Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 286: 23


$\sqrt{2}-\sqrt[4] 8+1$

Work Step by Step

$\int^{\sqrt{2}}_{1}\frac{s^2+\sqrt{s}}{s^2}ds$ =$\int^{\sqrt{2}}_{1}(1+s^{\frac{-3}{2}})ds$ =$[s-\frac{2}{\sqrt{s}}]_1^{\sqrt{2}}$ =$(\sqrt{2}-\frac{2}{\sqrt{\sqrt{2}}})-(1-\frac{2}{\sqrt{1}})$ =$\sqrt{2}-2^\frac{3}{4}+1$ =$\sqrt{2}-\sqrt[4] 8+1$
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