Answer
1
Work Step by Step
$\int^{e}_{1} \int^{2 lnx }_{ln x} dy dx$
=$\int^{e}_{1} ln x dx$
=$[xln x-x]^e_1$
=(e-e)-(0-1)=1
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: 6
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