2005a_x1b_sols

2005a_x1b_sols - Spring 2005 Math 152 Exam 1B: Solutions c...

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Spring 2005 Math 152 Exam 1B: Solutions Mon, 21/Feb c ± 2005, Art Belmonte For specificity, lengths are in centimeters unless stated otherwise. Graphs appear at the bottom of the right column to conserve space. 1. (d) First compute an antiderivative, then apply the FTC. Let u = ln xd v = xdx du = 1 x dx v = 1 2 x 2 . Then R x ln = 1 2 x 2 ln x - R 1 2 = 1 2 x 2 ln x - 1 4 x 2 . Hence R 3 1 x ln = ± 1 2 x 2 ln x - 1 4 x 2 ² ³ ³ 3 1 = ± 9 2 ln 3 - 9 4 ² - ± - 1 4 ² = 9 2 ln 3 - 2. 2. (b) We have dV = Adx = π r 2 = π y 2 = π ( e x ) 2 = π e 2 x . Therefore V = R 3 0 π e 2 x = 1 2 π e 2 x ³ ³ 3 0 = π 2 ( e 6 - 1 ) 621 . 13 cm 3 . 3. (e) Find where the curves intersect, draw a sketch, then set up and compute the integral for the volume. Now x 2 = 2 x implies x 2 - 2 x = 0or x ( x - 2 ) = 0 whence x = 0 , 2. (See figure at bottom of column.) We have = = ( π r 2 o - π r 2 i ) = π ± ( 2 x ) 2 - ( x 2 ) 2 ² = π ( 4 x 2 - x 4 ) . The volume is V = π R 2 0 4 x 2 - x 4 = π ± 4 3 x 3 - 1 5 x 5 ² ³ ³ 2 0 = π ± 32 3 - 32 5 ² = 32 π ± 5 - 3 15 ² = 64 15 π 13 . 40 cm 3 . 4. (c) We have R 2 1 x x + 1 = R 2 1 1 - 1 x + 1 = ( x - ln | x + 1 | ) ³ ³ 2 1 = ( 2 - ln 3 ) - ( 1 - ln 2 ) = 1 + ln 2 - ln 3 = 1 + ln 2 3 .
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2005a_x1b_sols - Spring 2005 Math 152 Exam 1B: Solutions c...

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