Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 15 - Section 15.2 - Double Integrals over General Regions - 15.2 Exercise - Page 1008: 5

Answer

$\frac{\sin \left(1\right)}{3}$

Work Step by Step

$\int _0^1\int _0^{s^2}\cos \left(x^3\right)dtds$ $\int _0^{x^2}\cos \left(s^3\right)dt=s^2\cos \left(s^3\right)$ Note: $\int \:x^2\cos \left(x^3\right)dx=\frac{1}{3}\sin \left(x^3\right)+C$ So, $\int _0^1s^2\cos \left(s^3\right)dx=\frac{\sin \left(1\right)}{3}$
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