Question:

Step 1: Analyze the given functions and their absolute values over the interval $0 \leq x \leq 1$. The region is defined by $|2x-1| \leq y \leq |x^2-x|$ for $0 \leq x \leq 1$. Let $y_1(x) = |2x-1|$ and $y_2(x) = |x^2-x|$. For $y_1(x)$: If $2x-1 \geq 0 \implies x \geq 1/2$, then $y_1(x) = 2x-1$. If $2x-1 < 0 \implies x < 1/2$, then $y_1(x) = -(2x-1) = 1-2x$. For $y_2(x)$: For $x \in [0,1]$, $x^2-x = x(x-1) \leq 0$. So, $y_2(x) = -(x^2-x) = x-x^2$. Step 2: Determine the interval(s) where the region is defined, i.e., $y_1(x) \leq y_2(x)$. We need to find $x \in [0,1]$ such that $|2x-1| \leq x-x^2$. Case 1: For $x \in [0, 1/2)$, $y_1(x) = 1-2x$. The inequality becomes $1-2x \leq x-x^2$. Rearranging gives $x^2-3x+1 \leq 0$. To find the roots of $x^2-3x+1=0$, we use the quadratic formula: $x = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(1)}}{2(1)} = \frac{3 \pm \sqrt{9-4}}{2} = \frac{3 \pm \sqrt{5}}{2}$. The parabola $x^2-3x+1$ opens upwards, so $x^2-3x+1 \leq 0$ for $x \in \left[\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right]$. The intersection with $x \in [0, 1/2)$ is $x \in \left[\frac{3-\sqrt{5}}{2}, \frac{1}{2}\right)$. Let $\alpha = \frac{3-\sqrt{5}}{2}$. (Note: $\alpha \approx 0.382$, $1/2 = 0.5$) Case 2: For $x \in [1/2, 1]$, $y_1(x) = 2x-1$. The inequality becomes $2x-1 \leq x-x^2$. Rearranging gives $x^2+x-1 \leq 0$. To find the roots of $x^2+x-1=0$, we use the quadratic formula: $x = \frac{-1 \pm \sqrt{1^2-4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{1+4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$. The parabola $x^2+x-1$ opens upwards, so $x^2+x-1 \leq 0$ for $x \in \left[\frac{-1-\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}\right]$. The intersection with $x \in [1/2, 1]$ is $x \in \left[\frac{1}{2}, \frac{-1+\sqrt{5}}{2}\right]$. Let $\beta = \frac{-1+\sqrt{5}}{2}$. (Note: $\beta \approx 0.618$) Thus, the region exists for $x \in [\alpha, \beta]$. The area $A$ is given by the integral of $y_2(x) - y_1(x)$ over this interval. Since the definition of $y_1(x)$ changes at $x=1/2$, we split the integral: $$A = \int_{\alpha}^{1/2} ((x-x^2)-(1-2x))dx + \int_{1/2}^{\beta} ((x-x^2)-(2x-1))dx$$ $$A = \int_{\alpha}^{1/2} (-x^2+3x-1)dx + \int_{1/2}^{\beta} (-x^2-x+1)dx$$ Step 3: Evaluate the first definite integral. Let $I_1 = \int_{\alpha}^{1/2} (-x^2+3x-1)dx$. The antiderivative of $-x^2+3x-1$ is $F_1(x) = -\frac{x^3}{3} + \frac{3x^2}{2} - x$. Evaluating $F_1(1/2)$: $$F_1(1/2) = -\frac{(1/2)^3}{3} + \frac{3(1/2)^2}{2} - \frac{1}{2} = -\frac{1}{24} + \frac{3}{8} - \frac{1}{2} = \frac{-1+9-12}{24} = -\frac{4}{24} = -\frac{1}{6}$$ Evaluating $F_1(\alpha)$: We