Question

Evaluate ${\int }_0^2|1-x|dx$

Answer 1

$1-x<0$ for $1<x\le 2$ and $\ge 0$ for $0<x\le 1$

$\Rightarrow I={\int }_0^1(1-x)dx-{\int }_1^2(1-x)dx$

$=x-\frac{x^2}{2}{|}_0^1-(x-\frac{x^2}{2}){|}_1^2=\frac{1}{2}-[0-\frac{1}{2}]=1$.