The area of the figure bounded by the curves y = |x – 1| and y = 3 –…

Required area

$=\int\limits_{-1}^0 3+x-(-x+1) d x+\int\limits_0^1 3-x-(-x+1) d x+\int\limits_1^2 3-x-(x-1) d x$

$=\int\limits_{-1}^0(2+2 x) d x+\int\limits_0^1 2 d x+\int\limits_1^2(4-2 x) d x$

= 4 sq. units.

Hence (3) is the correct answer.