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: 36

Answer

$\frac{\pi^{3}}{24}$

Work Step by Step

$\int^{\pi/2}_{0}\theta^{2}d\theta=\frac{(\frac{\pi}{2})^{3}}{3}-\frac{0^{3}}{3}=\frac{\pi^{3}}{24}$
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