#### Answer

$\frac{9}{2}$

#### Work Step by Step

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

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Published by
Pearson

ISBN 10:
0-32187-896-5

ISBN 13:
978-0-32187-896-0

$\frac{9}{2}$

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