Let $g\left(x\right)={\int }_0^{x}f\left(t\right)dt$ , where  $\frac{1}{2}\le f\left(t\right)\le 1,t\in [0,1]$ and $0\le f\left(t\right)\le \frac{1}{2}$  for t $\in (1,2]$, then 

$\begin{array}{l}g\left(2\right)={\int }_0^{2}f\left(t\right)dt\\ ={\int }_0^{1}f\left(t\right)dt+{\int }_1^{2}f\left(t\right)dt\\ as\frac{1}{2}\le f\left(t\right)\le 1for0<x<1\\ \Rightarrow {\int }_0^1\frac{1}{2}dt\le {\int }_0^{1}f\left(t\right)dtK,{\int }_0^{1}dt\\ \Rightarrow \frac{1}{2}\le {\int }_0^{1}f\left(t\right)dt\le \mathrm{1...........}\left(i\right)\\ as0\le f\left(t\right)\le \frac{1}{2}for1<t\le 2\\ \Rightarrow {\int }_1^{2}0dt\le {\int }_1^{2}f\left(t\right)\le {\int }_1^2\frac{1}{2}dt\\ 0\le {\int }_1^{2}f\left(t\right)dt\le \frac{1}{2}\mathrm{...............}\left(ii\right)\\ AddingEqs.\left(i\right)and\left(ii\right).weget\\ \frac{1}{2}\le g\left(2\right)\le \frac{3}{2}\end{array}$