Solucao Calculo 5ed SEC.7.8

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(1)Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. . 1. (a) Since. 4 x. x e. 4. dx has an infinite interval of integration, it is an improper integral of Type I.. 1. (b) Since y=sec x has an infinite discontinuity at x= , 2. /2. sec xdx is a Type II improper integral. 0 2. (c) Since y=. x has an infinite discontinuity at x=2 , (x2)(x3). 0. x 2. x 5x+6. dx is a Type II improper. integral. 0. . (d) Since. . 1. dx has an infinite interval of integration, it is an improper integral of Type I.. 2. x +5. 2. (a) Since y=1/(2x1) is defined and continuous on 1,2 , the integral is proper. 1. 1 1 1 dx is a Type II improper integral. (b) Since y= has an infinite discontinuity at x= , 2x1 2 0 2x1 . (c) Since. sin x dx has an infinite interval of integration, it is an improper integral of Type I.. . 2. 1+x. 2. (d) Since y=ln (x1) has an infinite discontinuity at x=1 , ln (x1)dx is a Type II improper integral. 1 3. 3. The area under the graph of y=1/x =x t. 3. between x=1 and x=t is. 1 2 t 1 2 1 1 2 x = t = 1 2t . So the area for 1 x 10 is 1 2 2 2 2 1 A(10)=0.50.005=0.495 , the area for 1 x 100 is A(100)=0.50.00005=0.49995 , and the area for 1 x 1000 is A(1000)=0.50.0000005=0.4999995 . The total area under the curve for x 1 is 1 1 2 lim A(t)=lim 1 2t = . 2 2 t t 3. /( ). A(t)= x dx= . /( ). 4. (a) 1.

(2) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. (b) The area under the graph of f from x=1 to x=t is F(t) =. t. t. 1. 1. f (x)dx= x. 1.1. 1 0.1 dx= x 0.1. t 1. = 10 t 0.11 =10 1t 0.1 and the area under the graph of g is. (. ). t. t. 1. 1. G(t) = g(x)dx= x. (. = 10 t 0.11 t 10 100. (. ). 0.9. 1 0.1 x 0.1. dx=. G(t) 2.59 5.85 15.12. 6. 7.49. 29.81. 10. 9. 90. 20. 9.9. 990. 4. 10 10 10. 1. ). F(t) 2.06 3.69 6.02. 10. t. (. (c) The total area under the graph of f is lim F(t)=lim 10 1t t . t . 0.1. ) =10 . ( 0.11) = .. The total area under the graph of g does not exist, since lim G(t)=lim 10 t t . . . t. 1. 5. I=. 1. . t . 2. (3x+1). 1 2. (3x+1). dx =. dx=lim. . t 1. 1 2. dx . Now. (3x+1). 1 1 2 du [ u=3x+1 , du=3dx ] 3 u 2.

(3) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. 1 1 +C= +C , 3u 3(3x+1). =. t. 1 3(3x+1). so I=lim. t . 0. . =lim. 1 t . 1 1 + 3(3t+1) 12. 0. 1 1 dx= lim dx= lim 6. t t 2x5 t 2x5 Divergent. =0+. 1 1 = . Convergent 12 12. 0. 1 ln 2x5 2. 1 1 ln 5 ln 2t5 2 2. =lim. t t . = .. 7. 1. 1. 1 dw = lim 2w t t. . . 1. 1 dw= lim 2 2w 2w t . t. 2 3 +2 2t = . Divergent. = lim. t . 8. . t. . x 2. 2. dx. =lim. t 0. (x +2). 0. = . 9.. y/2. e. 1 2 t. dy=lim. y/2. e. 2. 2. 1 2. =. dx=lim. t . (x +2). 0+. t 4. 4. x. . 1. 1 2. t. =. 2. x +2. 0. 1 lim 2 t . 1 2. +. t +2. 1 2. 1 . Convergent 4. dy=lim. t . y/2 t. 2e. t/2. =lim (2e. 4. 2. 2. 2. +2e )=0+2e =2e. t . .. Convergent 1. 10.. e. . 11.. . 0. . . 2t. 1. dt= lim. 2t. e. x x 0 x 2. dt= lim. x . . 1 2t e 2. 1 x. = lim. x . . 1 2 1 2x e+ e = . Divergent 2 2. x x 2 2 1+x = 1+x + 1+x and dx dx 0 dx. x 2 1+x = lim dx t . 1 2 ln 1+x 2. (. ). 0. = lim. t t . 0. 1 2 ln 1+t 2. (. ). = . Divergent 3.

