Calculus (3rd Edition)

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

Chapter 6 - Applications of the Integral - Chapter Review Exercises - Page 319: 9


$3\sqrt 2 -1 $

Work Step by Step

The point of intersection can be computed as: $\dfrac{\sin x}{\cos x}=1 \\ \tan x=1$ The area $(A)$ can be calculated as: $A=\int_{0}^{\pi/4} [\cos x -\sin x] \ dx +\int_{\pi/4}^{5 \pi/4} [\sin x -\cos x] \ dx \\= [\sin x+\cos x]_{0}^{\pi/4}+[-\cos x -\cos x]_{\pi/4}^{5 \pi/4}\\=\sqrt 2-1 +\sqrt 2+\sqrt 2 \\=3\sqrt 2 -1 $
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