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: 31

Answer

=$\frac{3\pi^2}{2}$

Work Step by Step

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