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: 35



Work Step by Step

Because $0\lt1/2$, by Equation (4): $\int_{0}^{1/2} t^2 dt=\frac{(\frac{1}{2})^3}{3}-\frac{0^3}{3}=\frac{\frac{1}{8}}{3}-\frac{0}{3}=\frac{1}{24}$
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