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: 22



Work Step by Step

$\int^{-1}_{-3}\frac{y^5-2y}{y^3}dy$ =$\int^{-1}_{-3}(y^2-2y^{-2})dy$ =$[\frac{y^3}{3}+2y^{-1}]^{-1}_{-3}$ =$(\frac{(-1)^3}{3}+\frac{2}{-1})-(\frac{-3^3}{3}+\frac{2}{-3})$ =$\frac{22}{3}$
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