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: 8

Answer

$\frac{5}{2}$

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.