Question

Consider $f\left(x\right)=minimum\left(x+2,\sqrt{4-x}\right),\forall x\le 4.$ If the area bounded by $y=f\left(x\right)$ and the $x$ -axis is $\frac{22}{k}$ square units, then the value of $k$ is

Answer 1

Answer: 3

Required area $={\int }_0^2\left[\left(4-y^2\right)-\left(y-2\right)\right]dy$
$={\int }_0^2\left(6-y^2-y\right)dy$
$={\left[6y-\frac{y^3}{3}-\frac{y^2}{2}\right]}_0^2=12-\frac{8}{3}-2$
$=\frac{22}{3}$ sq. units