University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.3 - The Definite Integral - Exercises - Page 309: 25



Work Step by Step

Using equation (2), we obtain $\int^{b}_{a}2sds= 2\times[\frac{b^{2}}{2}-\frac{a^{2}}{2}]$ $=2\times\frac{b^{2}-a^{2}}{2}=b^{2}-a^{2}$
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