Area Bounded by Minimum of Two Functions

To find the value of $9\alpha$, we first analyze the given functions $f(x)$ and $g(x)$:

$f(x) = x^2 + \frac{5}{12}$

$g(x) = \begin{cases} 2\left(1 - \frac{4|x|}{3}\right), & |x| \le \frac{3}{4} \\ 0, & |x| > \frac{3}{4} \end{cases}$

Notice that both $f(x)$ and $g(x)$ are even functions, i.e., $f(-x) = f(x)$ and $g(-x) = g(x)$. Consequently, the region bounded by $y = \min\{f(x), g(x)\}$ and the $x$-axis over the domain $|x| \le \frac{3}{4}$ is symmetric with respect to the $y$-axis.

Therefore, the area $\alpha$ is given by: $\alpha = 2 \int_{0}^{3/4} \min\{f(x), g(x)\} \, dx$

For $x \ge 0$, the functions simplify to: $f(x) = x^2 + \frac{5}{12}$ $g(x) = 2 - \frac{8x}{3} \quad \text{for } 0 \le x \le \frac{3}{4}$

To find the point of intersection of $f(x)$ and $g(x)$ in the interval $\left[0, \frac{3}{4}\right]$, we equate the two functions: $x^2 + \frac{5}{12} = 2 - \frac{8x}{3}$

Multiplying the entire equation by $12$ yields: $12x^2 + 5 = 24 - 32x$ $12x^2 + 32x - 19 = 0$

Factoring the quadratic equation: $(2x - 1)(6x + 19) = 0$

Since $x \ge 0$, the relevant root is: $x = \frac{1}{2}$

Now we compare the values of $f(x)$ and $g(x)$ in the two sub-intervals:

1. For $x \in \left[0, \frac{1}{2}\right]$: At $x = 0$, $f(0) = \frac{5}{12}$ and $g(0) = 2$, so $f(x) \le g(x)$. Thus, $\min\{f(x), g(x)\} = f(x) = x^2 + \frac{5}{12}$.

2. For $x \in \left[\frac{1}{2}, \frac{3}{4}\right]$: At $x = \frac{3}{4}$, $f\left(\frac{3}{4}\right) = \frac{47}{48}$ and $g\left(\frac{3}{4}\right) = 0$, so $g(x) \le f(x)$. Thus, $\min\{f(x), g(x)\} = g(x) = 2 - \frac{8x}{3}$.

Now, we calculate the area in the first quadrant, denoted by $A_1$: $A_1 = \int_{0}^{1/2} \left(x^2 + \frac{5}{12}\right) dx + \int_{1/2}^{3/4} \left(2 - \frac{8x}{3}\right) dx$

Evaluating the first integral $I_1$: $I_1 = \left[ \frac{x^3}{3} + \frac{5x}{12} \right]_{0}^{1/2} = \frac{1}{24} + \frac{5}{24} = \frac{6}{24} = \frac{1}{4}$

Evaluating the second integral $I_2$: $I_2 = \left[ 2x - \frac{4x^2}{3} \right]_{1/2}^{3/4} = \left(2 \cdot \frac{3}{4} - \frac{4}{3} \cdot \frac{9}{16}\right) - \left(2 \cdot \frac{1}{2} - \frac{4}{3} \cdot \frac{1}{4}\right)$ $I_2 = \left(\frac{3}{2} - \frac{3}{4}\right) - \left(1 - \frac{1}{3}\right) = \frac{3}{4} - \frac{2}{3} = \frac{1}{12}$

Summing the two integrals gives $A_1$: $A_1 = I_1 + I_2 = \frac{1}{4} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3}$

Using symmetry, the total area $\alpha$ is: $\alpha = 2 A_1 = 2 \times \frac{1}{3} = \frac{2}{3}$

Finally, we compute $9\alpha$: $9\alpha = 9 \times \frac{2}{3} = 6$