The area of the region bounded by $y^2 = 9x,x = 2,x = 4$ and the x-axis in the first quadrant, is

$16 - 4\sqrt{2}$ sq. units

The correct answer is Option (4) → $16 - 4\sqrt{2}$ sq. units

Region: \(y^2=9x\Rightarrow y=3\sqrt{x}\) (first quadrant). Area = \(\int_{2}^{4}3\sqrt{x}\,dx\).

\(\displaystyle \int_{2}^{4}3\sqrt{x}\,dx=3\int_{2}^{4}x^{1/2}\,dx=3\left[\frac{2}{3}x^{3/2}\right]_{2}^{4}=2\left(4^{3/2}-2^{3/2}\right)\)

\(\displaystyle =2\left(8-2^{3/2}\right)=16-2^{5/2}=16-4\sqrt{2}\)