prob7sol

# prob7sol - Math 151 Spring 2008 Problem Set # 7 Solutions...

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Math 151 Spring 2008 Problem Set # 7 Solutions 1. y 2 1 -1.0 -2 -1 -0.5 0 0 0.0 1 2 z 0.5 x 1.0 -1 -2 f ( x )= e x Therequ iredvo lumeis π Z ln(2) ln(2) ¡ e x ¢ 2 dx = π Z ln(2) ln(2) e 2 x dx = π Ã 1 2 e 2 x ¯ ¯ ¯ ¯ ln(2) ln(2) ! = π 2 ³ e 2ln(2) e ´ = π 2 ¡ 2 2 2 2 ¢ = 15 π 8 2. y 1 -1 0 0 4 1 z 5 6 -1 7 8 9 10 x 11 f ( x )=sin( x 4 ) π Z 3 π 2 π sin 2 ³ x 4 ´ dx = π Z 3 π 2 π 1 cos ³ x 2 ´ 2 = π μ x 2 sin ³ x 2 ´¯ ¯ ¯ 3 π 2 π = π μ 3 π 2 sin μ 3 π 2 π +sin( π ) = π ³ π 2 +1 ´ 1

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3. The required volume is π Z e 3 4 e 2 4 μr x x 2 +4 2 dx = π Z e 3 4 e 2 4 x x 2 dx. We have π Z x x 2 dx = π 2 ln ¡ x 2 ¢ . Thus, π Z e 3 4 e 2 4 x x 2 dx = π 2 ln ¡ e 3 ¢ π 2 ln ¡ e 2 ¢ = π 2 . 4. The required volume is π Z 1 1 Ã r 1 x 2 +1 ! 2 dx = π Z 1 1 1 x 2 dx = π ³ arctan ( x ) | 1 1 ´ = π ³ π 4 ³ π 4 ´´ = π 2 2 .
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## This note was uploaded on 04/21/2008 for the course MATH 151 taught by Professor Geveci during the Winter '08 term at San Diego State.

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prob7sol - Math 151 Spring 2008 Problem Set # 7 Solutions...

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