Answer
$8ln8 -16+e$
Work Step by Step
$\int$1 to \ln8\int0 to \lny e^x+y
=\int1 to \ln8(e^x+y)x=0 to x=y dy
=\int1 to \ln8e^y(y-1)dy
=[(y-1)\inte^ydy-\int(d(y-1)/dy)\inte^ydy)dy]
=[(y-2)e^y]1 to ln8
=8ln8-16+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 875: 21
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