Question

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

• A. $3$
• B. $4$
• C. $6$
• D. $2$

Answer 1

Answer: (B)

$4$

Consider the curves $y=|x-1|$ & $y=3-|x|$

when $x<0$

$3-(-x)=(-x+1)$

$\Rightarrow 3+x=-x+1$

$\Rightarrow 2x=-2$

$x=-1$

Put $x=-1$ in $y=|x-1|$ we have

$y=2$

When $x>0$

$3-x=x-1$

$\Rightarrow 3+1=2x$

$x=2$

put $x=2$ in $y=|x-1|$ we have

$y=1$

Thus pt. of intersection is $(-1,2),(2,1)$

$\therefore$ required area $={\int }_{-1}^0(3+x+x-1)dx+{\int }_0^1(3-x+x-1)dx$

$+{\int }_1^{2}3-x-x+1dx$

$\Rightarrow {\int }_{-1}^0(2x+2)dx+{\int }_0^{1}2dx+{\int }_1^2(4-2x)dx$

$\Rightarrow {\left(\frac{2x^2}{2}\right)}^1+(2x{)}_{-1}^0+(2x{)}_0^1+(4x{)}_1^2-{\left(\frac{2x^2}{2}\right)}_1^2$

$\Rightarrow -1+2(-(-1\left)\right)+2+4(2-1)-(2^2-1^2)$

$\Rightarrow -1+2+2+4-3=4$