. For instance, in Eample 4 we matched the 4 in the given integral with the 2 a 2 in the formula, and the rest of the integral followed from the differential. t Practice Problem Find [Hint: Use formula 5 with 9t 4 dt. 3t 2.] Solution on page 456 EXAMPLE 5 USING A TABLE WITH A SUBSTITUTION Find e 2t e t dt.

21 30434_ch06_p qd /6/03 8:09 AM Page CHAPTER 6 INTEGRATION TECHNIQUES AND DIFFERENTIAL EQUATIONS Solution Looking in the table of integrals under Forms Involving e a and ln, none of the formulas looks anything like this integral. However, replacing e t by would make the denominator of our integral into, so formula 9 might help. This formula with a and b is With the substitution formula 9 becomes or d ln C Differential of e t e t (et ) dt e t ln e t C e t Ecept for the negative sign, this is the given integral. Multiplying through by gives the final answer. e 2t e 2t e t d e t dt e t dt et ln (e t ) C e t dt et ln (e t ) C Formula 9 with a and b Reduction Formulas Sometimes we must apply a formula several times to simplify an integral in stages. EXAMPLE 6 USING A REDUCTION FORMULA Find 3 e d. Solution In the integral table, we find formula 2. n e a d a n e a n a n e a d Formula 2. With n 3 and a, the left side fits our integral

22 30434_ch06_p qd /6/03 8:09 AM Page INTEGRATION USING TABLES 455 The right-hand side of this formula involves a new integral, but with a lower power of. We will apply formula 2 several times, each time reducing the power of until we eliminate it completely. Applying formula 2 with n 3 and a, 3 e d 3 e 3 2 e d 3 e 3 2 e 2 e d 3 e 3 2 e 6 e d The power has been reduced 3 e 3 2 e 6 e 0 e d 3 e 3 2 e 6e 6 e d 3 e 3 2 e 6e 6e C e ( ) C After one application Applying formula 2 again (now with n 2) to the last integral above Multiplying out Using formula 2 a third time (now with n ) Now solve this last integral by the formula e a d a ea C The solution, after three applications of formula 2 Factoring We used formula 2 three times, reducing the 3 in steps, first down to 2, then to, and finally to 0, at which point we could solve the integral easily. If the power of in the integral had been higher, more applications of formula 2 would have been necessary. Formulas such as 2 and 22 are called reduction formulas, since they epress an integral in terms of a similar integral but with a smaller power of. Graphing Calculator Eploration Some advanced graphing calculators can find indefinite integrals. For eample, the Teas Instruments TI-89 graphing calculator finds the integrals in Eamples 5 and 6 as follows.

23 30434_ch06_p qd /6/03 8:09 AM Page CHAPTER 6 INTEGRATION TECHNIQUES AND DIFFERENTIAL EQUATIONS F Tools F2 Algebra F3 Calc F4 F5 F6 Other PrgmI0 Clean Up F Tools F2 Algebra F3 Calc F4 F5 F6 Other PrgmI0 Clean Up e -2.t dt e -t + ln( e t +)-e -t -t ( e^(-2t)/( e^(-t)+),t) MAIN RAD AUTO FUNC /30 For Eample 5 integrals answers ( 3 e - ) (- 3 d (^3 ^(-),) )e- e MAIN RAD AUTO FUNC /30 For Eample 6 With some algebra, you can verify that these answers agree with those found in Eamples 5 and 6, even though they do not look the same. (Notice that this calculator omits the arbitrary constant of integration.) 6.2 Section Summary To find a formula that fits a given integral, we look for the formula whose left-hand side most closely matches the most complicated part of the integral. Then we choose constants (and possibly a substitution) to make the formula fit eactly. Although we have been using a very brief table, the technique is the same with a more etensive table. Many integral tables have been published, some of which are booklength, containing several thousand formulas.* Solution to Practice Problem Formula 5 with the substitution 3t 2, d 6t dt, and a becomes 6t dt ln 3t 2 C 9t 3t 2 2 Dividing each side by 6 gives the answer: dt ln 3t 2 C 9t 2 3t 2 * A useful table of integrals containing more than 400 formulas is found in CRC Standard Mathematical Tables, CRC Press, Boca Raton, Florida.

