2006a_x1a_sols

2006a_x1a_sols - Spring 2006 Math 152 Exam 1A: Solutions c...

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Spring 2006 Math 152 Exam 1A: Solutions Mon, 20/Feb c ± 2006, Art Belmonte 1. (c) We have f ave = 1 b - a ± b a f ( x ) dx = 1 4 - 1 ± 4 1 ( x 2 - 1 ) 1 / 2 xdx = ² 1 3 ³² 1 2 2 3 ³ ( x 2 - 1 ) 3 / 2 ´ ´ ´ 4 1 = 1 9 · 15 15 - 0 = 5 3 15. 2. (d) Use integration by parts. First compute an antiderivative, then apply the FTC. Let u = xd v = e - 2 x du = v =- 1 2 e - 2 x . Then ± xe - 2 x 1 2 - 2 x + ± 1 2 e - 2 x 1 2 - 2 x - 1 4 e - 2 x 1 4 ( 2 x + 1 ) e - 2 x . Hence ± 1 0 - 2 x = ² - 1 4 ( 2 x + 1 ) e - 2 x ³ ´ ´ 1 0 = ² - 3 4 e - 2 ³ - ² - 1 4 ³ = 1 - 3 e - 2 4 . 3. (b) Use trigonometric substitution. Let x = 5sin θ .Then = 5cos θ d θ andwehavethetable x 0 5 θ 0 π/ 2 .So µ 5 0 25 - x 2 = µ 2 0 θ · θ d θ = 25 2 µ 2 0 1 + cos 2 θ d θ = 25 2 ² θ + 1 2 sin 2 θ ³ ´ ´ ´ 2 0 = 25 4 π - 0 = 25 4 π. [ Alternatively , the integral ± 5 0 25 - x 2 represents the area in the first quadrant under the curve y = 25 - x 2 ,part of the circle x 2 + y 2 = 25 = 5 2 . This quarter-circular area is 1 4 π r 2 = 1 4 π ( 5 ) 2 = 25 4 π .] 4. (d) 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 ² x 2 - 1 ³ + B ² x 2 + x ³ + C ² x 2 - x ³ 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 .
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2006a_x1a_sols - Spring 2006 Math 152 Exam 1A: Solutions c...

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