Answer
$e-1$
Work Step by Step
We reverse the order of integration to get:
$I= \int_0^{1} \int_0^{3y^2} e^{y^3} dx dy$
or, $= \int_0^{1} 3y^2e^{y^3} dy $
Set $a = y^3 \implies da=3y^2 dy$
Now, $I=\int_0^{1} e^{a} da=[e^a]_0^{1}=e^1-e^0=e-1$
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 15.2 - Double Integrals over General Regions - Exercises 15.2 - Page 883: 52
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