Let $A_1$ be the bounded area enclosed by the curves $y=x^2+2,; x+y=8$ and y-axis lies in the first quadrant. Let $A_2$ be the bounded area enclosed by the curves $y=x^2+2,; y^2=x,; x=2$ and y-axis that lies in the first quadrant. Then $A_1-A_2$ is equal to:

$A_2=\int_0^2(x^2+2),dx-\frac{2}{3}(2\sqrt{2})$

$=\left[\frac{x^3}{3}+2x\right]_0^2-\frac{4\sqrt{2}}{3}$

$=\frac{8}{3}+4-\frac{4\sqrt{2}}{3}=\frac{20}{3}-\frac{4\sqrt{2}}{3}$

$A_1-A_2=\frac{22}{3}-\left(\frac{20}{3}-\frac{4\sqrt{2}}{3}\right)$

$=\frac{2}{3}+\frac{4\sqrt{2}}{3}=\frac{2}{3}(2\sqrt{2}+1)$