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: 19

Answer

$-\frac{8}{3}$

Work Step by Step

$\int_{1}^{-1} (r+1)^2 dr=\int_{1}^{-1} (r^2+2r+1) dr=[\frac{r^3}{3}+r^2+r]_{1}^{-1}=[\frac{(-1)^3}{3}+(-1)^2+(-1)]-[\frac{1^3}{3}+1^2+1]=[\frac{-1}{3}+1-1]-[\frac{1}{3}+1+1]=-\frac{1}{3}-\frac{7}{3}=-\frac{8}{3}$
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