The difference between the greatest and least values of the function $F\left(x\right)={\int }_0^x (t+1)\mathrm{d}t\text{ on }[1,3]$ is

$F^{\mathrm{\prime }}\left(x\right)=x+1>0\text{ for }x\in [1,3]$ Therefore, the greatest value is F(3) and the least value is F(1). The required difference is ${\int }_0^3 (t+1)dt-{\int }_0^1 (t+1)dt={\int }_1^3 (t+1)dt=6.$