The area (in sq. units) of the region described by $$\{(x,y) : y^2 \leq 2x$$ and $$y \geq 4x - 1\}$$ is

The region is defined by the inequalities $$ y^2 \leq 2x $$ and $$ y \geq 4x - 1 $$. To find the area, we first determine the points of intersection between the parabola $$ y^2 = 2x $$ and the line $$ y = 4x - 1 $$.

Substitute $$ x = \frac{y^2}{2} $$ from the parabola into the line equation:

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

Simplify:

$$ y = 2y^2 - 1 $$

Rearrange into a quadratic equation:

$$ 2y^2 - y - 1 = 0 $$

Solve using the quadratic formula. The discriminant is:

$$ D = (-1)^2 - 4 \cdot 2 \cdot (-1) = 1 + 8 = 9 $$

So,

$$ y = \frac{1 \pm \sqrt{9}}{4} = \frac{1 \pm 3}{4} $$

This gives two solutions:

$$ y = \frac{1 + 3}{4} = 1 \quad \text{and} \quad y = \frac{1 - 3}{4} = -\frac{1}{2} $$

Find the corresponding $$ x $$ values using $$ x = \frac{y^2}{2} $$:

For $$ y = 1 $$:

$$ x = \frac{(1)^2}{2} = \frac{1}{2} $$

For $$ y = -\frac{1}{2} $$:

$$ x = \frac{\left(-\frac{1}{2}\right)^2}{2} = \frac{\frac{1}{4}}{2} = \frac{1}{8} $$

Thus, the points of intersection are $$ \left( \frac{1}{2}, 1 \right) $$ and $$ \left( \frac{1}{8}, -\frac{1}{2} \right) $$.

Rewrite the inequalities to express $$ x $$ in terms of $$ y $$:

From $$ y^2 \leq 2x $$, we get $$ x \geq \frac{y^2}{2} $$.

From $$ y \geq 4x - 1 $$, solve for $$ x $$:

$$ 4x \leq y + 1 \quad \Rightarrow \quad x \leq \frac{y + 1}{4} $$

For these inequalities to hold simultaneously, we need $$ \frac{y^2}{2} \leq \frac{y + 1}{4} $$. Solve this inequality:

Multiply both sides by 4:

$$ 4 \cdot \frac{y^2}{2} \leq y + 1 \quad \Rightarrow \quad 2y^2 \leq y + 1 $$

Rearrange:

$$ 2y^2 - y - 1 \leq 0 $$

The quadratic $$ 2y^2 - y - 1 = 0 $$ has roots $$ y = 1 $$ and $$ y = -\frac{1}{2} $$, and since the coefficient of $$ y^2 $$ is positive, the quadratic is less than or equal to zero between the roots. Thus, $$ y \in \left[ -\frac{1}{2}, 1 \right] $$.

In this interval, for each $$ y $$, $$ x $$ ranges from $$ \frac{y^2}{2} $$ to $$ \frac{y + 1}{4} $$. The area is the integral:

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

Simplify the integrand:

$$ \frac{y + 1}{4} - \frac{y^2}{2} = \frac{y}{4} + \frac{1}{4} - \frac{y^2}{2} $$

So,

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

Split the integral:

$$ \text{Area} = \frac{1}{4} \int_{-\frac{1}{2}}^{1} y dy + \frac{1}{4} \int_{-\frac{1}{2}}^{1} 1 dy - \frac{1}{2} \int_{-\frac{1}{2}}^{1} y^2 dy $$

Compute each integral separately.

First integral:

$$ \int_{-\frac{1}{2}}^{1} y dy = \left[ \frac{y^2}{2} \right]_{-\frac{1}{2}}^{1} = \frac{(1)^2}{2} - \frac{\left(-\frac{1}{2}\right)^2}{2} = \frac{1}{2} - \frac{\frac{1}{4}}{2} = \frac{1}{2} - \frac{1}{8} = \frac{4}{8} - \frac{1}{8} = \frac{3}{8} $$

Second integral:

$$ \int_{-\frac{1}{2}}^{1} 1 dy = \left[ y \right]_{-\frac{1}{2}}^{1} = 1 - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{3}{2} $$

Third integral:

$$ \int_{-\frac{1}{2}}^{1} y^2 dy = \left[ \frac{y^3}{3} \right]_{-\frac{1}{2}}^{1} = \frac{(1)^3}{3} - \frac{\left(-\frac{1}{2}\right)^3}{3} = \frac{1}{3} - \frac{-\frac{1}{8}}{3} = \frac{1}{3} - \left( -\frac{1}{24} \right) = \frac{1}{3} + \frac{1}{24} = \frac{8}{24} + \frac{1}{24} = \frac{9}{24} = \frac{3}{8} $$

Now substitute back:

$$ \text{Area} = \frac{1}{4} \cdot \frac{3}{8} + \frac{1}{4} \cdot \frac{3}{2} - \frac{1}{2} \cdot \frac{3}{8} $$

Calculate each term:

$$ \frac{1}{4} \cdot \frac{3}{8} = \frac{3}{32} $$

$$ \frac{1}{4} \cdot \frac{3}{2} = \frac{3}{8} = \frac{12}{32} $$

$$ -\frac{1}{2} \cdot \frac{3}{8} = -\frac{3}{16} = -\frac{6}{32} $$

Sum them:

$$ \frac{3}{32} + \frac{12}{32} - \frac{6}{32} = \frac{3 + 12 - 6}{32} = \frac{9}{32} $$

Hence, the area is $$ \frac{9}{32} $$ square units. Comparing with the options, Option A matches.

Hence, the correct answer is Option A.