Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - 6.1 Area Between Two Curves - Exercises - Page 286: 11



Work Step by Step

Since $2\geq \sec^2 x$ on $ -\pi/4\leq x\leq \pi/4$ Then \begin{align*} A&= \int_{-\pi / 4}^{\pi / 4}\left(2-\sec ^{2} x\right) d x\\ &=\left.(2 x-\tan x)\right|_{-\pi / 4} ^{\pi / 4}\\ &=\left(\frac{\pi}{2}-1\right)-\left(-\frac{\pi}{2}+1\right)\\ &=\pi-2 \end{align*}
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