Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.2 - The Definite Integral - 5.2 Exercises - Page 390: 64


$\pi^2 \leq \int_{\pi}^{2\pi} (x-2sin~x)~dx \leq \frac{5\pi^2}{3}+\sqrt{3}~\pi$

Work Step by Step

On the interval $\pi \leq x \leq 2\pi$: $\pi \leq (x-2sin~x) \leq \frac{5\pi}{3}+\sqrt{3}$ Therefore, by Property 8: $\pi(2\pi-\pi) \leq \int_{\pi}^{2\pi} (x-2sin~x)~dx \leq (\frac{5\pi}{3}+\sqrt{3})(2\pi-\pi)$ $\pi^2 \leq \int_{\pi}^{2\pi} (x-2sin~x)~dx \leq \frac{5\pi^2}{3}+\sqrt{3}~\pi$
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