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: 3



Work Step by Step

$\int^{1}_{-2} \int^{-y^2}_{y-2}dx dy $ =$\int^{1}_{-2} (-y^2-y+2)dy$ =$[\frac{-y^3}{3}-\frac{y^2}{2}+2y]^1_{-2}$ =$(\frac{-1}{3}-\frac{1}{2}+2)-(\frac{8}{3}-2-4)$ $ =\frac{9}{2}$
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