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: 6

Answer

$A_T=2\sqrt2$

Work Step by Step

$A_T = A_1 + A_2 $ $A_1=\int_0^{\frac{\pi}{4}}(cos(x)-sin(x))dx$ $A_1=[sin(x)+cos(x)]_0^{\frac{\pi}{4}}$ $A_1=[\frac{1}{\sqrt2}+\frac{1}{\sqrt2}]-[0+1]$ $A_1=\sqrt 2-1$ $A_2=\int_{\frac{\pi}{4}}^{\pi}(sin(x)-cos(x))dx$ $A_2=[-cos(x)-sin(x)]_{\frac{\pi}{4}}^{\pi}$ $A_2=[-(-1)-0]-[-\frac{1}{\sqrt2}-\frac{1}{\sqrt2}]$ $A_2=1+\sqrt 2$ $A_T=(\sqrt2-1)+(1+\sqrt2)=2\sqrt2$
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