Consider the functions $f, g : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^{2} + \frac{5}{12}$ and $g(x) = \begin{cases} 2\left(1 - \frac{4|x|}{3}\right), & |x| \leq \frac{3}{4}, \\ 0, & |x| > \frac{3}{4}. \end{cases}$ If $\alpha$ is the area of the region $\left\{(x, y) \in \mathbb{R} \times \mathbb{R} : |x| \leq \frac{3}{4}, 0 \leq y \leq \min\{f(x), g(x)\}\right\},$ then the value of $9\alpha$ is ______________.

$f(x) = x^2 + \dfrac{5}{12}$ $g(x) = \begin{cases} 2\left(1 - \dfrac{4|x|}{3}\right), & |x| \le \dfrac{3}{4}, \\ 0, & |x| > \dfrac{3}{4}. \end{cases}$ $\alpha = \text{area where } |x|\le \dfrac{3}{4},\ 0\le y\le \min\{f(x),g(x)\}; \quad \text{find } 9\alpha$ $f, g \text{ even} \Rightarrow \alpha = 2\int_0^{3/4} \min\{f,g\}\,dx$ $f=g:\ 12x^2+32x-19=0 \Rightarrow x=\dfrac{1}{2}$ $x=0:\ f=\dfrac{5}{12}<2=g \Rightarrow \min = f \text{ on } \left[0,\tfrac12\right]$ $x=\dfrac{3}{4}:\ g=0<f \Rightarrow \min = g \text{ on } \left[\tfrac12,\tfrac34\right]$ $\int_0^{1/2}\left(x^2+\dfrac{5}{12}\right)dx = \left[\dfrac{x^3}{3}+\dfrac{5x}{12}\right]_0^{1/2} = \dfrac{1}{24}+\dfrac{5}{24} = \dfrac{1}{4}$ $\int_{1/2}^{3/4}\left(2-\dfrac{8x}{3}\right)dx = \left[2x-\dfrac{4x^2}{3}\right]_{1/2}^{3/4} = \dfrac{3}{4}-\dfrac{2}{3} = \dfrac{1}{12}$ $\alpha = 2\left(\dfrac{1}{4}+\dfrac{1}{12}\right) = 2\cdot\dfrac{1}{3} = \dfrac{2}{3}$ $9\alpha = 9\cdot\dfrac{2}{3} = 6$ $\boxed{9\alpha = 6}$