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

The region is bounded by the parabola $y^2 = 8x$ and the vertical line $x = 2$.

The parabola $y^2 = 8x$ opens to the right with vertex at the origin and can be rewritten as $x = \frac{y^2}{8}$.

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To find the intersection points, substitute $x = 2$ into the parabola equation:

$y^2 = 8(2)$

$y^2 = 16$

$y = \pm 4$

The curves intersect at $(2, 4)$ and $(2, -4)$.

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At any height $y$ between $-4$ and $4$:

Right boundary: $x = 2$

Left boundary: $x = \frac{y^2}{8}$

Horizontal width: $2 - \frac{y^2}{8}$

The area is:

$\text{Area} = \int_{-4}^{4} \left(2 - \frac{y^2}{8}\right) dy$

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$\int_{-4}^{4} \left(2 - \frac{y^2}{8}\right) dy$

$= \left[2y - \frac{y^3}{24}\right]_{-4}^{4}$

At $y = 4$:

$2(4) - \frac{4^3}{24}$

$= 8 - \frac{64}{24}$

$= 8 - \frac{8}{3}$

At $y = -4$:

$2(-4) - \frac{(-4)^3}{24}$

$= -8 + \frac{8}{3}$

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$\left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right)$

$= 8 - \frac{8}{3} + 8 - \frac{8}{3}$

$= 16 - \frac{16}{3}$

$= \frac{48 - 16}{3}$

$= \frac{32}{3}$

Therefore, the area of the region is $\frac{32}{3}$ square units.