Answer
$\frac{9}{2}$
Work Step by Step
$\int^{3}_{0} \int^{2x-x^2}_{-x}dydx$
=$\int^{3}_{0}(3x-x^2)dx$
=$[\frac{3}{2}x^2-\frac{1}{3}x^3]^3_0$
=$\frac{27}{2}-9$
=$\frac{9}{2}$
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: 14
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