Answer
=$4$
Work Step by Step
$\int^2_0\int^1_{-1}(x-y)dydx$
=$\int^2_0[xy-\frac{1}{2}y^2]^1_{-1}dx$
=$\int^2_02xdx$
=$[x^2]^2_0$
=$4$
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: 2
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