Question

Consider $f\left(x\right)=minimum\left(x + 2 , \sqrt{4 - x}\right), \, \forall x\leq 4.$ 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

Required area $=\displaystyle \int _{0}^{2} \left[\left(4 - y^{2}\right) - \left(y - 2\right)\right] d y$
$=\displaystyle \int _{0}^{2} \left(6 - y^{2} - y\right) d y$
$=\left[6 y - \frac{y^{3}}{3} - \frac{y^{2}}{2}\right]_{0}^{2}=12-\frac{8}{3}-2$
$=\frac{22}{3}$ sq. units