1. Interpret the region

We need the area of

$$\{(x,y): |x^2-2| \le y \le x\}.$$

So for a given $x$, the vertical strip exists only when

$$|x^2-2| \le x.$$

Then the area is

$$A=\int (\text{upper curve} - \text{lower curve})\,dx =\int \left(x-|x^2-2|\right)dx$$

over all $x$ satisfying $|x^2-2|\le x$.

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2. Find the interval where the region exists

Solve

$$|x^2-2|\le x.$$

Since the left side is nonnegative, we must have $x\ge 0$.

For $x\ge 0$, use

$$-x \le x^2-2 \le x.$$

This gives two inequalities:

• $x^2-2 \le x \implies x^2-x-2\le 0 \implies (x-2)(x+1)\le 0,$ so with $x\ge 0$, we get $0\le x\le 2.$

• $x^2-2 \ge -x \implies x^2+x-2\ge 0 \implies (x+2)(x-1)\ge 0,$ so with $x\ge 0$, we get $x\ge 1.$

Combining,

$$1\le x\le 2.$$

So the required region lies between $x=1$ and $x=2$.

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3. Handle the absolute value

On $[1,2]$, the expression $x^2-2$ changes sign at

$$x=\sqrt{2}.$$

Thus,

• for $1\le x\le \sqrt{2}$, $|x^2-2|=2-x^2;$
• for $\sqrt{2}\le x\le 2$, $|x^2-2|=x^2-2.$

Hence

$$A=\int_1^{\sqrt2}\left[x-(2-x^2)\right]dx+\int_{\sqrt2}^2\left[x-(x^2-2)\right]dx.$$

So

$$A=\int_1^{\sqrt2}(x^2+x-2)dx+\int_{\sqrt2}^2(-x^2+x+2)dx.$$

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4. Evaluate the integrals

First integral:

$$\int (x^2+x-2)dx=\frac{x^3}{3}+\frac{x^2}{2}-2x.$$

Therefore,

$$I_1=\left[\frac{x^3}{3}+\frac{x^2}{2}-2x\right]_1^{\sqrt2}.$$

Using $(\sqrt2)^2=2$ and $(\sqrt2)^3=2\sqrt2$,

$$I_1=\left(\frac{2\sqrt2}{3}+1-2\sqrt2\right)-\left(\frac13+\frac12-2\right) =1-\frac{4\sqrt2}{3}+\frac76 =\frac{13}{6}-\frac{4\sqrt2}{3}.$$

Second integral:

$$\int (-x^2+x+2)dx=-\frac{x^3}{3}+\frac{x^2}{2}+2x.$$

Thus,

$$I_2=\left[-\frac{x^3}{3}+\frac{x^2}{2}+2x\right]_{\sqrt2}^2.$$

At $x=2$,

$$-\frac{8}{3}+2+4=\frac{10}{3}.$$

At $x=\sqrt2$,

$$-\frac{2\sqrt2}{3}+1+2\sqrt2=1+\frac{4\sqrt2}{3}.$$

Hence,

$$I_2=\frac{10}{3}-\left(1+\frac{4\sqrt2}{3}\right) =\frac{7}{3}-\frac{4\sqrt2}{3}.$$

So,

$$A=I_1+I_2 =\left(\frac{13}{6}-\frac{4\sqrt2}{3}\right)+\left(\frac{7}{3}-\frac{4\sqrt2}{3}\right).$$

Simplify:

$$A=\frac{13}{6}+\frac{14}{6}-\frac{8\sqrt2}{3} =\frac{27}{6}-\frac{8\sqrt2}{3} =\frac92-\frac{8\sqrt2}{3}.$$

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5. Compute $6A+16\sqrt2$

$$6A=6\left(\frac92-\frac{8\sqrt2}{3}\right)=27-16\sqrt2.$$

Therefore,

$$6A+16\sqrt2=(27-16\sqrt2)+16\sqrt2=27.$$

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6. Final answer

$$\boxed{27}$$

The derived answer matches the stored correct answer.