Question

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

Answer 1

Given,
$I={\int }_1^4[|x-1|+|x-2|+|x-4|]dx={\int }_1^4(x-1)dx+{\int }_1^2-(x-2)dx+{\int }_2^4(x-2)dx+{\int }_1^4-(x-4)dx={[\frac{x^2}{2}-x]}_1^4+{[-\frac{x^2}{2}+2x]}_1^2+{[\frac{x^2}{2}-2x]}_2^4+{[-\frac{x^2}{2}+4x]}_1^4=(\frac{16}{2}-4-\frac{1}{2}+1)+(-2+4+\frac{1}{2}-2)+(\frac{16}{2}-8-2+4)+(-\frac{16}{2}+16+\frac{1}{2}-4)=(5-\frac{1}{2})+\frac{1}{2}+2+4+\frac{1}{2}=11+\frac{1}{2}=\frac{23}{2}$