The area (in sq. units) of the region bounded by the parabola $y^2 = 8x$ and the line $x = 2$ is

$\frac{32}{3}$

The correct answer is Option (1) → $\frac{32}{3}$

Given: Area bounded by the parabola $y^2 = 8x$ and the line $x = 2$.

From $y^2 = 8x$, rewrite as $x = \frac{y^2}{8}$.

Limits: When $x = 2$, from $y^2 = 8 \cdot 2 = 16$, so $y = \pm 4$.

Required area: $A = \int_{-4}^{4} \left(2 - \frac{y^2}{8} \right) dy$

$= \int_{-4}^{4} \left(2 - \frac{y^2}{8} \right) dy = \left[2y - \frac{y^3}{24} \right]_{-4}^{4}$

$= \left(2 \cdot 4 - \frac{4^3}{24} \right) - \left(2 \cdot (-4) - \frac{(-4)^3}{24} \right)$

$= (8 - \frac{64}{24}) - (-8 + \frac{64}{24})$

$= 8 - \frac{8}{3} + 8 - \frac{8}{3} = 16 - \frac{16}{3} = \frac{48 - 16}{3} = \frac{32}{3}$