(4) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. . (. 0. ). (. . ). (. ). 12. I= 2v dv=I +I = 2v dv+ 2v dv , but . 4. I = lim 1. 2. . 4. 0. 0. 1 5 v 5. 2v. t . 1. 4. 1 5 t = . Since I is divergent, I is divergent, and there is no 1 5. 2t+. =. t t . need to evaluate I . Divergent 2. . 2. x. xe. 13.. 0. dx= xe . dx= lim. t . Therefore,. e. x. 14.. 2 x. xe. 0. 2 x. xe. 3. 0. dx= x e . 3. dx= lim. . 2 x. t . t . 0. t. 1 2. t . ( e 1) = 12 1= 12 , and ( e 1) = 12 (1)= 12 . t. 2. 2. 1 1 + =0 . Convergent 2 2 3. . 2 x. dx+ x e. 3. dx , and. 0. 1 x e 3. . 15. sin d =lim 2. =lim. =lim. e. dx=. 1 2. t. 2. . t. x. 2. xe. dx . 0. 2. x. 1 2. dx=lim. 0. 2. x. dx+ xe. 1 2. 2. xe. 0. t . x. 2. x. 2. x. xe. . 0. 3. 0. 1 1 = + t 3 3. 2 sin d =lim t. t . t . t. lim e. t . cos . t 2. 3. = . Divergent. =lim (cos t+1) . This limit does not exist, so the t . integral is divergent. Divergent t 1 1 1 2 (1+cos 2 )d =lim + sin 2 16. cos d =lim 0 02 2 4 t t 1 1 1 sin 2t for all t , but t as t . Divergent 4 4 2. t. =lim. 0 t . 1 1 t+ sin 2t = since 2 4. 17. . 1. t. x+1 2. x +2x. t . dx =lim. 1 (2x+2) 2 2. x +2x = . Divergent 1. dx=. t 1 1 2 2 lim ln (x +2x) = lim ln (t +2t)ln 3 1 2 2 t t . 4.

(5) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. 18. . t. . dz. t 0. 2. z +3z+2. 0. 1 1 z+1 z+2. . =lim. =lim. t+1 t+2. ln. t . t . 1 2. ln. t. z+1 z+2. ln. dz=lim. 0. =ln 1+ln 2=ln 2 . Convergent. 19. . 5s. se. t. 5s. ds =lim. se. =lim. . t 0. 0. t . =. . ds=lim. t . t. 1 5s 1 5s se e 5 25. 1 5t 1 5t 1 te e + 5 25 25. =00+. 0. by integration by parts with u=s. 1 [ by l’Hospital’s Rule] 25. 1 . Convergent 25. 20. 6. 6. re. r/3. re. dr = lim. r/3. t t. . r/3. t . 2. 2. r/3 6. 3re 9e. dr= lim. t/3. t/3. by integration by parts with u=r. t 2. = lim (18e 9e 3te +9e )=9e 0+0 [ by l’Hospital’s Rule] t 2. =9e . Convergent . ( ln x ). ln x dx=lim 21. x t 1 Divergent . 22. . e. x. e 0. x. 0. t. 2. 2. . 1. x. dx= e dx+ e dx , x. . dx=lim. t . 0. x t. e. =lim. 0. t . t. (by substitution with u=ln x , du=dx/x ) =lim. t . 0. e. . x. dx= lim. t . e. x 0 t. = lim. t . ( ln t ) 2. 2. = .. ( 1et ) =1 , and. . ( 1e ) =1 . Therefore, e x dx=1+1=2 . Convergent . 23.. 5.

(6) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. . . . . 0 2. x. 9+x. Now . 6. dx=. . . 9+x. 2. x dx 9+x. . 2. x. 6. . dx+. u=x. 0. 9+x. 6. =. du=3x dx. 2. x. dx=2. 3 2. 6. . 2. x. 9+x. 0. 1 du 3 9+u. 6. dx [ since the integrand is even].. =. u=3v du=3dv. 2. . so 2. 0. 2. =. 9+9v. 1 1 1 1 = tan v+C= tan 9 9 . 1 (3dv) 3. u 3. 1 dv 2 9 1+v. 1 1 +C= tan 9. 3. x 3. +C ,. t 2. x. 9+x. 6. dx =2lim. . t 0. 2. x. 9+x. 6. 1 1 =2lim tan t 9. dx=2lim. t . 3. t. 3. 1 1 tan 9. x 3. 0. 2 . Convergent = = 9 2 9. t 3. ( 2) .. 3. 24. Integrate by parts with u=ln x , dv=dx/x du=dx/x , v=1/ 2x . t. t. ln3x dx =lim ln3x dx=lim. . 2. ln x. +. t 1 2x x 1 ln t 1 1 1 =lim +0 + = 2 4 2 2 4 t 4t t 1/t ln t 1 =lim =lim =0 . Convergent since lim t t 2 t 2t t 2t 2 1. t 1. t. 1. x. 1 2. 1 dx 1. x. 3. 2. 25. Integrate by parts with u=ln x , dv=dx/x du=dx/x , v=1/x . . t. ln x dx = lim ln x dx=lim 2. 2. ln x 1 x x. t x = 00+0+1=1 ln t 1/t since lim =lim =0 . Convergent t t t 1 1. x. t 1. . t. =lim. 1 t . . ln t 1 +0+1 t t. 6.

