prelim2_fa09solutions - Math 1910, Prelim 2 October 29,...

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Math 1910, Prelim 2 October 29, 2009 Solutions 1) a) Letting u = ln( x ) we have du = dx x , x = 1 u = 0, x = 2 u = ln(2). Thus Z 2 1 log 2 ( x ) x dx = 1 ln(2) Z 2 1 ln( x ) x dx = Z ln(2) 0 udu = 1 ln(2) ± 1 2 u 2 ² ln(2) 0 = ln(2) 2 b) Letting u = 3 - e x we have du = - e x dx , x = 0 u = 2, x = ln(2) u = 1. Therefore Z ln(2) 0 e x (3 - e x ) 2 dx = Z 1 2 - du u 2 = ± 1 u ² 1 2 = 1 1 - 1 2 = 1 2 c) Letting u = e 2 x we have e 4 x = u 2 and du = 2 e 2 x dx . So Z e 2 x 1 + e 4 x dx = 1 2 Z du 1 + u 2 = 1 2 tan - 1 ( u ) + C = 1 2 tan - 1 ( e 2 x ) + C d) Use integration by parts. Z ln 2 ( x ) dx = Z 1 · ln 2 ( x ) dx = x ln 2 ( x ) - Z x · 2 ln( x ) 1 x dx = x ln 2 ( x ) - 2 Z ln( x ) dx = x ln 2 ( x ) - 2 Z 1 · ln( x ) dx = x ln 2 ( x ) - 2( x ln( x ) - Z x · 1 x dx ) = x ln 2 ( x ) - 2 x ln( x ) + 2 Z dx = x ln 2 ( x ) - 2 x ln( x ) + 2 x + C e) Write x +1 x 3 + x = A x + Bx + C x 2 +1 . Then x + 1 = A ( x 2 + 1) + ( Bx + C ) x = ( A + B ) x 2 + Cx + A . This shows that
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prelim2_fa09solutions - Math 1910, Prelim 2 October 29,...

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