finalpractice_soln

# finalpractice_soln - Solution to MAT21B Practice Final Exam...

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Solution to MAT21B Practice Final Exam 1. (5 points each) Evaluate the following integrals. (a) R ( x + 1) 2 e x dx Solution: Using integration by parts twice, we have Z ( x + 1) 2 e x dx = ( x + 1) 2 e x - Z 2 ( x + 1) e x dx = ( x + 1) 2 e x - 2 ( x + 1) e x - 2 Z e x dx = ( x + 1) 2 e x - [2 ( x + 1) e x - 2 e x ] + C = e x ( x 2 + 1 ) + C (b) R x +2 x 3 - 3 x 2 +2 x dx Solution: x + 2 x 3 - 3 x 2 + 2 x = x + 2 x ( x - 2) ( x - 1) = A x + B x - 2 + C x - 1 By the Heaviside method, A = x + 2 ( x - 2) ( x - 1) | x =0 = 2 2 = 1 B = x + 2 x ( x - 1) | x =2 = 4 2 = 2 C = x + 2 x ( x - 2) | x =1 = 3 - 1 = - 3 . So, Z x + 2 x 3 - 3 x 2 + 2 x dx = Z 1 x + 2 x - 2 + - 3 x - 1 dx = ln | x | + 2 ln | x - 2 | - 3 ln | x - 1 | + C (c) R 6 x - 5 (3 x 2 - 5 x +7) 2 / 3 dx Solution: Let u = 3 x 2 - 5 x + 7, then du = 6 x - 5 and Z 6 x - 5 (3 x 2 - 5 x + 7) 2 / 3 dx = Z du u 2 / 3 = 3 u 1 3 + C = 3 ( 3 x 2 - 5 x + 7 ) 1 3 + C 1

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(d) R sin 3 x cos 4 xdx Solution: Let u = cos x , then du = - sin xdx , and Z sin 3 x cos 4 xdx = - Z ( 1 - u 2 ) u 4 du = Z u 6 - u 4 du = u 7 7 - u 5 5 + C = cos 7 x 7 - cos 5 x 5 + C (e) R e 1 ln x x dx Solution: Z e 1 ln x x dx = Z e 1 1 2 ln x x dx = Z 1 0 1 2 udu , by setting u = ln x = u 2 4 | 1 0 = 1 4 .
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