(7) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. . 26.. t. x arctanx dx=lim x arctanx dx . Let u=arctanx , dv= 22. t 0. xdx. 22. (1+x ) 0 (1+x ) 1 1/2 2xdx , and v= = 22 2 2 (1+x ) 1+x 1 1 x arctanx dx = arctanx + dx 22 2 22 2 2 (1+x ) 1+x (1+x ). 22. . Then du=. (1+x ). dx 2. ,. 1+x. x=tan , 2. dx=sec d. 2. 1 arctanx 1 sec d = + 2 2 2 2 2 1+x (sec ) 1 arctanx 1 2 = + cos d 2 2 2 1+x 1 arctanx sin cos = + + +C 2 2 4 4 1+x 1 x 1 arctanx 1 + arctanx+ +C = 2 2 4 4 2 1+x 1+x. It follows that . x arctanx dx 0. 22. =lim. t . . (1+x ). =lim. t . . 1 arctanx 1 1 x + arctanx+ 2 2 2 4 4 1+x 1+x 1 arctant 1 1 t + arctant+ 2 2 2 4 4 1+t 1+t. t 0. 1 . =0+ +0= 4 2 8. Convergent. 27. There is an infinite discontinuity at the left endpoint of 0,3 .. 7.

(8) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. 3. 0. 3. dx dx =lim =lim 2 x +t + x x t 0. 3. =. t. +. (2. 3 2 t ) =2 3 . Convergent. t 0. t 0. 28. There is an infinite discontinuity at the left endpoint of 0,3 . 3. 3. 0. dx =lim + x x t 0. . dx x. t. 2 x. =lim. 3/2. +. t 0. 2 2 +lim = . Divergent + t 3. 3. = t. t 0. 29. There is an infinite discontinuity at the right endpoint of 1,0 . 0. t. dx =lim dx =lim 1. 2. 2. . t 0 1. x. x. 1 x. . t 0. t. =lim. 1. 1 1 + t 1. . . t 0. = . Divergent. 9. 30.. 3. t. dx. 1. dx. =lim. t 9 1. x9 0. 3. 31.. . 4. 2. dx. . x=. 0. x9 =lim. 2. x. +. 4. 0. . 1 t 9. dx x. 4. , but. . dx. 2. x. 4. 3. =lim. t. x 3. . t 0. =lim. 2. dx. =lim 2. 1x. . t 1. 0. dx. =lim 2. 1. 3t. t 0. . 1. t. 0. . t 1. 1x. t 1. 1/5. (x1). 1. 3. 1. . = . Divergent. 24. 1. t. dx=lim. 1/5. (x1). t 1 0. dx=lim. . t 1. 5 4/5 (x1) 4. . Convergent 2. 1/5. dx= (x1). 0. 0. . sin x =lim sin t=. 33. There is an infinite discontinuity at x=1 .. (x1). 1. . t. 33. 1/5. 3 2/3 3 (t9) (4) =06=6 . Convergent 2 2. =. 0. . dx. t. 3 2/3 (x9) 2. t 9. 3. 1. 32.. 3. 33. 0 t. =lim. 0. . t 1. 1/5. dx+ (x1). dx . Here. 1. 5 4/5 5 (t1) 4 4. =. 5 and 4. 8.

(9) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. 33. 1/5. (x1). 33. dx=lim. (x1). 1/5. (x1). 33. 5 4/5 (x1) 4. dx=lim. + t 1 t. 1 33. 1/5. +. t 1. t. 34. f (y)=1/(4y1) has an infinite discontinuity at y=. 1/4. 1 . 4. 1. 1. 1 1 dy= lim ln 4y1 lim 4 + t 4y1 + t ( 1/4 ) t ( 1/4 ) 1 1 = lim + 4 ln 3 4 ln (4t1) = t ( 1/4 ). 1 dy = 4y1. 1. . so. +. t 1. 5 75 +20= . Convergent 4 4. dx=. 0. 1. 1/4. 5 5 4/5 16 (t1) =20 . Thus, 4 4. =lim. t. 1. 1 1 dy diverges, and hence, dy diverges. Divergent 4y1 4y1 0. 35. . /2. . /2. 0. 0. /2. 0. t. sec xdx = sec xdx+ sec xdx . sec xdx= lim sec xdx = lim. . ln sec x+tan x. t. = lim ln sec t+tan t = . Divergent. 0. t /2 4. 36.. 0. 2. 0. dx 2. x +x6. t /2 0. . t /2. 4. 2. 4. dx dx dx = = + , and 0 (x+3)(x2) 0 (x2)(x+3) 2 (x2)(x+3) t. dx = lim 0 (x2)(x+3) t 2. 1 lim = 5 t 2. 1/5 1/5 x2 x+3 ln. t2 t+3. dx [partial fractions =lim. 1 ln 5. . t 2. ln. 2 3. . 1. t 0. = .Divergent. 1. 37. There is an infinite discontinuity at x=0 .. x2 x+3. 0. e x. x. e 1. . dx=. 1. 1. e x. x. e 1. . dx+. 0. e x. x. e 1. dx .. 9.

