University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 321: 63

Answer

$\dfrac{\pi\sqrt 2}{2}$

Work Step by Step

The area of the rectangle curve is=$(\sqrt 2)(\pi/4]=\dfrac{\pi\sqrt 2}{4}$ Need to find the area of the rectangle curve. $A_1=-\int_{-\pi/4}^{0} \sec \theta \tan \theta d \theta$ This implies that $[-\sec \theta]_{-\pi/4}^{0}=\sqrt 2-1$ Thus, the area of shaded region is: $\dfrac{\pi\sqrt 2}{4}+(\sqrt 2-1)$ Now, $A_2=\int_{\pi/4}^{0} \sec \theta \tan \theta d \theta$ This implies that $[\sec \theta]_{\pi/4}^{0}=\sqrt 2-1$ Thus, the area of shaded region is: $A_2=\dfrac{\pi\sqrt 2}{4}-(\sqrt 2-1)$ Total area:$\int_{\pi/4}^{0} \sec \theta \tan \theta d \theta+ \dfrac{\pi\sqrt 2}{4}-(\sqrt 2-1)=\dfrac{\pi\sqrt 2}{2}$
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