prob3practicesol

prob3practicesol - Math 151 Spring 2008 Problem Set # 3...

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Math 151 Spring 2008 Problem Set # 3 Integration by Parts Solutions of the Practice Problems Remark: An arbitrary constant can be added to an antiderivative. 2. Set u = x 2 and dv = e x/ 3 dx so that du =2 xdx and v = Z e x/ 3 dx = 3 e x/ 3 . Therefore, Z x 2 e x/ 3 dx = Z udv = uv Z vdu = x 2 ³ 3 e x/ 3 ´ +3 Z e x/ 3 2 xdx = 3 x 2 e x/ 3 +6 Z xe x/ 3 dx. We set u = x and dv = e x/ 3 dx so that du = dx and v = 3 e x/ 3 . Therefore, Z xe x/ 3 = Z udv = uv Z vdu = x ³ 3 e x/ 3 ´ Z e x/ 3 dx = 3 xe x/ 3 9 e x/ 3 . Thus, Z x 2 e x/ 3 dx = 3 x 2 e x/ 3 Z xe x/ 3 dx = 3 x 2 e x/ 3 ³ 3 xe x/ 3 9 e x/ 3 ´ = 3 x 2 e x/ 3 18 xe x/ 3 54 e x/ 3 (a constant can be added). 3. Set u = x and dv =cos( x/ 2) so that du = dx and v = Z cos ( x/ 2) dx =2sin( x/ 2) . Therefore, Z x cos μ 1 2 x dx = Z udv = uv Z vdu = x (2 sin ( x/ 2)) 2 Z sin ( x/ 2) dx x sin ( x/ 2) + 4 cos ( x/ 2) . 1
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6. We set u = x 2 and dv =cosh( x ) dx so that du =2 xdx and v = Z cosh ( x ) dx =s inh( x ) . Thus. Z x 2 cosh ( x ) dx = Z udv = uv Z vdu = x 2 sinh ( x ) 2 Z x sinh ( x ) dx. As in Problem 5, Z x sinh ( x ) dx = x cosh ( x ) sinh ( x ) . Therefore, Z x 2 cosh ( x ) dx = x 2 sinh ( x ) 2 Z x sinh ( x ) dx = x 2 sinh ( x ) 2( x cosh ( x ) sinh ( x )) = x 2 sinh ( x ) 2 x cosh ( x )+2sinh( x ) .
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This homework help was uploaded on 04/21/2008 for the course MATH 151 taught by Professor Geveci during the Winter '08 term at San Diego State.

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prob3practicesol - Math 151 Spring 2008 Problem Set # 3...

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