Assignment 4 Solutions

Assignment 4 Solutions - MATH 138 Calculus 2, Solutions to...

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Unformatted text preview: MATH 138 Calculus 2, Solutions to Assignment 4 1: Let R be the region 0 x 4 , 2 sin 2 x y 1. (a) Find the area of R by integrating with respect to x . Solution: The area is A = Z / 4 x =0 1- 2 sin 2 x dx = Z / 4 cos 2 x dx = h 1 2 sin 2 x i / 4 = 1 2 . (b) Find the area of R by integrating with respect to y . Solution: We provide two solutions. For the first solution, note that for 0 x 4 we have y = 2 sin 2 x y = 1- cos 2 x cos 2 x = 1- y x = 1 2 cos- 1 (1- y ) . The region R is given by 0 y 1, 0 x 1 2 cos- 1 (1- y ), and the area is also given by A = Z 1 y =0 1 2 cos- 1 (1- y ) dy . Make the substitution t = 1- y , dt =- dy to get A = Z t =1- 1 2 cos- 1 t dt = Z 1 1 2 cos- 1 t dt. Integrate by parts using u = cos- 1 t , du =- dt 1- t 2 , v = 1 2 t , dv = 1 2 dt to get A = 1 2 t cos- 1 t + Z 1 2 tdt 1- t 2 1 = 1 2 t cos- 1 t- 1 2 1- t 2 1 = 1 / 2 . For the second solution, note that for 0 x / 4 we have y = 2 sin 2 x 1 2 y = sin 2 x q 1 2 y = sin x x = sin- 1 q 1 2 y . The region R is given by 0 y 1, 0 x sin- 1 q 1 2 y , and the area is A = Z 1 y =0 sin- 1 q 1 2 y dy . Make the substitution t = q 1 2 y , 2 t 2 = y , 4 tdt = dy to get A = Z 1 / 2 t =0 4 t sin- 1 t dt. Now integrate by parts using u = sin- 1 t , du = dt 1- t 2 , v = 2 t 2 and dv = 4 tdt to get A = 2 t 2 sin- 1 t- Z 2 t 2 dt 1- t 2 1 / 2 = 4- Z 1 / 2 2 t 2 dt 1- t 2 ....
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This note was uploaded on 09/18/2008 for the course MATH MAth 138 taught by Professor Marshman during the Spring '08 term at Waterloo.

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Assignment 4 Solutions - MATH 138 Calculus 2, Solutions to...

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