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

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