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: 21



Work Step by Step

$\int^{1}_{\sqrt{2}}(\frac{u^7}{2}-\frac{1}{u^5})du$ =$\int^{1}_{\sqrt{2}}(\frac{u^7}{2}-u^{-5})du$ =$[\frac{u^8}{16}-\frac{1}{4(u)^4}]^1_{\sqrt{2}}$ =$(\frac{1^8}{16}-\frac{1}{4(1)^4})-(\frac{\sqrt{2}^8}{16}+\frac{1}{4(\sqrt{2})^4})$ =$\frac{-3}{4}$
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