know $\alpha^2-3\alpha+1=0$, which means $\alpha^2 = 3\alpha-1$. Also, $\alpha^3 = 3\alpha^2 - \alpha = 3(3\alpha-1) - \alpha = 9\alpha-3-\alpha = 8\alpha-3$. $$F_1(\alpha) = -\frac{\alpha^3}{3} + \frac{3\alpha^2}{2} - \alpha = -\frac{8\alpha-3}{3} + \frac{3(3\alpha-1)}{2} - \alpha$$ $$F_1(\alpha) = \left(-\frac{8}{3} + \frac{9}{2} - 1\right)\alpha + \left(1-\frac{3}{2}\right) = \left(\frac{-16+27-6}{6}\right)\alpha - \frac{1}{2} = \frac{5}{6}\alpha - \frac{1}{2}$$ So, $I_1 = F_1(1/2) - F_1(\alpha) = -\frac{1}{6} - \left(\frac{5}{6}\alpha - \frac{1}{2}\right) = -\frac{1}{6} - \frac{5}{6}\alpha + \frac{3}{6} = \frac{2}{6} - \frac{5}{6}\alpha = \frac{1}{3} - \frac{5}{6}\alpha$. Step 4: Evaluate the second definite integral. Let $I_2 = \int_{1/2}^{\beta} (-x^2-x+1)dx$. The antiderivative of $-x^2-x+1$ is $F_2(x) = -\frac{x^3}{3} - \frac{x^2}{2} + x$. Evaluating $F_2(1/2)$: $$F_2(1/2) = -\frac{(1/2)^3}{3} - \frac{(1/2)^2}{2} + \frac{1}{2} = -\frac{1}{24} - \frac{1}{8} + \frac{1}{2} = \frac{-1-3+12}{24} = \frac{8}{24} = \frac{1}{3}$$ Evaluating $F_2(\beta)$: We know $\beta^2+\beta-1=0$, which means $\beta^2 = 1-\beta$. Also, $\beta^3 = \beta(1-\beta) = \beta-\beta^2 = \beta-(1-\beta) = 2\beta-1$. $$F_2(\beta) = -\frac{\beta^3}{3} - \frac{\beta^2}{2} + \beta = -\frac{2\beta-1}{3} - \frac{1-\beta}{2} + \beta$$ $$F_2(\beta) = \left(-\frac{2}{3} + \frac{1}{2} + 1\right)\beta + \left(\frac{1}{3} - \frac{1}{2}\right) = \left(\frac{-4+3+6}{6}\right)\beta + \left(\frac{2-3}{6}\right) = \frac{5}{6}\beta - \frac{1}{6}$$ So, $I_2 = F_2(\beta) - F_2(1/2) = \left(\frac{5}{6}\beta - \frac{1}{6}\right) - \frac{1}{3} = \frac{5}{6}\beta - \frac{1}{6} - \frac{2}{6} = \frac{5}{6}\beta - \frac{3}{6} = \frac{5}{6}\beta - \frac{1}{2}$. Step 5: Calculate the total area $A$. $$A = I_1 + I_2 = \left(\frac{1}{3} - \frac{5}{6}\alpha\right) + \left(\frac{5}{6}\beta - \frac{1}{2}\right)$$ $$A = \frac{2}{6} - \frac{5}{6}\alpha + \frac{5}{6}\beta - \frac{3}{6} = \frac{-1 - 5\alpha + 5\beta}{6}$$ Substitute the values of $\alpha = \frac{3-\sqrt{5}}{2}$ and $\beta = \frac{-1+\sqrt{5}}{2}$: $$A = \frac{1}{6} \left(-1 - 5\left(\frac{3-\sqrt{5}}{2}\right) + 5\left(\frac{-1+\sqrt{5}}{2}\right)\right)$$ $$A = \frac{1}{6} \left(\frac{-2 - 15 + 5\sqrt{5} - 5 + 5\sqrt{5}}{2}\right)$$ $$A = \frac{1}{12} (-22 + 10\sqrt{5}) = \frac{-11+5\sqrt{5}}{6}$$ Step 6: Calculate the final expression $(6A+11)^2$. First, calculate $6A+11$: $$6A+11 = 6 \left(\frac{-11+5\sqrt{5}}{6}\right) + 11$$ $$6A+11 = (-11+5\sqrt{5}) + 11$$ $$6A+11 = 5\sqrt{5}$$ Now, square the result: $$(6A+11)^2 = (5\sqrt{5})^2 = 5^2 \times (\sqrt{5})^2 = 25 \times 5 = 125$$ The final answer is $\boxed{125}$.