The area of the shaded region in the given figure is $\frac{2}{3}$. then m is equal to :

4

The correct answer is Option (4) → 4

Point of intersection

$y=mx$

$y^2=16x$

so $m^2x^2=16x$

$m^x-16=0$

$x=\frac{16}{m^2}$

Area = $\int\limits_{0}^{\frac{16}{m^2}}-mx+4\sqrt{x}dx$

$=\left[\frac{8}{3}x^{\frac{3}{2}}-\frac{mx^2}{2}\right]_{0}^{\frac{16}{m^2}}=\frac{2}{3}$

$\frac{8×64}{3×m^3}-\frac{256}{m^3}=\frac{2}{3}$

$⇒m=4$