! ln 2xdx = (x ln 2x - x) 3 1 = (3 ln 6-3) - (ln 2-1)

Transcription

1 7. e - d Le u = and dv = e - d. Then du = d and v = -e -. e - d = (-e - ) - (-e - )d = -e - + e - d = -e - - e - 9. e 2 d = e d = 2 e 2 2d = 2 e u du Le u = 2, hen du = 2 d. = 2 eu = 2 e2.! ( - )e d Le u = ( - ) and dv = e d. Then du = d and v = e. ( - )e d = ( - )e - e d = ( - )e - e = e - 4e. Thus,! ( - )e d = (e - 4e ) = (e - 4e) - (-4) = -e ! ln 2d Le u = ln 2 and dv = d. Then du = d ln 2d = (ln 2)() - d and v =. = ln 2 - Thus,! ln 2d = ( ln 2 - ) = ( ln 6 - ) - (ln 2 - ) d = u du = ln u = ln(2 + ) Subsiuion: u = 2 + du = 2d [Noe: Absolue value no needed, since 2 +.] 7. ln d = udu = u2 (ln )2 = 2 2 Subsiuion: u = ln du = d 46 CHAPTER 7 ADDITIONAL INTEGRATION TOPICS

2 9. ln d = /2 ln d Le u = ln and dv = /2 d. Then du = d and v = 2 /2. /2 ln d = 2 /2 ln - 2 d /2 = 2 /2 ln - 2 /2 d 2. ( ) 6 ( + )d = 2 /2 ln /2 Le u = + and dv = ( ) 6. Then du = d and v = ( ) 6 ( " )7 ( " )7 ( + )d = ( + ) - d 7 7 ( + )( " )7 ( " )8 = ( " ) ( + ) 2 ( ) 2 d = ( 2 - ) 2 d = ( )d = A Since f() = ( - )e < on [, ], he inegral represens he negaive of he area beween he graph of f and he -ais from = o =. 27. y = ln 2 A The inegral represens he area beween he curve y = ln 2 and he -ais from = o = e d Le u = 2 and dv = e d. Then du = 2 d and v = e. 2 e d = 2 e - e (2)d = 2 e - 2e d e d can be compued by using inegraion-by-pars again. EXERCISE 7-47

3 Le u = and dv = e d. Then du = d and v = e. e d = e - e d = e - e and 2 e d = 2 e - 2(e - e ) = 2 e - 2e + 2e. e a d. e! = ( )e Le u = and dv = e a d. Then du = d and v = ea a. e a d = ea a ln 2 d - ea a d = ea a - ea a 2 Le u = ln and dv = d d 2. Then du = and v =!. ln 2 d = (ln ) "! $ # % - - d = - ln + d 2 = - ln - e ln Thus,! 2 d = "! ln! $ e ln e # % = -! e e! "! ln! $ # % [Noe: ln e =.]! 2. ln( + 4)d Le = + 4. Then d = d and ln( + 4)d = ln d. = - 2 e Now, le u = ln and dv = d. Then du = d and v =.! ln d = ln - # " $ d = ln - d = ln - Thus, ln( + 4)d = ( + 4) ln( + 4) - ( + 4) and 2! ln( + 4)d = [( + 4) ln( + 4) - ( + 4)] 2 = 6 ln (4 ln 4-4) = 6 ln 6-4 ln e -2 d Le u = and dv = e -2 d. Then du = d and v = e -2. e -2 d = e -2 - e -2 d = e -2 - e CHAPTER 7 ADDITIONAL INTEGRATION TOPICS

4 9. ln( + 2 )d Le = + 2. Then d = 2 d and ln( + 2 )d = ln( + 2 ) d = ln d 2 = ln d. 2 Now, for ln d, le u = ln, dv = d. Then du = d and v =.! ln d = ln - # 2 " $ d = ln - d = ln - Therefore, ln( + 2 )d = 2 ( + 2 )ln( + 2 ) - 2 ( + 2 ). 4. e ln( + e )d Le = + e. Then d = e d and e ln( + e )d = ln d. Now, as shown in Problems and, ln d = ln -. Thus, e ln( + e )d = ( + e )ln( + e ) - ( + e ). 4. (ln ) 2 d Le u = (ln ) 2 and dv = d. Then du = 2 ln d and v =. (ln ) 2 d = (ln ) 2 2 ln - d = (ln ) 2-2ln d ln d can be compued by using inegraion-by-pars again. As shown in Problems and, ln d = ln -. Thus, (ln ) 2 d = (ln ) 2-2( ln - ) = (ln ) 2-2 ln (ln )d Le u = (ln ) and dv = d. Then du = (ln ) 2. d and v =. (ln )d = (ln ) - (ln ) 2 d = (ln ) - (ln )d Now, using Problem 9, (ln 2 )d = (ln ) 2-2 ln + 2. Therefore, (ln )d = (ln ) - [(ln ) 2-2 ln + 2] = (ln ) - (ln ) ln - 6. EXERCISE 7-49

