Answer
$\frac{3}{2}$
Work Step by Step
$\int^{2}_{-1}$ $\int^{\frac{\pi}{2}}_{0}$ y sinx dx dy =$\int^{2}_{-1}$ $[-ycosx]^{\frac{\pi}{2}}_{0}$ dy=$\int^{2}_{-1}$ ydy =$[\frac{1}{2}y^2]_{-1}^2 =\frac{3}{2}$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 15: Multiple Integrals - Section 15.1 - Double and Iterated Integrals over Rectangles - Exercises 15.1 - Page 874: 11
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