Let f be a twice differentiable function defined in [-3,3] such that $f\left(0\right)=-4,f\text{'}\left(3\right)=0,f\text{'}\left(-3\right)=12$  and $f\text{'}\text{'}\left(x\right)\ge -2\forall x\in \left[-3,3\right]$. Let $g\left(x\right)={\int }_0^{x}f\left(t\right)dt$. If the maximum value of g(x) is M, then  $\frac{M}{8}$ is _________.

  $$\begin{gathered} f\text{'}\left(x\right)\ge -2\forall x\in \left[-3,3\right] \\ {\left.f\text{'}\text{'}\left(x\right)dx\ge -2\underset{-3}{\overset{3}{\int }}dx\Rightarrow f\text{'}\left(x\right)\right]}_{-3}^3\ge -12\Rightarrow f\text{'}\left(3\right)-f\text{'}\left(-3\right)\ge -12 \\ \Rightarrow 0-12\ge -12\Rightarrow \end{gathered}$$
   
 equality sign holds
 $$\begin{gathered} \therefore f\text{'}\text{'}\left(x\right)=-2\forall x\in \left[-3,3\right] \\ f\text{'}\left(x\right)=-2x+c\Rightarrow f\text{'}\left(3\right)=c-6\Rightarrow c-6=0 \\ c=6 \\ \therefore f\text{'}\left(x\right)=-2x+6\Rightarrow f\left(x\right)=-x^2+6x+k. \\ f\left(0\right)=-4\Rightarrow k=-4\therefore f\left(x\right)=6x-x^2-4 \end{gathered}$$
                  
   $g\left(x\right)=\frac{-x^3}{3}+3x^2-4x$  in [-3,3]

$g\text{'}\left(x\right)=-x^2+6x-4$
$g\left(-3\right)=48$  Which is maximum value in [-3,3]