Calculus 8th Edition

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

Chapter 4 - Integrals - Review - Exercises - Page 349: 15



Work Step by Step

Given $$\int_{1}^{9}\frac{\sqrt{u}-u^2}{u}du$$ Since \begin{align*} \int_{1}^{9}\frac{\sqrt{u}-2u^2}{u}du&=\int_{1}^{9}(u^{-1/2}-2u)du\\ &=2\sqrt{u}-u^2\bigg|_{1}^{9}\\ &=4-80=-76 \end{align*}
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