The area of the region bounded by the curves $y=|x-1|$ and $y=3-|x|$ is:

Find points of intersection:$|x-1|=3-|x|$ Case 1: $x\ge 1⟹x-1=3-x⟹2x=4⟹x=2,y=1$.Case 2: $0\le x<1⟹1-x=3-x⟹1=3$ (Impossible).Case 3: $x<0⟹1-x=3+x⟹-2=2x⟹x=-1,y=2$.The curves are V-shaped. The region forms a rectangle/parallelogram.Vertices: $(-1,2)$, $(1,0)$, $(2,1)$, $(0,3)$.Area can be found using the shoelace formula or splitting into triangles.Alternatively, drawing the graph shows a square of diagonal length $\sqrt{2^2+2^2}=\sqrt{8}$.Area $=4$.