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: 8



Work Step by Step

$\int_{1}^{32} (x^{-6/5})dx=[\frac{x^{-6/5+1}}{-6/5+1}]_{1}^{32}=[\frac{x^{-1/5}}{-1/5}]_{1}^{32}=[\frac{32^{-1/5}}{-1/5}]-[\frac{1^{-1/5}}{-1/5}]=\frac{-5}{2}+5=\frac{5}{2}$
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