Thomas' Calculus 13th Edition

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



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, $I=\int_0^{1} e^{a} da=[e^a]_0^{1}=e^1-e^0=e-1$
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