Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 443: 14

Answer

$A= {e^{\pi /2}} - 2$

Work Step by Step

$$\eqalign{ & {\text{Let the functions}} \cr & y = \cos x,{\text{ }}y = {e^x},{\text{ }}x = \frac{\pi }{2} \cr & {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr & \cr & {\text{We can find the area integrating with respect to }}x \cr & {\text{Use }}A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx,{\text{ with }}f\left( x \right) \geqslant g\left( x \right) \cr & {\text{Let }} \cr & f\left( x \right) = {e^x}{\text{, }}g\left( x \right) = \cos x \cr & x = a = 0{\text{ and }}x = b = \frac{\pi }{2} \cr & {\text{Therefore}}{\text{,}} \cr & \underbrace {A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx}_ \Downarrow \cr & A = \int_0^{\pi /2} {\left( {{e^x} - \cos x} \right)} dx \cr & {\text{Integrating}} \cr & A = \left[ {{e^x} - \sin x} \right]_0^{\pi /2} \cr & A = \left[ {{e^{\pi /2}} - \sin \left( {\frac{\pi }{2}} \right)} \right] - \left[ {{e^0} - \sin \left( 0 \right)} \right] \cr & {\text{Simplifying}} \cr & A = {e^{\pi /2}} - 1 - 1 + 0 \cr & A = {e^{\pi /2}} - 2 \cr} $$
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