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: 4



Work Step by Step

$\int^{2}_{0} \int^{y-y^2}_{-y}dxdy$ =$\int^{2}_{0}(2y-y^2)dy$ =$[y^2-\frac{y^3}{3}]^2_0$ =$4-\frac{8}{3}$ =$\frac{4}{3}$
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