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: 26



Work Step by Step

Using equation (2), we have $\int^{b}_{a}3tdt=3\times[\frac{b^{2}}{2}-\frac{a^{2}}{2}]=3\times\frac{b^{2}-a^{2}}{2}$ $=\frac{3}{2}(b^{2}-a^{2})$
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