If the area of the region $\left\{\left(x,y\right):\left|x^2-2\right|\le y\le x\right\}$ is $\mathrm{A}$, then $6A+16\sqrt{2}$ is equal to ______________

Given,

$\left|x^2-2\right|\le y \& y\le x$

Now on finding the intersection point we get, $\left(x,y\right)\equiv \left(1,1\right) \& \left(2,2\right)$

Now plotting the curve of the above function we get,

Now from above diagram required bounded area will be,

$\mathrm{A}=\left[{\int }_1^{\sqrt{2}}\left(x-\left(2-x^2\right)\right)dx+{\int }_{\sqrt{2}}^2\left(x-\left(x^2-2\right)\right)dx\right]$

$\Rightarrow A={\left[\frac{x^2}{2}-2x+\frac{x^3}{3}\right]}_1^{\sqrt{2}}+{\left[\frac{x^2}{2}-\frac{x^3}{3}+2x\right]}_{\sqrt{2}}^2$

$\Rightarrow A=\left[1-2\sqrt{2}+\frac{2\sqrt{2}}{3}\right]-\left[\frac{1}{2}-2+\frac{1}{3}\right]+\left[2-\frac{8}{3}+4\right]-\left[1-\frac{2\sqrt{2}}{3}+2\sqrt{2}\right]$

$\Rightarrow \mathrm{A}=\frac{9}{2}-\frac{8\sqrt{2}}{3}$

$\Rightarrow 6\mathrm{A}=27-16\sqrt{2}$

$\Rightarrow 6\mathrm{A}+16\sqrt{2}=27$