Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 16 - Multiple Integration - 16.2 Double Integrals over More General Regions - Exercises - Page 859: 51


$e^2 -2e+1 $

Work Step by Step

The iterated integral can be calculated as: $\iint_{D} f(x,y) d A=\int_0^1 \int_{0}^{1} e^{x+y} dx dy\\=\int_0^1 [e^{x+y}]_0^1 \ dy \\=\int_0^1 [e^{1+y}-e^y] \ dy \\=[e^{y+1}-e^y]_0^1 \\=e^2 -e^1-(e^1-e^0) \\=e^2-e-e+1 \\=e^2 -2e+1 $
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