24 30434_ch06_p qd /6/03 8:09 AM Page INTEGRATION USING TABLES Eercises For each integral, state the number of the integration formula (from the inside back cover) and the values of the constants a and b so that the formula fits the integral. (Do not evaluate the integral.) d 2 (5 ) d d 2 d d 4 d Find each integral by using the integral table on the inside back cover d 2 [Hint: Use formula 6 with a 3.] d [Hint: Use formula 5 with a 5.] 9. 2 (2 ) d [Hint: Use formula 2 with a 2, b.] 0. 2 d [Hint: Use formula 9 with a, b 2.]. [Hint: Use formula 9.] d 2. [Hint: Use formula 3.] d 3. (2 )( ) d 4. ( )( 2) d d 2 d z z dz z 2 z dz e 2 d 99 ln d ln d (ln )2 d ( 3) d ( 3) d z z z 9 z dz 4 4 dz d 6 e d 30. t dt dt 2t 2t 4 e 9 e e t e 2t e e dt 33. 2t dt t d d 6 d 3 d e e t 2t (e (e t )(e t ) dt t )(e t ) dt e/2 d e d e e 4 d 4 d For each definite integral: a. Evaluate it using the table of integrals on the inside back cover. (Leave answers in eact form.) b. Use a graphing calculator to verify your answer to part (a) d d d 2 2 d d d 3 2 3

25 30434_ch06_p qd /6/03 8:09 AM Page CHAPTER 6 INTEGRATION TECHNIQUES AND DIFFERENTIAL EQUATIONS Find each integral by whatever means are necessary (either substitution or tables) d 2 4 d d 4 d d d 56. d 2 2 d Find each integral. [Hint: Separate each integral into two integrals, using the fact that the numerator is a sum or difference, and find the two integrals by two different formulas.] 57. (3 )( ) d d ( ) d d 6. [Hint: After separating into two d integrals, find one by a formula and the other by a substitution.] 62. [Hint: After separating into two 2 d integrals, find one by a formula and the other by a substitution.] APPLIED EXERCISES 63. BUSINESS: Total Sales A company s sales rate is 2 e million sales per month after months. Find a formula for the total sales in the first months. [Hint: Integrate the sales rate to find the total sales and determine the constant C so that total sales are zero at time 0.] 64. GENERAL: Population The population of a city is epected to grow at the rate of / 9 thousand people per year after years. Find the total change in population from year 0 to year BIOMEDICAL: Gene Frequency Under certain circumstances, the number of generations necessary to increase the frequency of a gene from 0. to 0.3 is n q 2 ( q) dq Find n (rounded to the nearest integer) BEHAVIORAL SCIENCE A subject in a psychology eperiment gives responses at the rate of t/ t correct answers per minute after t minutes. a. Find the total number of correct responses from time t 0 to time t 5. b. Verify your answer to part (a) using a graphing calculator. 67. BUSINESS: Cost The marginal cost function for a computer chip manufacturer is MC() / 2, and fied costs are $2000. Find the cost function. 68. SOCIAL SCIENCE: Employment An urban job placement center estimates that the number of residents seeking employment t years from now will be t/(2t 4) million people. a. Find the average number of job seekers during the period t 0 to t 0. b. Verify your answer to part (a) using a graphing calculator.

26 30434_ch06_p qd /6/03 8:09 AM Page IMPROPER INTEGRALS IMPROPER INTEGRALS Introduction In this section we define integrals over intervals that are infinite in length. Such improper integrals have many applications, such as in the Application Preview on page 435 at the beginning of this chapter. Limits as Approaches The notation : ( approaches infinity ) means that takes on arbitrarily large values. Limits Approaching : means: : means: takes values arbitrarily far to the right on the number line. takes values arbitrarily far to the left on the number line Evaluating limits as approaches positive or negative infinity is simply a matter of thinking about large and small numbers. The reciprocal of a large number is a small number. For eample, Similarly: lim : 0 2 lim : ,000,000 lim e : e 0 One over a million is one one-millionth As the denominator approaches infinity (with the numerator constant), the value approaches zero

