Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.3 - The Definite Integral - Exercises 5.3 - Page 275: 35

Answer

$\frac{1}{24}$

Work Step by Step

$\int_{0}^{\frac{1}{2}}t^2dt$ =$\frac{(\frac{1}{2})^3}{3}$ =$\frac{1}{24}$
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