(10) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. 0. . 1. t. e. x. dx=lim. x. e 1. . . e. x. x. e 1. t 0 1. ln e 1. . 1. t. =lim. 1. t 0. ln e 1 ln e 1. . = ,. t 0. 1. so. t. x. dx=lim. 1. . e. x. x. e 1. 1. . dx is divergent. The integral. e. dx also diverges since. x. e 1. 0. 1. x. 1. . e. x. dx=lim. x. e 1 Divergent 0. +. . e x. e 1. t 0 t. 2. x. 3/2. 1. x. ln e 1. dx=lim +. t. t 0. t. ln e1 ln e 1. =lim +. = .. t 0. 2. x3 x3 x3 dx= dx+ dx and 38. 0 2x3 0 2x3 3/2 2x3 x3 1 2x6 1 3 1 3 2x3 dx= 2 2x3 dx= 2 1 2x3 dx= 2 x 4 ln 2x3 +C , so 3/2. x3 dx= lim 2x3. 0. 1 2x3ln 2x3 4. . t 3/2. t. = . Divergent. 0. 39. 2. I = z ln zdz=lim. 2. z ln zdz=lim. 2. 0. +t. t 0. z. 2. 2. +. 2. 3. (3ln z1). 3. t 0. t. 8 1 3 8 8 1 8 8 1 3 (3ln 21) t (3ln t1) = ln 2 lim t (3ln t1) = ln 2 L . 9 9 3 9 9 3 9 9 +. =lim +. t 0. t 0. 3ln t1. 3. Now L=lim t (3ln t1) =lim +. +. t 0. t 0. t. 3. 3/t. =lim +. t 0. 3/t. 4. =lim +. t 0. ( t 3) =0 . Thus, L=0 and I= 83 ln 2 89. .. Convergent 40. Integrate by parts with u=ln x , dv=dx 1. 0. 1. ln x ln x dx=lim dx = lim +t + x x t 0 t 0 =. lim +. ( 2. /. x du=dx/x , v=2 x . 1. 2 x ln x 2 1. t. t. dx. x. =lim +. 2 t ln t4. x. 1 t. t 0. t ln t4+4 t ) =4. t 0. 10.

(11) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. since lim. ln t. t ln t=lim +. +. t 0. t 0. t. 1/t. =lim. 1/2. +. t 0. t. 3/2. =lim /2. +. ( 2 t ) =0 . Convergent. t 0. 41.. 1. Area = e dx= lim x. t . . e. x 1 t. t. =e lim e =e t . 42.. . x/2. Area = e. x/2 t. dx=2lim. e. t . 2. 2. t/2. =2lim e t . +2e=2e. 43.. . Area =. . . . 2 2. x +9. . dx=2 2. 0. 1 2. dx. x +9 11.

(12) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. t. . =4lim. 1 2. t 0. =. 1 1 x tan 3 3. dx=4lim. t . x +9. t 0. 4 4 2 1 t lim tan 0 = = 3 3 t 3 2 3. 44.. . . t. x. Area =. 0. 2. t . . x. 1 2 ln x +9 2. (. dx. 2. t 0. x +9. =lim =. dx=lim. x +9. ). t 0. 1 2 lim ln t +9 ln 9 = 2 t . (. ). Infinite area 45.. /2. Area. = sec xdx= lim 0. = lim. t. 2. sec. 0. 2. xdx. t ( /2) t. . t ( /2). tan x = lim (tan t0) 0. . t ( /2). = Infinite area. 12.

(13) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. 46.. 0. 0. 1 dx= lim +t x+2. Area = . 2. 1 dx x+2. t 2. = lim. 2 x+2 +. t 2. 0 t. (2. = lim +. 2 2 t+2. ). t 2. =2 2 0=2 2 47. (a) t. g(x)dx. t. 1. 2 5 10 100 1000 10 , 000. 0.447453 0.577101 0.621306 0.668479 0.672957 0.673407 2. g(x)=. sin x 2. . It appears that the integral is convergent.. x. . 2. 2. (b) 1 sin x 1 0 sin x 1 0. sin x 2. x. . 1 2. x. . Since. 1 dx is convergent 1. 2. x. . (Equation 2 with p=2>1 ),. 1. 2. sin x 2. dx is convergent by the Comparison Theorem.. x. (c) 13.

(14) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. . Since. f (x)dx is finite and the area under g(x) is less than the area under. f (x) on any interval 1,t ,. 1 . g(x)dx must be finite; that is, the integral is convergent. 1. 48. (a) t. g(x)dx. t. 2. 5 10 100 1000 10 , 000 g(x)=. 3.830327 6.801200 23.328769 69.023361 208.124560. 1 . It appears that the integral is divergent. x 1. 1 1 < (b) For x 2 , x > x 1. . Since x x 1 . 2. . 2. 1 1 dx is divergent (Equation 2 with p= 1 ), 2 x. 1 dx is divergent by the Comparison Theorem. x 1. (c) Since. . f (x)dx is infinite and the area under g(x) is greater than the area under. f (x) on any interval. 2 . 2,t ,. g(x)dx must be infinite; that is, the integral is divergent. 2 . 2. 49. For x 1 ,. cos x 2. 1+x. . 1 2. 1+x. <. 1 2. x. .. 1 dx is convergent by Equation 2 with p=2>1 , so 1. 2. x. 14.

