y=|x−1| ......(i) and y=3−|x| ........(ii) Solving the Eqs. (i) and (ii) if x≥1, then 3−x=x−1⇒x=2 if 0≤x≤1, then 3−x=1−x (which is not possible) Also, if x<0, then 3+x=1−x⇒x=−1 Thus, required area =
2
∫
−1
(3−|x|−|x−1|)dx =
0
∫
−1
[3+x−(1−x)]dx+
1
∫
0
[(3−x)−(1−x)]dx+
2
∫
1
[(3−x)−(x−1)]dx =
0
∫
−1
(2+2x)dx+2‌
1
∫
0
1dx+
2
∫
1
(4−2x)dx =[2x+x2]−10+[4x−x2]12 =1+2+1=4 sq units