5 47. " e ln( 2 e e )d = " 2 ln d = 2 " ln d By Eample 4, ln d = ln. Therefore 2 e e " ln d = 2( ln ) = 2[e ln e e] 2[ln ] = " ln(e 2 )d = " 2 d = = (Noe: ln(e 2 ) = 2 ln e = 2.) 2. y = ln, 4 y = a A =! [-( ln )]d +! ( ln )d =! (ln )d +! ( ln )d.46 Now, ln d is found using inegraion-by-pars. Le u = ln and dv = d. Then du = d and v =.! ln d = ln - # " $ d = ln - d = ln - Thus, " A = ln! + 2! # 2 2 $.46 " % + # 2 2! 2! ln + $ " = ln +! # 2 2 $.46 " % + # 2 2!! ln $ 4 %.46 (.8 -.) + ( ) = %.46. y = - e, y = a A =! ( - e )d +!.27 [-( - e )]d.27 =! ( - e )d +!.27 (e - )d Now, e d is found using inegraion-by-pars. Le u = and dv = e d. Then, du = d and v = e. e d = e - e d = e - e Thus, A = ( - [e - e.27 ]) + (e - e - ).27 (.42 - ) + (2.7 - [-.42]) CHAPTER 7 ADDITIONAL INTEGRATION TOPICS

6 . Marginal profi: P'() = 2 - e -. The oal profi over he firs years is given by he definie inegral:! (2 - e- )d =! 2d -! e- d We calculae he second inegral using inegraion-by-pars. Le u = and dv = e - d. Then du = d and v = -e - e - d = -e - - -e - d = -e - - e - = -e - [ + ] Thus, Toal profi = 2 + (e - [ + ]) 2 + (.4 - ) = 24.4 To he neares million, he oal profi is $24 million. 7. P'() P'() = 2 - e - Toal Profi The oal profi for he firs five years (in millions of dollars) is he same as he area under he marginal profi funcion, P'() = 2 - e -, from = o =. 9. From Secion 7-2, Fuure Value = e rt! f()e-r d. Now r =.8, T =, f() = - 2. Thus, FV = e (.8)! ( - 2)e-.8 d = e.4! e-.8 d - 2e.4! e-.8 d. We calculae he second inegral using inegraion-by-pars. Le u =, dv = e -.8 d. Then du = d and v = e!.8!.8. e -.8 d = e!.8!.8 Thus, we have: T - e!.8!.8 d = -2.e-.8 - e!.8.64 = -2.e e -.8 FV = e.4 e!.8-2e.4 [-2.e e -.8 ]!.8 = -2, + 2,e.4-2e.4 [-62.e e ] = -2, + 2,e.4 + 4,7 -,2e.4 =,2-8,7e.4,278 or $,278 EXERCISE 7-44

7 6. Gini Inde = 2! ( - e- )d = 2! d - 2! e- d We calculae he second inegral using inegraion-by-pars. Le u =, dv = e - d. Then du = d, v = e -. e - d = e - - e - d = e - - e - Therefore, 2! d - 2! e- d = 2-2[e - - e - ] = - 2[ - + (e - )] = - 2e y The area bounded by y = and he Lorenz curve 6. y = e (-) divided by he area under he curve y = from = o = is he inde of income y = y = e (-) concenraion, in his case.264. I is a. measure of he concenraion of income he closer o zero, he closer o all he income being equally disribued; he closer o one, he closer o all he income being concenraed in a. few hands. 6. S'() = -4e., S() = 2, S() = -4e. d = -4e. d Le u = and dv = e. d. Then du = d and v = e.. = e. e. d = e. - e. d = e. - e. Now, S() = -4e. + 4e. Since S() = 2,, we have 2, = 4, C =,6 2 Thus, S() =,6 + 4e. - 4e. To find how long he company will coninue o manufacure his compuer, solve S() = 8 for. The company will manufacure he compuer for monhs. 67. p = D() = 9 - ln( + 4); p = $2.89. To find, solve 9 - ln( + 4) = 2.89 ln( + 4) = = e 6.9 (ake he eponenial of boh sides), Now,, CS =!, (D() - p )d =! [9 - ln( + 4) ]d =,,! 6.9d -! ln( + 4)d 442 CHAPTER 7 ADDITIONAL INTEGRATION TOPICS

