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Assignment 4 Solutions

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

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MATH 138 Calculus 2, Solutions to Assignment 4 1: Let R be the region 0 x π 4 , 2 sin 2 x y 1. (a) Find the area of R by integrating with respect to x . Solution: The area is A = π/ 4 x =0 1 - 2 sin 2 x dx = π/ 4 0 cos 2 x dx = 1 2 sin 2 x π/ 4 0 = 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 π 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 = 1 y =0 1 2 cos - 1 (1 - y ) dy . Make the substitution t = 1 - y , dt = - dy to get A = 0 t =1 - 1 2 cos - 1 t dt = 1 0 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 + 1 2 t dt 1 - t 2 1 0 = 1 2 t cos - 1 t - 1 2 1 - t 2 1 0 = 1 / 2 . For the second solution, note that for 0 x π/ 4 we have y = 2 sin 2 x ⇐⇒ 1 2 y = sin 2 x ⇐⇒ 1 2 y = sin x ⇐⇒ x = sin - 1 1 2 y . The region R is given by 0 y 1, 0 x sin - 1 1 2 y , and the area is A = 1 y =0 sin - 1 1 2 y dy . Make the substitution t = 1 2 y , 2 t 2 = y , 4 t dt = dy to get A = 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 t dt to get A = 2 t 2 sin - 1 t - 2 t 2 dt 1 - t 2 1 / 2 0 = π 4 - 1 / 2 0 2 t 2 dt 1 - t 2 .
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