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: 31



Work Step by Step

Because $\pi\lt2\pi$, by Equation (2): $\int_{\pi}^{2\pi} \theta d\theta=\frac{(2\pi)^2}{2}-\frac{(\pi)^2}{2}=\frac{4\pi^2}{2}-\frac{\pi^2}{2}=\frac{3\pi^2}{2}$
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