Answer
$\frac{4}{3}$
Work Step by Step
$\int^{2}_{0} \int^{y-y^2}_{-y}dxdy$
=$\int^{2}_{0}(2y-y^2)dy$
=$[y^2-\frac{y^3}{3}]^2_0$
=$4-\frac{8}{3}$
=$\frac{4}{3}$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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
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