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 15.3 - Area by Double Integration - Exercises 15.3 - Page 887: 7



Work Step by Step

$\int^{1}_{0} \int^{2y-y^2}_{y^2}dx dy$ =$\int^{1}_{0}(2y-2y^2)dy$ =$[y^2-\frac{2}{3}y^3]_0^1$ =$\frac{1}{3}$
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