27 30434_ch06_p qd /6/03 8:09 AM Page CHAPTER 6 INTEGRATION TECHNIQUES AND DIFFERENTIAL EQUATIONS These eamples illustrate the following general rules. lim : 0 n lim : ea 0 (n 0) (a 0) As approaches infinity, over to a positive power approaches zero As approaches infinity, e to a negative number times approaches zero EXAMPLE EVALUATING LIMITS a. lim Using the first rule in the bo above b: b 0 2 b. lim Because the approaches zero b: 3 b 3 b c. lim Because the e 2b approaches zero b: (e2b 5) 5 Similar rules hold for approaching negative infinity. lim : 0 n lim e a 0 : (for integer n 0) (a 0) As approaches negative infinity, over to a positive integer approaches zero As approaches negative infinity, e to a positive number times approaches zero (because the eponent is approaching ) Graphing Calculator Eploration a. Define y and y 2 e / 2 and use the TABLE feature of your calculator to evaluate these functions at -values like, 2, 3, 5, 0, 00, and 000. Which of the limit rules above do the results verify? [Note: An answer such as 3E 5 means ] b. Change y 2 to be y 2 e and use the TABLE to evaluate y and y 2 at negative -values like, 2, 3, 5, 0, and 00. Which limit rules do the results verify?

28 30434_ch06_p qd /6/03 8:09 AM Page IMPROPER INTEGRALS 46 Some quantities become arbitrarily large, and so do not have limits. (For a limit to eist, it must be finite.) The following limits do not eist: lim : n lim : e a lim ln : (n 0) (a 0) As approaches infinity, to a positive power has no limit As approaches infinity, e to a positive number times has no limit As approaches infinity, the natural logarithm of has no limit EXAMPLE 2 FINDING WHETHER A LIMIT EXISTS a. lim does not eist. b: b3 b. lim ( b ) does not eist. b: Because b 3 becomes arbitrarily large as b approches infinity Because b becomes arbitrarily large as b approaches infinity Practice Problem Evaluate the following limits (if they eist). b: b a. lim b. lim 3 b 3 Solutions on page 469 b: Improper Integrals Adefinite integral over an interval of infinite length is an improper integral. As a first eample, we evaluate the improper integral which gives the area under the curve y d, 2 from etending arbitrarily far to the right. 2 y y 2 d

29 30434_ch06_p qd /6/03 8:09 AM Page CHAPTER 6 INTEGRATION TECHNIQUES AND DIFFERENTIAL EQUATIONS EXAMPLE 3 EVALUATING AN IMPROPER INTEGRAL Evaluate d. 2 Solution To integrate to infinity, we first integrate over a finite interval, from to some number b (think of b as some very large number), and then take the limit as b approaches. First integrate from to b. b b Using the Power Rule d 2 d ( ) b 2 b b b at b at Then take the limit of this answer as b :. This gives the answer: lim b: b d 2 Evaluating and simplifying Limit as b : (the approaches zero) b Integral from to equals Since the limit eists, we say that the improper integral is convergent. Geometrically, this procedure amounts to finding the area under the curve from to some number b, shown on the left below, and then letting b : to find the area arbitrarily far to the right. y Integrating to b y Integrating to b d 2 d 2 b 2 3 4

30 30434_ch06_p qd /6/03 8:09 AM Page IMPROPER INTEGRALS 463 Improper Integrals Integrating to If f is continuous and nonnegative for a, we define f() d lim a b b:a f() d provided that the limit eists. The improper integral is said to be convergent if the limit eists, and divergent if the limit does not eist. y a f() d a y f() It is possible to define improper integrals for functions that take negative values, and even for discontinuous functions, but we shall not do so in this book, since most applications involve functions that are positive and continuous. 2 Practice Problem 2 Evaluate Solution on page 469 d. 2 EXAMPLE 4 FINDING WHETHER AN INTEGRAL DIVERGES Evaluate d. Solution Integrating up to b: b b d /2 d 2 /2 b Integrating by the Power Rule 2 b 2 2 b 2 Evaluating Letting b approach infinity: lim 2 b 2 does not eist b: Therefore, d is divergent Because b becomes infinite as b : The integral cannot be evaluated