(15) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. . . 2. cos x. dx is convergent by the Comparison Theorem.. 2. 1+x. 1. x. 2+e 2 1 x 50. For x 1 , > [ since e >0 ] > . x x x . 1. 1 dx is divergent by Equation 2 with p=1 1 , so x. 1. x. 2+e x. dx is divergent by the Comparison Theorem. 2x. 1. 2x. 51. For x 1 , x+e >e >0. . . 2x. 2x. x+e 2x. e. dx=lim. t . 1. 1 2x e 2. t 1 t . . on 1, ) .. e. 2x. dx. 1. =e. 1 2t 1 2 1 2 e + e = e . Therefore, 2 2 2. =lim. by the Comparison Theorem,. 2x. 1. . x+e. . 2x. e. dx is convergent, and. 1. is also convergent. . x. 52. For x 1 , 0<. x. <. 1+x. 6. =. x. 6. x x. 3. =. 1 2. x. .. 1 dx is convergent by Equation 2 with p=2>1 , so 1. 2. x. . 1. x 1+x. dx is convergent by the Comparison Theorem. 6. 1 1 53. on x sin x x. /2. 0, 2 +. But ln t as t 0 , so. since 0 sin x 1 .. 0. /2. 0. /2. dx dx =lim =lim ln x x x + t + t 0. /2 t. .. t 0. dx is divergent, and by the Comparison Theorem, x. /2. 0. dx is also x sin x. divergent. x. 54. For 0 x 1 , e 1. x. e. x. . 1 . x 15.

(16) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. 1. 1. 1 1 dx=lim dx=lim 2 x +t + x x. 0. t 0. 1. 1 t. =lim +. t 0. ( 22 t ) =2 is convergent. Therefore, . x. e. x. 0. t 0. dx is. convergent by the Comparison Theorem. . . 55.. . 1. dx = x (1+x) 0. 0. 1. dx + x (1+x) 1. t. dx =lim +t x (1+x). dx +lim x (1+x) t 1. t 0. dx . Now x (1+x). 2 2udu dx = [ u= x ,x=u ,dx=2udu ] 2 x (1+x) u 1+u 1 1 du =2tan u+C=2tan x +C , =2 2 1+u. . (. ). so . dx =lim 2tan + x (1+x) t 0. 0. =lim. 1. 4. 2 +. t 0. 1. x. t. +lim. t . 2tan. 1. 2tan. 1. x. +lim. t. t . t 1. 2tan. 1. t 2. 4. =. 0+2 2. 2. . = 2. . . 56.. . dx 2. dx 2. x 4. x. =. . dx. =. x 4. x. 2. . . 3. 2. x. 2. x 4. . 3. dx. +. 3. x. 2. =lim. x 4. +. t 2. t. t. dx x. 2. x 4. +lim. t . 3. dx x. 2. . Now. x 4. 2sec tan d [ x=2sec , where 0 < /2 or <3 /2 ] 2sec 2tan . 1 1 1 = +C= sec 2 2. 1 x +C , so 2. . 2. dx x. 2. =lim +. t 2. 1 1 sec 2. 1 x 2. 3. +lim. t t . 1 1 sec 2. 1 x 2. t 3. x 4. 16.

(17) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. =. 1 1 sec 2 1. 57. If p=1 , then 0. 3 2. 2. 1 2. . 1. dx x=lim. p. dx. 0+. x =lim. + t 0 t. 1 1 sec 2. 3 2. 4. =. 1. ln x = . Divergent. t. +. t 0. If p 1 , then 1. 1. dx x 0. p. =lim. dx x. p. (note that the integral is not improper if p<0). +t. t 0. p+1. =lim +. t 0. 1. x p+1. t. =lim +. t 0. 1. If p>1 , then p1>0 , so t. p1. 1. If p<1 , then p1<0 , so t. p1. 1 1 p. 1. 1 t. p1. +. as t 0 , and the integral diverges. 1. 0 as t 0 and +. 0. p. dx. x =. 1 1 p. ( 1t1 p). lim +. t 0. Thus, the integral converges if and only if p<1 , and in that case its value is . 58. Let u=ln x . Then du=dx/x. e. =. 1 . 1 p. 1 . 1 p. . dx x ( ln x ). p. du. =. 1. u. p. . By Example 4, this converges to. 1 if p>1 p1. and diverges otherwise. 59. First suppose p=1 . Then 1. 1. 1. ln x ln x x ln xdx= x dx=lim x dx=lim +t + 0 0 p. t 0. 1 2 (ln x) 2. t 0. 1 t. =. 1 2 lim (ln t) = , so the integral diverges. 2 + t 0. Now suppose p 1 . Then integration by parts gives p+1. p. p+1. p+1. x x x x ln xdx= p+1 ln x p+1 dx= p+1 ln x x 2 +C . If p<1 , then p+1<0 , so ( p+1) p. 1. p+1. p+1. x x x ln xdx=lim ln x 2 p+1 + 0 ( p+1) t 0 If p>1 , then p+1>0 and p. 1. = t. 1. ( p+1 ). 2. . 1 p+1. lim. t +. t 0. p+1. ln t. 1 p+1. = .. 17.

