Let  $f:[0,4]\to R$  is continuous function such that $\left|f\left(x\right)\right|=6\left|(x-n+1)(x-n)\right|$  where  $n\in N \forall x\in [n-1,n]$ and  $g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt-\underset{x}{\overset{4}{\int }}f\left(t\right)dt$  has maximum at  $x=2$, then  $\underset{1/2}{\overset{4}{\int }}f\left(t\right)dt$  is/are