Answer
$\frac{2}{\pi}$
Work Step by Step
$\int^{1}_{0}\int^{\pi}_{0}y \cos xy dxdy$
=$\int^{1}_{0}[sinxy]^{\pi}_0 dy$
=$\int^{1}_{0}\sin \pi y dy$
=$[\frac{-1}{\pi}cos \pi y]_0^1$
=$\frac{-1}{\pi}(-1-1)$
=$\frac{2}{\pi}$
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 875: 24
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