The area of the region bounded by the curves $\mathrm{y}+\left|\mathrm{x}\right|=3\mathrm{and}\mathrm{y}=|\mathrm{x}-1|$.

$\mathrm{Since},\mathrm{y}=3-\left|\mathrm{x}\right|=\{\begin{array}{l}3+\mathrm{x},\mathrm{x}<0\\ 3-\mathrm{x},\mathrm{x}\ge 0\end{array}$

$\mathrm{y}=|\mathrm{x}-1|=\{\begin{array}{l}\mathrm{x}-1,\mathrm{x}\ge 0\\ 1-\mathrm{x},\mathrm{x}<0\end{array}$

$\therefore$Required area

$\begin{array}{l}={\int }_{-1}^0 \left[\right(3+\mathrm{x})-(1-\mathrm{x}\left)\right]\mathrm{dx}+{\int }_0^1 \left[\right(3-\mathrm{x})-(1-\mathrm{x}\left)\right]\mathrm{dx}+{\int }_1^2 \left[\right(3-\mathrm{x})-(\mathrm{x}-1\left)\right]\mathrm{dx}\\ ={\int }_{-1}^0 (2+2\mathrm{x})\mathrm{dx}+{\int }_0^1 2\mathrm{dx}+{\int }_1^2 (4-2\mathrm{x})\mathrm{dx}={\left[2\mathrm{x}+{\mathrm{x}}^2\right]}_{-1}^0+[2\mathrm{x}{]}_0^1+{\left[4\mathrm{x}-{\mathrm{x}}^2\right]}_1^2\\ =-(-2+1)+2+(4-3)=1+3=4\text{ squnits }\end{array}$