University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 21

Answer

$-\frac{3}{4}\quad $

Work Step by Step

\begin{aligned} \int_{\sqrt{2}}^{1}\left(\frac{u^{7}}{2}-\frac{1}{u^{5}}\right) d u&=\int_{\sqrt{2}}^{1}\left(\frac{u^{7}}{2}-u^{-5}\right) d u\\ &= \frac{1}{16}u^{8}+ \frac{1}{4}u^{-4}\bigg|_{\sqrt{2}}^{1}\\ &= \frac{1}{16}+ \frac{1}{4}- 1-\frac{1}{16}\\ &=-\frac{3}{4}\quad \end{aligned}
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