{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Assignment 4 Solutions

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

This preview shows pages 1–2. Sign up to view the full content.

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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}