Thomas' Calculus 13th Edition

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

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 =$[\frac{1}{2}y^2]_{-1}^2 =\frac{3}{2}$
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