q9-sol-1042-sp08

q9-sol-1042-sp08 - Calculus II-1042 Solution of Quiz #9...

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Calculus II—1042 Solution of Quiz #9 Fri., 03/28/08 1. Find the volume of the solid generated by revolving the infinite region in the 1st quadrant between y = e - x and x - axis about the x - axis y=e -x 0 x y Solution: Area of cross section is A ( x ) = π r 2 = π ( e - x ) 2 = πe - 2 x . Integral for the volume with limits fron x = 0 to x = is V = Z 0 π e - 2 x dx, which is an improper integral with the upper limit as a problem point. First write the integral as a limit them compute it using Z e kx dx = e kx k V = Z 0 π e - 2 x dx = lim t →∞ Z t 0 πe - 2 x dx = lim t →-∞ ± π · e - 2 x - 2 | t 0 = lim t →∞ ± π · e - 2 t - 2 - π · e 0 - 2 = 0 + π 2 = π 2 2. Use The Direct Comparison Test or Limit Comparison Test to determine whether the integral Z π 2 + cos x x 3 dx is convergent or divergent. Solution: Since - 1 cos x 1 = 1 x 3 2 + cos x x 3 3 x 3 (cross the function 1 x 3 in the left side, which can’t be used for the direct Comparison Test)
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This homework help was uploaded on 04/18/2008 for the course MATH 1042 taught by Professor Dr.z during the Spring '08 term at Temple.

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q9-sol-1042-sp08 - Calculus II-1042 Solution of Quiz #9...

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