(18) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. 1. x. 1. p. ln xdx =. 1 p+1. . 2. ( p+1). 0. 1. =. 1. +. 2. 2. ( p+1). ( p+1). 1 ln t1/( p+1) = ( p+1 ) 2 ( p+1) t. lim +. t 0. lim t. 1/t. lim +. t 0. ( p+1)t. ( p+2 ). 1. p+1. =. 2. +. ( p+1). t 0. 1. Thus, the integral converges to . 1 p+1. 2. if p>1 and diverges otherwise.. ( p+1) 60. (a) n=0 : . t. n x. x e. x. e. dx =lim. dx=lim. t 0. 0. t . x t. e. 0. t. e +1 =0+1=1. =lim. t . n=1 : . t. n x. x e. dx=lim. x. xe. t 0. 0. x. dx . To evaluate xe dx , we’ll use integration by parts. x. x. with u=x , dv=e dx du=dx , v=e x. x. x. x. .. x. x. So xe dx=xe e dx=xe e +C=(x1)e +C and t. lim. x t. x. xe. t 0. =lim (x1)e. dx. t . 0. t. t. t . t. te e +1. =lim (t1)e +1 =lim. t . =00+1 [ use l’Hospital’s Rule] =1 n=2 : . t. n x. x e. dx=lim. t 0. 0. 2 x. x e. 2 x. dx . To evaluate x e dx , we could use integration by parts. again or Formula 97. Thus, t. lim. 2 x. x e. t 0. dx. =lim. t . 2 x t. x e. t. +2lim. 0. x. xe. t 0. dx. =0+0+2(1) [ use l’Hospital’s Rule and the result for n=1 ] =2 18.

(19) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. n=3 : . t. n x. x e. dx =lim. 3 x. x e. t 0. 0. 3 x t. dx=lim. t . x e. t. +3lim. 0. 2 x. x e. t 0. dx. =0+0+3(2) [ use l’Hospital’s Rule and the result for n=2 ] =6 . (b) For n=1 , 2 , and 3 , we have. n x. x e. dx=1 , 2 , and 6 . The values for the integral are equal to the. 0 . factorials for n , so we guess. n x. x e. dx=n! .. 0 . (c) Suppose that. . k x. x e. dx=k! for some positive integer k . Then. 0 k+1 x. evaluate x k+1 x. x. t. lim. k+1. k+1 x. e dx=x. k x. t 0. e dx=lim. k+1 x t. e dx =lim. x. =lim. t. t t . e. x. e dx . To. . So. k x. e +(k+1) x e dx and t. +(k+1)lim. 0. k. k+1 x. x. t 0 x. , dv=e dx du=(k+1)x dx , v=e. k+1 x. e (k+1)x e dx=x. k+1 x. x. x 0. e dx , we use parts with u=x. t. k+1 x. k x. x e. dx. t 0. k+1 t. e +0 +(k+1)k!=0+0+(k+1)!=(k+1)! ,. so the formula holds for k+1 . By induction, the formula holds for all positive integers. (Since 0!=1 , the formula holds for n=0 , too.) . . 0. 61. (a) I= xdx= xdx+ xdx , and . . 0. . t. xdx=lim xdx=lim t 0. 0. t . 1 2 x 2. t. =lim. 0 t . 1 2 t 0 = , so I is 2. divergent. t. (b). xdx= t. 1 2 x 2. . t. t. t. 1 2 1 2 = t t =0 , so lim xdx=0 . Therefore, xdx lim xdx . t 2 2 t t t t . 2. M 4 3/2 3 kv 62. Let k= so that v= k v e dv . Let I denote the integral and use parts to integrate I . 2RT 0 2. 2. kv. Let =v , d =ve. 2. 1 kv : dv d =2vdv , = e 2k. 19.

(20) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. 2. t. (. 2. 2. ). 1 2 kv 1 kv 1 1 2 kt I =lim ve + ve dv= lim t e + lim 0 k 0 2k 2k t k t t 1 1 1 = 0 (01)= 2 2 2k 2k 2k. Thus, v=. 4 3/2 1 2 2 k = = 2 1/2 2k (k ) M/ ( 2RT ) . 63. Volume = 1. t. 2. 1 x. 1/2. =. dx x = lim. dx= lim. 2. t 1. t . 2 2 RT M. = lim. 1. t 0. 8RT . M. =. t. 1 x. 2. 1 kv e 2k. 1. t . 1 t. = < .. . . 64. Work =. R. t. GMm r. GMm dr=lim 2. dr=lim. 2. t R 24. t . r. t. =GMm. R. t . 1 1 + t R. =. GMm , where R 6. kg, m= mass of satellite =10 kg, R= radius of Earth =6.37

(21) 10 m, and 11. G= gravitational constant =6.67

(22) 10 11. Therefore, Work =. 1 r 3. M= mass of Earth =5.98

(23) 10. 2. N m / kg. 24. 3. 5.98

(24) 10 10. 6.67

(25) 10. GMm. 6. 10. 6.26

(26) 10. J.. 6.37

(27) 10 t . 65. Work = F dr=lim. GmM dr=lim GmM. t R. R. provides the work, so. r. 2. t . 1 2 GmM mv =. v = 0 2 0 R. 1 1 R t. =. GmM . The initial kinetic energy R. 2GM . R. R. . 66. y(s)=. s. 2r 2. x(r)dr and x(r)= 2. r s. 1 2 (Rr). 2. 20.

