Thomas' Calculus 13th Edition

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


$2 ln 2 -\frac{1}{2}$

Work Step by Step

$\int^{2}_{1} \int^{ln y}_{1-y}1dxdy$ =$\int^{2}_{1}[x]^{ln x}_{1-y}dy$ =$\int^{2}_{1}(lny-1+y)dy=[yln y-2y+\frac{y^2}{2}]_1^2$ =$2 ln 2 -\frac{1}{2}$
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