Animated Solution for Mathematics - Definite Integration: If the area of the region $\{(x,y):|x^2-2|\le y\le x\}$ is $A$, then $6A+16\sqrt{2}$ is equal to ______________

Imagine you are standing on the Cartesian plane, looking at two distinct mathematical entities: a simple, elegant line $y=x$ and a more complex, mirrored parabola $y=|x^2-2|$. Your task is to find the area trapped between them.

The function $y=|x^2-2|$ is a classic 'W' shape, formed by taking the parabola $y=x^2-2$ and reflecting its negative portion above the $x$-axis. The line $y=x$ cuts through this shape, creating a bounded region that we must measure.

Before we can integrate, we must know where our region begins and ends. We look for the intersection points by equating the two functions.

For the left side of the modulus, where $x<\sqrt{2}$, we solve $2-x^2=x$, which simplifies to $x^2+x-2=0$. Factoring this, we find $(x+2)(x-1)=0$. Since our region is in the first quadrant, we discard $x=-2$ and keep $x=1$.

For the right side, where $x\ge \sqrt{2}$, we solve $x^2-2=x$, leading to $x^2-x-2=0$. Factoring this gives $(x-2)(x+1)=0$. Again, we discard the negative root and keep $x=2$. Our boundaries are set: $x=1$ to $x=2$.

Now, we set up our integral for the area $A={\int }_1^2(x-|x^2-2|)dx$. The modulus is the gatekeeper here; it changes its definition at $x=\sqrt{2}$.

We must split the integral into two parts:

For $I_1$, we integrate $x^2+x-2$, yielding: Substituting the limits, we find $I_1=\frac{13}{6}-\frac{4\sqrt{2}}{3}$.

For $I_2$, we integrate $-x^2+x+2$, yielding: Substituting these limits, we find $I_2=\frac{7}{3}-\frac{4\sqrt{2}}{3}$.

Combining these, we get:

The final step is to evaluate the expression $6A+16\sqrt{2}$. Substituting our value for $A$, we get:

Distributing the $6$, we get $27-16\sqrt{2}+16\sqrt{2}$. The irrational terms cancel out with beautiful precision, leaving us with the final answer: 27.