Answer
$\frac{1}{2}$
Work Step by Step
$\int\int_R e^{x-y}dA$
=$\int_0^{ln2}\int_0^{ln2} e^{x-y}dydx$
=$\int^{ln2}_{0}[-e^{x-y}]^{ln2}_0 dx$
=$\int^{ln2}_{0}(-e^{x-ln2}+e^x)dx$
=$[-e^{x-ln2}+e^x]^{ln2}_0$
=$\frac{1}{2}$
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 15.1 - Double and Iterated Integrals over Rectangles - Exercises 15.1 - Page 875: 19
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