The area (in sq.units) of region bounded by y^2 = 9x, x=2, x =4 and t…

The correct answer is Option (4) → $4[4 - \sqrt{2}]$

Given curve: y² = 9x

Bounded by: x = 2 , x = 4 , and the x-axis in the first quadrant

⇒ y = 3√x (since y ≥ 0)

Required area:

$A = \int_{2}^{4} (3\sqrt{x} - 0)\,dx$

$A = 3\int_{2}^{4} x^{1/2}\,dx$

$A = 3\left[\frac{2}{3}x^{3/2}\right]_{2}^{4}$

$A = 2\left(4^{3/2} - 2^{3/2}\right)$

$A = 2(8 - 2\sqrt{2})$

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