1. We need the area of the region

$$\{(x,y): |2x-1| \le y \le |x^2-x|,\, 0\le x\le 1\}.$$

So for each $x\in[0,1]$, the vertical thickness is

$$|x^2-x|-|2x-1|,$$

but only where the upper curve is above the lower curve.

2. Simplify the expressions on $[0,1]$:

• Since $x\in[0,1]$, we have

$$x^2-x=x(x-1)\le 0$$

so

$$|x^2-x|=-(x^2-x)=x-x^2=x(1-x).$$

• Also,

$$|2x-1|= \begin{cases} 1-2x, & 0\le x\le \tfrac12,\\ 2x-1, & \tfrac12\le x\le 1. \end{cases}$$

Thus we need to solve

$$|2x-1| \le x-x^2.$$

3. Case 1: $0\le x\le \tfrac12$

Then $|2x-1|=1-2x$. Inequality becomes

$$1-2x \le x-x^2.$$

Rearrange:

$$0\le 3x-x^2-1$$ $$x^2-3x+1\le 0.$$

Its roots are

$$\frac{3\pm\sqrt{5}}{2}.$$

Hence in $[0,\tfrac12]$, the valid part is

$$x\in\left[\frac{3-\sqrt5}{2},\frac12\right].$$

4. Case 2: $\tfrac12\le x\le 1$

Then $|2x-1|=2x-1$. Inequality becomes

$$2x-1 \le x-x^2.$$

Rearrange:

$$0\le 1-x-x^2$$ $$x^2+x-1\le 0.$$

Its roots are

$$\frac{-1\pm\sqrt5}{2}.$$

Hence in $[\tfrac12,1]$, the valid part is

$$x\in\left[\frac12,\frac{\sqrt5-1}{2}\right].$$

So overall the region exists for

$$x\in\left[\frac{3-\sqrt5}{2},\frac{\sqrt5-1}{2}\right].$$

5. Compute the area by splitting at $x=\tfrac12$:

$$A=\int_{(3-\sqrt5)/2}^{1/2}\Big[(x-x^2)-(1-2x)\Big]dx +\int_{1/2}^{(\sqrt5-1)/2}\Big[(x-x^2)-(2x-1)\Big]dx.$$

Simplify integrands:

$$(x-x^2)-(1-2x)=3x-x^2-1,$$ $$(x-x^2)-(2x-1)=1-x-x^2.$$

Thus

$$A=\int_{(3-\sqrt5)/2}^{1/2}(3x-x^2-1)\,dx +\int_{1/2}^{(\sqrt5-1)/2}(1-x-x^2)\,dx.$$

6. A symmetry substitution makes this easier. Let

$$u=x-\frac12.$$

Then

$$|2x-1|=2|u|,$$

and

$$x-x^2=\frac14-u^2.$$

So the condition becomes

$$2|u|\le \frac14-u^2.$$

For $u\ge 0$:

$$2u\le \frac14-u^2 \iff u^2+2u-\frac14\le 0.$$

The positive endpoint is

$$a=\frac{\sqrt5-2}{2}.$$

Hence

$$A=2\int_0^a \left(\frac14-u^2-2u\right)du.$$

Now,

$$\int \left(\frac14-u^2-2u\right)du=\frac{u}{4}-\frac{u^3}{3}-u^2.$$

So

$$A=2\left[\frac{u}{4}-\frac{u^3}{3}-u^2\right]_0^a.$$

With

$$a=\frac{\sqrt5-2}{2},$$

this simplifies to

$$A=\frac{5\sqrt5-11}{6}.$$

7. Now compute:

$$6A+11 = 6\cdot \frac{5\sqrt5-11}{6}+11 = 5\sqrt5.$$

Therefore

$$(6A+11)^2=(5\sqrt5)^2=125.$$

So the required integer is

$$\boxed{125}.$$