10Bquiz3sol

# 10Bquiz3sol - Z 2 x x 3-x 2 x-1 dx Solution We use the...

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Quiz 3 Solution for MATH 10 B, 2010 1 . Evaluate the integral Z 1 0 arctan xdx . (Note that d dx arctan x = 1 1 + x 2 ). Solution. Using the integration by parts with u = tan - 1 x , v 0 = 1, u 0 = 1 1 + x 2 , v = x the integral is Z 1 0 tan - 1 xdx = h x tan - 1 x i 1 0 - Z 1 0 x 1 + x 2 dx = tan - 1 ( 1 ) - ± 1 2 ln ( 1 + x 2 ) ² 1 0 = π 4 - ³ 1 2 ln 2 - 1 2 ln 1 ´ = π 4 - 1 2 ln 2

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2 . (a) Find the partial fraction decomposition of 2 x x 3 - x 2 + x - 1 Solution. We set 2 x x 3 - x 2 + x - 1 = 2 x ( x - 1 )( x 2 + 1 ) = A x - 1 + Bx + C x 2 + 1 Then, that means 2 x ( x - 1 )( x 2 + 1 ) = A ( x 2 + 1 ) + ( Bx + C )( x - 1 ) ( x - 1 )( x 2 + 1 ) or, 2 x = A ( x 2 + 1 ) + ( Bx + C )( x - 1 ) x = 1 : 2 = 2 A A = 1 x = 0 : 0 = 1 - C C = 1 x = - 1 : - 2 = 2 + ( - B + 1 )( - 2 ) = 2 B B = - 1 Thus, 2 x x 3 - x 2 + x - 1 = 1 x - 1 + - x + 1 x 2 + 1 (b) Find the integral
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Unformatted text preview: Z 2 x x 3-x 2 + x-1 dx Solution. We use the partial fraction decomposition that we found in part (a) : Z 2 x x 3-x 2 + x-1 dx = Z 1 x-1 +-x + 1 x 2 + 1 dx = Z 1 x-1 dx-Z x x 2 + 1 dx + Z 1 x 2 + 1 dx = ln | x-1 |-1 2 ln ( x 2 + 1 ) + tan-1 x + C...
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10Bquiz3sol - Z 2 x x 3-x 2 x-1 dx Solution We use the...

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