Answer
$\dfrac{1-e^{-9}}{2}$
Work Step by Step
$\int_{0}^{3}\int_{0}^{y}e^{-y^2}\,dx\, dy=\int_{0}^{3}\biggl[xe^{-y^2}\biggr]_{0}^{y}\, dy=\int_{0}^{3}ye^{-y^2}\,dy=\biggr[-\dfrac{e^{-y^2}}{2}\biggr]_{0}^{3}=\dfrac{1-e^{-9}}{2}$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 15 - Multiple Integrals - 15.2 Double Integrals over General Regions - 15.2 Exercises - Page 1048: 9
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