\(\int_1^5(|x-3|+|1-x|) d x=\)

\(\int_1^5[|x-3|+|1-x|] d x\)
\(=\int^5|x-3| d x+\int^5|1-x| d x\)
\(=\int_1^3|x-3| d x+\int_3^5|x-3| d x+\int_1^5|1-x| d x\)
\(=\int_1^3(3-x) d x+\int_3^5(x-3) d x+\int_1^5(x-1) d x\)
\(=\left[3 x-\frac{x^2}{2}\right]_1^3+\left[\frac{x^2}{2}-3 x\right]_3^5+\left[\frac{x^2}{2}-x\right]_1^5\)
\(=\left(3 \times 3-\frac{9}{2}\right)-\left(3 \times 1-\frac{1}{2}\right)+\left(\frac{5 \times 5}{2}-3 \times 5\right)\)
\(-\left(\frac{3 \times 3}{2}-3 \times 3\right)+\left(\frac{5 \times 5}{2}-5\right)-\left(\frac{1}{2}-1\right)\)
\(=\frac{9}{2}-\frac{5}{2}-\frac{5}{2}+\frac{9}{2}+\frac{15}{2}+\frac{1}{2}\)
\(=\frac{9-5-5+9+15+1}{2}=\frac{24}{2}=12\)