2006a_x1b_sols - Spring 2006 Math 152 Exam 1B Solutions c 2006 Art Belmonte Mon 20\/Feb 1(b Use integration by parts First compute an antiderivative then

2006a_x1b_sols - Spring 2006 Math 152 Exam 1B Solutions c...

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Spring 2006 Math 152 Exam 1B: Solutions Mon, 20/Feb c 2006, Art Belmonte 1. (b) Use integration by parts. First compute an antiderivative, then apply the FTC. Let u = ln ( 2 x ) d v = dx du = 2 2 x dx = 1 x dx v = x . Then ln ( 2 x ) dx = x ln ( 2 x ) - 1 dx = x ln ( 2 x ) - x = x ( ln ( 2 x ) - 1 ) . Hence e 1 ln ( 2 x ) dx = x ( ln ( 2 x ) - 1 ) e 1 = ( e ( ln ( 2 e ) - 1 )) - ( ln 2 - 1 ) = e ( ln 2 + 1 - 1 ) - ln 2 + 1 = e ln 2 - ln 2 + 1. 2. (e) We’ll integrate the rational function via partial fractions. Split the integrand into a sum of partial fractions. 1 x ( x - 1 ) ( x + 1 ) = A x + B x - 1 + C x + 1 1 = A x 2 - 1 + B x 2 + x + C x 2 - x 0 x 2 + 0 x + 1 = ( A + B + C ) x 2 + ( B - C ) x + - A Equate coefficients of like terms. Thus 1 = - A , whence A = - 1. Next B - C = 0 implies C = B . Substituting for A and C in A + B + C = 0 yields 2 B - 1 = 0, whence B = 1 2 = C . Therefore, 1 x ( x - 1 ) ( x + 1 ) = - 1 x + 1 2 x - 1 + 1 2 x + 1 . Integrate term-by-term. Recall that x > 1. Hence 1 x ( x - 1 ) ( x + 1 ) dx = - 1 x + 1 2 x - 1 + 1 2 x + 1 dx = - ln x + 1 2 ln ( x - 1 ) + 1 2 ln ( x + 1 ) + C = ln x 2 - 1 x + C via the properties of logarithms. 3. (d) When the curves y = x 2 and y = x intersect, their y -coordinates are equal. Thus x 2 = x implies x 4 = x . Hence 0 = x 4 - x = x ( x 3 - 1 ) whence x = 0 , 1. Since 1 4 2 = 1 16 < 1 2 = 1 4 , we conclude that y = x 2 lies below y = x on [0 , 1]. Therefore the area of the region is given by 1 0 x - x 2 dx .

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