Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - Review - Exercises - Page 349: 18



Work Step by Step

Given $$\int_0^2y^2\sqrt{1+y^3}dy$$ Let $u=1+y^3\ \rightarrow \ \ du=3y^2dy$ at $y=0\to u=1$ at $y=2\to u=9$ \begin{align*} \int_0^2y^2\sqrt{1+y^3}dy&=\frac{1}{3}\int_1^9u^{1/2}du\\ &=\frac{2}{9}u^{3/2}\bigg|_{1}^{9}\\ &=\frac{52}{9} \end{align*}
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