The area (in sq. units) of the region bounded by the parabola $y^2 = 4x$ and the line $x = 1$ is

The region is bounded by the parabola $y^2 = 4x$ and the vertical line $x = 1$.

The parabola $y^2 = 4x$ opens to the right.

To find intersection points, substitute $x = 1$ into the parabola equation:

$y^2 = 4(1)$

$y^2 = 4$

$y = \pm 2$

The curves intersect at $(1, 2)$ and $(1, -2)$.

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From the parabola equation $y^2 = 4x$, we can write:

$x = \frac{y^2}{4}$

For any value of $y$ between $-2$ and $2$:

Right boundary: $x = 1$

Left boundary: $x = \frac{y^2}{4}$

The horizontal width of a strip is $1 - \frac{y^2}{4}$.

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The area is:

$A = \int_{-2}^{2} \left(1 - \frac{y^2}{4}\right) dy$

$A = \left[y - \frac{y^3}{12}\right]_{-2}^{2}$

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At $y = 2$:

$2 - \frac{(2)^3}{12}$

$= 2 - \frac{8}{12}$

$= 2 - \frac{2}{3}$

$= \frac{4}{3}$

At $y = -2$:

$-2 - \frac{(-2)^3}{12}$

$= -2 - \frac{-8}{12}$

$= -2 + \frac{2}{3}$

$= -\frac{4}{3}$

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$A = \frac{4}{3} - \left(-\frac{4}{3}\right)$

$A = \frac{4}{3} + \frac{4}{3}$

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

Therefore, the area of the region is $\frac{8}{3}$ square units.