Answer
$\frac{22}{3}$
Work Step by Step
$\int^{-1}_{-3}\frac{y^5-2y}{y^3}dy$
=$\int^{-1}_{-3}(y^2-2y^{-2})dy$
=$[\frac{y^3}{3}+2y^{-1}]^{-1}_{-3}$
=$(\frac{(-1)^3}{3}+\frac{2}{-1})-(\frac{-3^3}{3}+\frac{2}{-3})$
=$\frac{22}{3}$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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
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