Answer
=$0$
Work Step by Step
$\int^3_0\int^0_{-2}(x^2y-2xy)dydx$
=$\int^3_0[\frac{x^2y^2}{2}-xy^2]^0_{-2}dx$
=$\int^3_0(4x-2x^2)dx$
=$[2x^2-\frac{2x^3}{3}]^3_0$
=$0$
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 15.1 - Double and Iterated Integrals over Rectangles - Exercises 15.1 - Page 874: 6
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