The area (in sq. units) of the region $$\{(x, y) \in R^2 | 4x^2 \le y \le 8x + 12\}$$ is

We have to find the area enclosed by the set of all points $$ (x , y) $$ that satisfy the two simultaneous inequalities $$4x^{2} \le y \le 8x+12.$$

Because $$y$$ is trapped between the two curves $$y = 4x^{2}$$ (parabola opening upward) and $$y = 8x+12$$ (straight line), the required area can be obtained by integrating the vertical “strip” height $$\bigl[\; (8x+12) - (4x^{2}) \bigr]$$ with respect to $$x$$ over the interval where the two curves intersect.

To discover that interval, we set the two expressions for $$y$$ equal to one another:

$$4x^{2} = 8x + 12.$$

Bringing every term to the left side gives

$$4x^{2} - 8x - 12 = 0.$$

Now we divide by $$4$$ so that the quadratic is in its simplest form:

$$x^{2} - 2x - 3 = 0.$$

This factors neatly:

$$ (x - 3)(x + 1) = 0. $$

Hence the solutions are

$$x = 3 \quad \text{or} \quad x = -1.$$ So the two curves intersect at $$x = -1$$ and $$x = 3.$$ These become the limits of integration, because for any $$x$$ between $$-1$$ and $$3$$ the inequality $$4x^{2} \le 8x + 12$$ is satisfied, while outside this range it is not.

Next, we write down the definite integral that represents the area:

$$ \text{Area} = \int_{x=-1}^{3} \bigl[(8x + 12) - (4x^{2})\bigr]\,dx. $$

First we simplify the integrand:

$$ (8x + 12) - (4x^{2}) = -4x^{2} + 8x + 12. $$

Now we integrate term by term, recalling the standard power rule $$\int x^{n}\,dx = \dfrac{x^{n+1}}{n+1} + C$$:

$$ \int -4x^{2}\,dx = -4 \cdot \frac{x^{3}}{3} = -\frac{4}{3}x^{3}, $$

$$ \int 8x\,dx = 8 \cdot \frac{x^{2}}{2} = 4x^{2}, $$

$$ \int 12\,dx = 12x. $$

Combining these antiderivatives, the integral of the whole expression is

$$ -\frac{4}{3}x^{3} \;+\; 4x^{2} \;+\; 12x. $$

We now evaluate this from $$x = -1$$ to $$x = 3$$:

First at $$x = 3$$:

$$ -\frac{4}{3}(3)^{3} + 4(3)^{2} + 12(3) \;=\; -\frac{4}{3}(27) + 4(9) + 36 $$

$$ = -\frac{108}{3} + 36 + 36 = -36 + 36 + 36 = 36. $$

Next at $$x = -1$$:

$$ -\frac{4}{3}(-1)^{3} + 4(-1)^{2} + 12(-1) = -\frac{4}{3}(-1) + 4(1) - 12 $$

$$ = +\frac{4}{3} + 4 - 12 = \frac{4}{3} + \frac{12}{3} - \frac{36}{3} = \frac{16}{3} - \frac{36}{3} = -\frac{20}{3}. $$

The definite integral is the value at $$3$$ minus the value at $$-1$$:

$$ \text{Area} = 36 \;-\;\left(-\frac{20}{3}\right) = 36 + \frac{20}{3} = \frac{108}{3} + \frac{20}{3} = \frac{128}{3}. $$

Thus the area of the given region is $$\dfrac{128}{3}$$ square units.

Hence, the correct answer is Option B.