2006a_x1b_sols

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 = 1 x v = x . Then ± ln ( 2 x ) = x ln ( 2 x ) - ± 1 = x ln ( 2 x ) - x = x ( ln ( 2 x ) - 1 ) . Hence ± e 1 ln ( 2 x ) = x ( ln ( 2 x ) - 1 ) ² ² 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 ³ x 2 - 1 ´ + B ³ x 2 + x ´ + C ³ 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 µ 1 x ( x - 1 x + 1 ) = µ - 1 x + 1 2 x - 1 + 1 2 x + 1 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 ³ 1 4 ´ 2 = 1 16 < 1 2 = ¹ 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 .
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This note was uploaded on 03/25/2008 for the course MATH 152 taught by Professor Teitler during the Spring '08 term at Texas A&M.

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

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