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: 37



Work Step by Step

Using equation (2), we have $\int^{2a}_{a}xdx= \frac{(2a)^{2}}{2}-\frac{a^{2}}{2}$ $=\frac{4a^{2}}{2}-\frac{a^{2}}{2}= \frac{3a^{2}}{2}$
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