Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Practice Exercises - Page 308: 32

Answer

$4\sqrt 2-2$
1579350479

Work Step by Step

Step 1. Graph the functions as shown in the figure. The question asks the total area of not only the enclosed region, so we need to find the sum of areas for three regions $A_1, A_2, A_3$. Step 2. The total area between the two functions in the interval of $[0,\frac{3\pi}{2}]$, after identifying the intersections at $x=\frac{\pi}{4},\frac{5\pi}{4}$, is given by $A=\int_0^{\pi/4}(cos(x)-sin(x))dx+\int_{\pi/4}^{5\pi/4}(sin(x)-cos(x))dx+\int_{5\pi/4}^{3\pi/2}(cos(x)-sin(x))dx=(sin(x)+cos(x))|_0^{\pi/4}+(-cos(x)-sin(x))|_{\pi/4}^{5\pi/4}+(sin(x)+cos(x))|_{5\pi/4}^{3\pi/2}=(\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2})-(0+1)+(\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2})-(-\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2})+(-1+0)-(-\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2})=4\sqrt 2-2$
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