Question:

Step 1: Identify the bounding curves and region definition. The region is defined by the inequalities: 1. $y \geq x^2$ 2. $y \geq (1-x)^2$ 3. $y \leq 2x(1-x)$ Let the three curves be $C_1: y=x^2$, $C_2: y=(1-x)^2$, and $C_3: y=2x(1-x)$. The region is bounded below by the maximum of $C_1$ and $C_2$, and bounded above by $C_3$. So, $y_{lower}(x) = \max(x^2, (1-x)^2)$ and $y_{upper}(x) = 2x(1-x)$. Step 2: Determine the function for the lower boundary and the relevant $x$-interval. First, find where $x^2 = (1-x)^2$: $$x^2 = 1 - 2x + x^2$$ $$0 = 1 - 2x \implies x = \frac{1}{2}$$ So, $x^2 \leq (1-x)^2$ for $x \leq \frac{1}{2}$, and $x^2 \geq (1-x)^2$ for $x \geq \frac{1}{2}$. Therefore, the lower boundary function $y_{lower}(x)$ is: $$y_{lower}(x) = \begin{cases} (1-x)^2 & \text{for } x \in \left(-\infty, \frac{1}{2}\right] \\ x^2 & \text{for } x \in \left[\frac{1}{2}, \infty\right) \end{cases}$$ Next, we need to find the $x$-interval where $y_{upper}(x) \geq y_{lower}(x)$. Case 1: For $x \in \left(-\infty, \frac{1}{2}\right]$. We need $2x(1-x) \geq (1-x)^2$. Since $1-x \geq 0$ for $x \leq 1$, we can divide by $(1-x)$ (assuming $x \neq 1$): $$2x \geq 1-x$$ $$3x \geq 1 \implies x \geq \frac{1}{3}$$ So, for $x \in \left(-\infty, \frac{1}{2}\right]$, the region exists when $x \in \left[\frac{1}{3}, \frac{1}{2}\right]$. Case 2: For $x \in \left[\frac{1}{2}, \infty\right)$. We need $2x(1-x) \geq x^2$. Since $x \geq 0$ for $x \in \left[\frac{1}{2}, \infty\right)$, we can divide by $x$: $$2(1-x) \geq x$$ $$2 - 2x \geq x$$ $$2 \geq 3x \implies x \leq \frac{2}{3}$$ So, for $x \in \left[\frac{1}{2}, \infty\right)$, the region exists when $x \in \left[\frac{1}{2}, \frac{2}{3}\right]$. Combining both cases, the region of integration for $x$ is $\left[\frac{1}{3}, \frac{2}{3}\right]$. Step 3: Set up the integral for the area $A$. The area $A$ is given by the integral of the difference between the upper and lower boundary functions over the determined $x$-interval: $$A = \int_{1/3}^{2/3} (y_{upper}(x) - y_{lower}(x)) dx$$ We split the integral at $x=\frac{1}{2}$ because $y_{lower}(x)$ changes definition there: $$A = \int_{1/3}^{1/2} (2x(1-x) - (1-x)^2) dx + \int_{1/2}^{2/3} (2x(1-x) - x^2) dx$$ Let's simplify the integrands: For the first integral: $2x(1-x) - (1-x)^2 = (1-x)(2x - (1-x)) = (1-x)(3x - 1) = 3x - 1 - 3x^2 + x = -3x^2 + 4x - 1$. For the second integral: $2x(1-x) - x^2 = x(2(1-x) - x) = x(2 - 2x - x) = x(2 - 3x) = 2x - 3x^2$. Notice the symmetry of the region about $x = \frac{1}{2}$. The two integrals will have the same value. We can compute one and multiply by 2. Let's calculate the first integral $I_1 = \int_{1/3}^{1/2} (-3x^2 + 4x - 1) dx$. Step 4: Evaluate the integral. $$I_1 = \int_{1/3}^{1/2} (-3x^2 + 4x - 1) dx$$ $$I_1 = \left[ -x^3 + 2x^2 - x \right]_{1/3}^{1/2}$$ Substitute the limits of integration: $$I_1 = \left( -\left(\frac{1}{2}\right)^3 + 2\left(\frac{1}{2}\right)^2 - \frac{1}{2} \right) - \left( -\left(\frac{1}{3}\right)^3 + 2\left(\frac{1}{3}\right)^2 - \frac{1}{3} \right)$$ $$I_1 = \left( -\frac{1}{8} + 2\left(\frac{1}{4}\right) - \frac{1}{2} \right) - \left( -\frac{1}{27} + 2\left(\frac{1}{9}\right) - \frac{1}{3} \right)$$ $$I_1 = \left( -\frac{1}{8} + \frac{1}{2} - \frac{1}{2} \right) - \left( -\frac{1}{27} + \frac{2}{9} - \frac{1}{3} \right)$$ $$I_1 = \left( -\frac{1}{8} \right) - \left( -\frac{1}{27} + \frac{6}{27} - \frac{9}{27} \right)$$ $$I_1 = -\frac{1}{8} - \left( -\frac{4}{27} \right)$$ $$I_1 = -\frac{1}{8} + \frac{4}{27}$$ To combine the fractions, find a common denominator, which is $8 \times 27 = 216$: $$I_1 = \frac{-27}{216} + \frac{32}{216} = \frac{5}{216}$$ The total area $A$ is $2 \times I_1$: $$A = 2 \times \frac{5}{216} = \frac{10}{216} = \frac{5}{108}$$ Step 5: Calculate the final value $540A$. We need to find $540A$: $$540A = 540 \times \frac{5}{108}$$ $$540A = \frac{540}{108} \times 5$$ Since $108 \times 5 = 540$: $$540A = 5 \times 5 = 25$$ The final answer is $\boxed{25}$.