1. We need the area of the region

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

So the region lies above both parabolas $y=x^2$ and $y=(1-x)^2$, and below the parabola

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

2. Since $y$ must be above both $x^2$ and $(1-x)^2$, the effective lower boundary is

$$y\ge \max\{x^2,(1-x)^2\}.$$

Thus area exists where

$$2x(1-x)\ge \max\{x^2,(1-x)^2\}.$$

3. Split according to which of $x^2$ and $(1-x)^2$ is larger.

• For $x\le \tfrac12$, we have $(1-x)^2\ge x^2$, so lower curve is $(1-x)^2$.
• For $x\ge \tfrac12$, we have $x^2\ge (1-x)^2$, so lower curve is $x^2$.

So we first find where the top curve meets these lower curves.

4. Intersections:

(i) Solve

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

Expanding,

$$2x-2x^2=1-2x+x^2$$ $$3x^2-4x+1=0$$ $$(3x-1)(x-1)=0.$$

So $x=\tfrac13$ or $x=1$. In the interval $x\le \tfrac12$, relevant point is

$$x=\tfrac13.$$

(ii) Solve

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

Then

$$2x-2x^2=x^2$$ $$3x^2-2x=0$$ $$x(3x-2)=0.$$

So $x=0$ or $x=\tfrac23$. In the interval $x\ge \tfrac12$, relevant point is

$$x=\tfrac23.$$

Hence the enclosed region exists for

$$\frac13\le x\le \frac23.$$

5. Therefore,

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

Simplify each integrand:

For $x\in[\tfrac13,\tfrac12]$,

$$2x(1-x)-(1-x)^2=(2x-2x^2)-(1-2x+x^2)=4x-3x^2-1.$$

For $x\in[\tfrac12,\tfrac23]$,

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

So

$$A=\int_{1/3}^{1/2}(4x-3x^2-1)\,dx+\int_{1/2}^{2/3}(2x-3x^2)\,dx.$$

6. Compute the first integral:

$$\int(4x-3x^2-1)dx=2x^2-x^3-x.$$

Thus

$$I_1=\left[2x^2-x^3-x\right]_{1/3}^{1/2}.$$

At $x=\tfrac12$:

$$2\cdot\frac14-\frac18-\frac12=\frac12-\frac18-\frac12=-\frac18.$$

At $x=\tfrac13$:

$$2\cdot\frac19-\frac1{27}-\frac13=\frac29-\frac1{27}-\frac13 =\frac6{27}-\frac1{27}-\frac9{27}=-\frac4{27}.$$

Hence

$$I_1=-\frac18+\frac4{27}=\frac{5}{216}.$$

7. Compute the second integral:

$$\int(2x-3x^2)dx=x^2-x^3.$$

Thus

$$I_2=\left[x^2-x^3\right]_{1/2}^{2/3}.$$

At $x=\tfrac23$:

$$\frac49-\frac8{27}=\frac{12-8}{27}=\frac4{27}.$$

At $x=\tfrac12$:

$$\frac14-\frac18=\frac18.$$

Hence

$$I_2=\frac4{27}-\frac18=\frac5{216}.$$

8. Therefore,

$$A=I_1+I_2=\frac5{216}+\frac5{216}=\frac{10}{216}=\frac5{108}.$$

9. Now,

$$540A=540\cdot \frac5{108}=5\cdot 25=25.$$

So,

$$\boxed{540A=25}.$$

10. Comparison with stored answer: Stored correct answer = 25, which matches our result.