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 307: 27

Answer

$\frac{8\sqrt 2-7}{6}$

Work Step by Step

Step 1. Graph the function and we can identify intersections at $x=0,1,\sqrt 2$. The total area is the sum of the two integrals with intervals $[0,1]$ and $[1,\sqrt 2]$. Step 2. Evaluate $A_1=\int_0^1(2-(2-x))dx=\int_0^1(x)dx=\frac{1}{2}x^2|_0^1=\frac{1}{2}$ Step 3. Evaluate $A_2=\int_1^{\sqrt 2}(2-x^2)dx=(2x-\frac{1}{3}x^3)|_1^{\sqrt 2}=(2\sqrt 2-\frac{1}{3}(\sqrt 2)^3)-(2-\frac{1}{3})=\frac{4\sqrt 2-5}{3}$ Step 4. Thus, we have the area as $A=|A_1|+|A_2|=\frac{8\sqrt 2-7}{6}$
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