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: 54


$\dfrac{\ln 17}{4}$

Work Step by Step

We reverse the order of integration to get: $I= \int_0^{2} \int_0^{y^3} \dfrac{1}{y^4+1} dx dy$ or, $= \int_0^{2} \dfrac{y^3}{y^4+1} dy$ Set $a =y^4+1 \implies da=4y^3 dy$ Now, $I=\dfrac{1}{4} \int_{1}^{17}\dfrac{da}{a}=\dfrac{\ln 17}{4}$
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