∫^2_0|x-1|dx=

The correct answer is Option (1) → 1

$\int\limits^{2}_{0}|x-1|dx$

$|x-1|\left\{\begin{matrix}1-x,&x<1\\x-1,&x≥1\end{matrix}\right.$

$=\int\limits^{2}_{0}1-xdx+\int\limits^{2}_{1}x-1dx$

$=\left[x-\frac{x^2}{2}\right]_0^1+\left[\frac{x^2}{2}-x\right]^{2}_{1}$

$=1-\frac{1}{2}+\frac{4}{2}-2+1-\frac{1}{2}$

$=2-1+2-2=1$ sq. unit