The minimum  value of $f\left(x\right)={\int }_0^4{e}^{|x-t|}·dt,$ $0\le x\le 4$ is

$f\left(x\right)={\int }_0^x{e}^{x-t}dt+{\int }_x^4{e}^{t-x}dt$

$=-{\left.e^{x-t}\right|}_0^x+{\left.e^{t-x}\right|}_x^4=-1+e^x+e^{4-x}-1=e^x+e^{4-x}-2$

By symmetry, the minimum value of $f\left(x\right)$ is $f\left(2\right)=2\left(e^2-1\right).$