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 $A=|A_1|+|A_2|=\frac{8\sqrt 2-7}{6}$
