Question

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

• A. $6$ sq. units
• B. $2$ sq. units
• C. $3$ sq. units
• D. $4$ sq. units

Answer 1

The correct option is D $4$ sq. units

Intersection point of $y=x-1$ and $y=3-x$ is $(2,1)$ and intersection point of $y=-x+1$ and $y=3+x$ is $(-1,2)$
$A=\underset{-1}{\overset{0}{\int }}\left[\right(3+x)-(-x+1\left)\right]dx+\underset{0}{\overset{1}{\int }}\left[\right(3-x)-(-x+1\left)\right]dx+\underset{1}{\overset{2}{\int }}\left[\right(3-x)-(x-1\left)\right]dx$
$=\underset{-1}{\overset{0}{\int }}(2+2x)dx+\underset{0}{\overset{1}{\int }}2dx+\underset{1}{\overset{2}{\int }}(4-2x)dx$
$=[2x+x^2{]}_{-1}^0+[2x{]}_0^1+[4x-x^2{]}_1^2$
$=0-(-2+1)+(2-0)+(8-4)-(4-1)$
$=1+2+4-3=4$ sq. units