(28) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. R. y(s) = lim +. t s. R. . 2. r(Rr) 2. R +. t. t s. . R. 3. r dr. . t s. +. r s. t. = lim. dr=lim. 2. 3. 2. 2. 2. dr. 2. r s. t. R. 2. r r s 2R t dr 2. 2. r 2Rr +R r. 2. 2. 2. r s +R. r. dr. 2. 2. r s. t. 2. = lim + I 12RI 2+R I 3 =L t s. 2. 2. 2. 2. 2. 2. 2. 2. For I : Let u= r s u =r s , r =u +s , 2r dr=2udu , so, omitting limits and constant of 1. integration, I = 1 =. 2 2 ( u +s ) u 1 3 2 1 2 2 2 2 du= ( u +s ) du= u +s u= u ( u +3s ). 3. u. 1 3. 2. 2. r s. (r 2s2+3s2) = 13. 2. 3. (r 2+2s2). 2. r s. r For I : Using Formula 44, I = 2 2 2. 2. s 2 2 . r s + ln r+ r s 2 1 du 1 2 2 2 2 For I : Let u=r s du=2r dr . Then I = = 2 u = r s . 3 3 2 u 2 Thus,. +. 1 3. +. 1 3. L = lim t s. = lim t s. lim. +. = 1 3. 1 3. 2. 2. R s. 2. 2. 2. ( R +2s ) 2R. R 2. 2. 2. ( t +2s ) 2R. t 2. t s 2. 2. 2. r 2. (r +2s ) 2R. R s. R s 2. 2. r s. 1 3. t s. =. 2. 2. 2. 2. 2. 2. ( R2+2s2) Rs2ln. ( R +2s ) Rs ln 2. 2. 2. R. 2. s 2 2 r s + ln r+ r s 2 2. 2. 2. 2. 2. 2. s 2 2 t s + ln t+ t s 2 2. 2. 2. 2. 2. 2. R+ R s. 2. +R 2. +R. 2. r s. +R. s 2 2 R s + ln R+ R s 2 2. 2. 2. t. 2. 2. R s 2. t s. 2. Rs ln s. R+ R s s. 21.

(29) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. 67. (a) /. (b) r(t)=F (t) is the rate at which the fraction F(t) of burntout bulbs increases as t increases. This could be interpreted as a fractional burnout rate. . (c). r(t)dt=lim. x . 0. F(x)=1 , since all of the bulbs will eventually burn out.. 68. . I = te dt=lim kt. s . 0. 1 k. 2. s . [ Formula 96, or parts] 0. 1 ks 1 ks se e 2 k k. =lim. s. kt. ( kt1 ) e. . . 1 k. .. 2. Since k<0 the first two terms approach 0 (you can verify that the first term does so with l’Hospital’s. ( 2). 2. Rule), so the limit is equal to 1/k . Thus, M=kI=k 1/k =1/k=1/(0.000121) 8264.5 years. . . 69. I=. a. t. 1. t . dx=lim. 2. x +1. 1. 2 a x +1 1. 1. dx=lim. t . t. tan x =lim a. 1 tan a<0.001 tan a> 0.001 a>tan 2 2. t . ( tan 1ttan 1a) = 2 tan 1a . I<0.001. 0.001 1000 . 2. 2. x. 4 0 1 = . 8 2 4 1 0 f (x)dx S8 = 2 3 [ f (0)+4 f (0.5)+2 f (1)+ +2 f (3)+4 f (3.5)+ f (4)] 1. (5.31717808) 0.8862 6. 70. f (x)=e. and x=. 2. x. 4x. Now x>4 x x<x 4 e <e. 2. x. e 4. 4x. dx< e 4. dx .. 22.

