Answer
$2 ln 2 -\frac{1}{2}$
Work Step by Step
$\int^{2}_{1} \int^{ln y}_{1-y}1dxdy$
=$\int^{2}_{1}[x]^{ln x}_{1-y}dy$
=$\int^{2}_{1}(lny-1+y)dy=[yln y-2y+\frac{y^2}{2}]_1^2$
=$2 ln 2 -\frac{1}{2}$
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 15.3 - Area by Double Integration - Exercises 15.3 - Page 887: 10
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