Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 15 - Multiple Integrals - Review - True-False Quiz - Page 1101: 4

Answer

True

Work Step by Step

$\int_{-1}^{1} \int_{0}^{1} e^{x^2+y^2} \sin y \, dx \, dy = \int_{0}^{1} e^{x^2} \, dx \int_{-1}^{1} e^{y^2} \sin y \, dy = \int_{0}^{1} e^{x^2} \, dx (0) = 0$. Note that $\int_{-1}^{1} e^{y^2} \sin y \, dy = 0$ because $\sin{y}$ is odd and integrating from $-1$ to $1$ will be $0$.
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