(30) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. 4x. 4. e. dx=lim. t . t. 1 4x e 4. . (. 4 . st. 1 16 0e =1 4. =. st. 71. (a) F(s)= f (t)e dt= e dt=lim 0. n . 0. if s>0 . Therefore F(s)=. / ( 4e ) 0.0000000281<0.0000001 , as desired.. ). 16. st. . sn. n. e. =lim. s. 0 n . e 1 + s s. . This converges to. 1 only s. 1 with domain { s|s>0} . s. (b) . . st. t st. F(s) = f (t)e dt= e e dt=lim 0. =lim. ( 1s ) n. . 1s. n . e (. t 1s ). n 0. 0. e. n. n . st. (c) F(s)= f (t)e dt=lim . Then F(s)=lim. n . F(s)=. 1 2. n. st. te. n 0. 0. t. st. e . s. 0. 1 1s. This converges only if 1s<0 s>1 , in which case F(s)= . n. 1 t ( 1s ) e 1s. dt=lim. 2. e. s. st. dt . Use integration by parts: let u=t , dv=e dt du=dt , v=. st. 1. 1 with domain { s|s>1} . s1. n. n. =lim. 0 n . se. . sn. 1 2 sn. +0+. se. 1 2. s. =. 1 2. st. e. s. .. only if s>0 . Therefore,. s. and the domain of F is { s|s>0} .. s. st. at. at st. 72. 0 f (t) Me 0 f (t)e Me e . Me. n. at st. t ( as ). e dt=lim M e n . 0. 0. dt=M lim. n . for t 0 . Now use the Comparison Theorem: 1 t ( as ) e as. n. =M lim. 0. n . 1 n ( as ) e 1 as. This is convergent only when as<0 s>a . Therefore, by the Comparison Theorem, . st. F(s)= f ( t ) e dt is also convergent for s>a . 0 . st. st. 73. G(s)= f (t)e dt . Integrate by parts with u=e /. /. st. , dv= f (t)dt du=se. , v= f (t) :. 0 23.

(31) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. . st n. n . st. sn. +s f (t)e dt=lim f (n)e. f (t)e. G(s)=lim. 0. f (0)+sF(s). n . 0 st. at. at st. t ( as ). But 0 f (t) Me 0 f (t)e Me e. and lim Me t . st. =0 for s>a . So by the Squeeze Theorem,. lim f (t)e =0 for s>a G(s)=0 f (0)+sF(s)=sF(s) f (0) for s>a .f}(0) for s>a .. t . 74. Assume without loss of generality that a<b . Then a. . . . a. u. t t. u a. f (x)dx+ f (x)dx = lim a. f (x)dx+lim f (x)dx a. f (x)dx+lim. = lim. t t. u. a. b. f (x)dx+ f (x)dx. u . a. b. u. t t. a. u b. f (x)dx+ f (x)dx+lim f (x)dx. = lim. a. b. t. a. t b. . t t. b. . . b. . f (x)dx+ f (x)dx b 2. x. 75. We use integration by parts: let u=x , dv=xe . x e 0. 2. 2 x. 2. dx = lim. t . = lim. t . + f (x)dx. f (x)dx+ f (x)dx. = lim b. . f (x)dx+ f (x)dx. = lim. =. b. 1 x xe 2. t / ( 2e ) t. 2. t. 1 + 0 2 1 + 2. . 2. 1 x dx du=dx , v= e . So 2. 2. x. e. dx. 0 . e 0. 2. x. 1 dx= 2. . 2. x. e. dx. 0. (The limit is 0 by l’Hospital’s Rule.) 76. 24.

(32) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. . 2. x. e. 2. x. dx is the area under the curve y=e. 2. x. for 0 x< and 0<y 1 . Solving y=e. for x , we get. 0 2. x. 2. 2. ln y=x ln y=x x= ln y . Since x is positive, choose x= ln y , and the area is. y=e. 1. represented by ln y dy . Therefore, each integral represents the same area, so the integrals are 0. equal. 2. 77. For the first part of the integral, let x=2tan dx=2sec d . 2. 1. . 2. x +4. 2sec x dx= d = sec d =ln sec +tan . From the figure, tan = , and 2sec 2 2. sec =. x +4 . So 2. . . t. 2. 1. I=. 2. . x +4. 0. C x+2. dx. = lim. x +4 x + 2 2. ln. t . Cln x+2. 0. 2. = lim. t +4 +t Cln (t+2) ( ln 1Cln 2 ) 2. ln. t . 2. = lim. t +4 +t. ln. t . 2 ( t+2 ). C. +ln 2. C. 2. = ln 2. Now L=lim. t . t+ t +4. ( t+2 ). C. lim. t+ t +4. t . ( t+2 ). 2. =lim. t . 1+t/ t +4 C ( t+2 ). C1. C. C1. +ln 2. 2. =. Clim ( t+2 ). C1. .. t . If C<1 , L= and I diverges. If C=1 , L=2 and I converges to 25.

(33) Stewart Calculus ET 5e 0534393217;7. Techniques of Integration; 7.8 Improper Integrals. 0. ln 2+ln 2 =ln 2 . If C>1 , L=0 and I diverges to . 78. . . I=. 0. x 2. x +1. C 3x+1. dx. =lim. t . =lim. t . =lim. t . 1 1 2 ln x +1 Cln (3x+1) 2 3. (. ). t 0. ( 2 ) 1/2ln (3t+1)C/3. ln t +1. ln. ( t2+1) 1/2 ( 3t+1 ). 2. =ln. C/3. t +1. lim. t . ( 3t+1 ). C/3. For C 0 , the integral diverges. For C>0 , we have 2. L=lim. t +1. =lim. t (3t+1)C/3 t . t. /. 2. t +1. C ( 3t+1 ). (C/3)1. For C/3<1C<3 , L= and I diverges. For C=3 , L= to .. =. 1 lim C t . 1 . ( C/3 ) 1. ( 3t+1 ). 1 1 and I=ln . For C>3 , L=0 and I diverges 3 3. 26.

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