Question

${\int }_1^4\{|x-1|+|x-2|+|x-3|\}dx$

Answer 1

${\int }_1^4\{|x-1|+|x-2|+|x-3|\}dx={\int }_1^2(|x-1|+|x-2|+|x-3|)dx+{\int }_2^3\{|x-1|+|x-2|+|x-3|\}dx+{\int }_3^4\{|x-1|+|x-2|+|x-3|\}dx={\int }_1^2\{x-1-(x-2)-(x-3\left)\right\}dx+{\int }_2^3(x-1+x-2-(x-3)dx+{\int }_3^4(x-1+x-2+x-3)dx={\int }_1^2(-x+4)dx+{\int }_2^{3}xdx+{\int }_3^4(3x-6)dx=[-\frac{x^2}{2}+4x{]}_1^2+[\frac{x^2}{2}{]}_2^3+[\frac{3x^2}{2}-6x{]}_3^4=(\frac{-2^2}{2}+8)-(\frac{-1}{2}+4)+\frac{1}{2}(3^2-2^2)+(\frac{3}{2}\times 4^2-6\times 4)-(\frac{3}{2}\times 3^2-6\times 3)=6-\frac{7}{2}+\frac{5}{2}+(24-24)-(-\frac{9}{2})=\frac{12-7+5}{2}+0+\frac{9}{2}=\frac{19}{2}$