Calculus 8th Edition

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

Chapter 4 - Integrals - Review - Exercises - Page 349: 16



Work Step by Step

Given $$\int_{0}^{1}(\sqrt[4]{u}+1)^2du$$ Since \begin{align*} \int_{0}^{1}(\sqrt[4]{u}+1)^2du&=\int_{0}^{1}( u^{1/2}+2u^{1/4}+1) du\\ &=\frac{2}{3}u^{3/2}+\frac{8}{5}u^{5/4}+u\bigg|_{0}^{1}\\ &=\frac{49}{15} \end{align*}
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