${\int }_0^4 \left\{\right|x-1|+|x-3\left|dx\right\}=$

$\begin{array}{l}I=I_1+I_2\\ I_1={\int }_0^1 -(x-1)dx+{\int }_1^4 (x-1)dx\\ =-\frac{1}{2}{\left[(x-1{)}^2\right]}_0^1+\frac{1}{2}{\left[(x-1{)}^2\right]}_1^4\\ I_1=\frac{1}{2}+\frac{1}{2}\cdot 9=5\\ I_2={\int }_0^3 -(x-3)dx+{\int }_3^4 (x-3)dx\\ =-\frac{1}{2}{\left[(x-3{)}^2\right]}_0^3+\frac{1}{2}{\left[(x-3{)}^2\right]}_3^4\\ I_2=\frac{9}{2}+\frac{1}{2}=5\\ \therefore I=I_1+I_2=5+5=1\overline{0}\end{array}$