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

Plotting the diagram of the given function we have,

Now by diagram we can see twice of $A_1$ and $A_2$ will be the required area,

So, required area $=2·\left(A_1+A_2\right)$

$\Rightarrow \alpha =\left({\int }_0^{\frac{1}{2}}\left(x^2+\frac{5}{12}\right)dx+\frac{1}{2}\left(\frac{3}{4}-\frac{1}{2}\right)·\frac{2}{3}\right)\times 2$

$\Rightarrow \alpha =\left({\left.\left(\frac{x^3}{3}+\frac{5x}{12}\right)\right|}_0^{\frac{1}{2}}+\frac{1}{2}\cdot \frac{1}{4}\cdot \frac{2}{3}\right)\times 2$

$\Rightarrow \alpha =\left(\frac{1}{4}+\frac{1}{12}\right)\times 2$

$\Rightarrow \alpha =\frac{2}{3}\Rightarrow 9\alpha =6$