Answer
=$\frac{3}{2}(5-e)$
Work Step by Step
$\int^{\ln2}_0\int^{\ln5}_1e^{2x+y} dydx$
=$\int^{\ln2}_0[e^{2x+y}]^{\ln5}_1dx$
=$\int^{\ln2}_0(5e^{2x}-e^{2x+1})dx$
=$[\frac{5}{2}e^{2x}-\frac{1}{2}e^{2x+1}]^{\ln2}_0$
=$\frac{3}{2}(5-e)$
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: 9
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