8 To calculae he second inegral, we firs le z = + 4 and dz = d o ge ln( + 4)d = ln zdz Then we use inegraion-by-pars. Le u = ln z and dv = dz. Then du = dz and v = z. z ln zdz = z ln z - z dz = z ln z - z z Therefore, and ln( + 4)d = ( + 4)ln( + 4) - ( + 4) CS = 6.9, - [( + 4)ln( + 4) - ( + 4)], 69 - (9.9 -.) $ p = 2.89 CS p = D() = The area bounded by he price-demand equaion, p = 9 - ln( + 4), and he price equaion, y = p = 2.89, from = o = =,, represens he consumers' surplus. This is he amoun saved by consumers who are willing o pay more han $ Average concenraion: = 2 ln( + ) ln( + )!! ( + ) 2 d = 4! ( + ) 2 ln( + ) ( + ) 2 d is found using inegraion-by-pars. Le u = ln( + ) and dv = ( + ) -2 d. Then du = + d = ( + )- d and v = -( + ) -. ln( + ) ln( + ) ( + ) 2 d = - + ln( + ) = ( + ) - ( + ) - d + ( + ) -2 ln( + ) d = - + Therefore, he average concenraion is: - +! 2 ln( + ) ( + ) 2 d = 4 ln( + ) " = 4! ln 6! $ # 6 6% - 4(-ln - ) = 4-2 ln 6-2 = ( - 2 ln 6) 2.88 ppm EXERCISE 7-44

9 7. N'() = ( + 6)e -.2, ; N() = 4 N() - N() =! N'()d; N() = 4 +! ( + 6)e-.2 d = 4 + 6! e-.2 d +! e-.2 d = 4 + 6(-4e -.2 ) +! e-.2 d = 64-24e -.2 +! e-.2 d Le u = and dv = e -.2 d. Then du = d and v = -4e -.2 ; e -.2 d = -4e e -.2 d = -4e -.2-6e -.2 Now,! e-.2 d = (-4e -.2-6e -.2 ) = -4e -.2-6e and N() = 8-4e -.2-4e -.2 To find how long i will ake a suden o achieve he 7 words per minue level, solve N() = 7: I will ake 8 weeks. By he end of he course, a suden should be able o ype N() = 8-4e -.2() - 6e -.2() 78 words per minue. 7. Average number of voers =! ( e-. )d =! (2 + 4)d -! e-. d e -. d is found using inegraion-by-pars. Le u = and dv = e -. d. Then du = d and v = e!.!. = -e-.. e -. d = -e e -. d = -e -. + e -. d = -e -. + e!. = -e -. - e -.!. Therefore, he average number of voers is:! (2 + 4)d -! e-. d = ( ) - (-e -. - e -. ) = ( + ) + (e-. + e -. ) = + (e -. + e -. ) - = e (housands) or 2, CHAPTER 7 ADDITIONAL INTEGRATION TOPICS

10 EXERCISE 7-4. Use Formula 9 wih a = b =. ( + ) d = ln + = ln +. Use Formula 8 wih a =, b =, c =, d = 2: ( + ) 2 ( + 2) d =! 2 "! (! 2 "! ) 2 ln = + 2 ln + +. Use Formula 2 wih a = 6 and b = : 2( " 2 # 6) d = 6 + # = 7. Use Formula 29 wih a = : 2( " 2) 6 + d = - " 2 ln + " 2 = -ln + " 2 9. Use Formula 7 wih a = 2 (a 2 = 4): d = ln Use Formula wih n = 2: 2 ln d = ln - (2 + ) 2 = ln - 9. Firs le u = e. Then du = e d and d = e du = u du. Thus, + e d = u( + u) du Now use Formula 9 wih a = b = : u( + u) du = ln u + u e = ln + e = ln e - ln + e = - ln + e EXERCISE