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: 19



Work Step by Step

Integrate with respect to r and plug in limits to evaluate: $=\frac{1}{3}(r+1)^3$ with limits from -1 to 1 $=-\frac{1}{3}[(1+1)^3-(-1+1)^3]=-\frac{1}{3}(8)=-8/3$
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