The area (in sq. units) of the region bounded by the curve $$x^2 = 4y$$ and the straight line $$x = 4y - 2$$ is:

We have the parabola $$x^{2}=4y$$, which can be rewritten as $$y=\dfrac{x^{2}}{4}$$, opening upward along the positive $$y$$-axis.

The straight line is given by $$x=4y-2$$. Solving this for $$y$$ gives $$y=\dfrac{x+2}{4}$$.

To find the enclosed region, we first locate the intersection points of the two curves. At their common points we must have equally valued ordinates, so we set

$$\dfrac{x^{2}}{4}=\dfrac{x+2}{4}.$$

Multiplying both sides by $$4$$ removes the denominators:

$$x^{2}=x+2.$$

Bringing all terms to one side yields

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

Factoring,

$$(x-2)(x+1)=0,$$

which gives the $$x$$-coordinates of intersection as $$x=2$$ and $$x=-1$$.

Substituting these back into either expression for $$y$$ (say, $$y=\dfrac{x^{2}}{4}$$):

For $$x=2$$, $$y=\dfrac{2^{2}}{4}=1.$$

For $$x=-1$$, $$y=\dfrac{(-1)^{2}}{4}=\dfrac14.$$

Thus the curves meet at the points $$(-1,\dfrac14)$$ and $$(2,1).$$

Next we decide which curve lies above the other between these $$x$$-values. Choosing a convenient test value, say $$x=0$$, gives

Parabola ordinate: $$y=\dfrac{0^{2}}{4}=0,$$

Line ordinate: $$y=\dfrac{0+2}{4}=0.5.$$

Since $$0.5>0$$, the line lies above the parabola in the interval $$-1\le x\le 2$$. Therefore, while integrating with respect to $$x$$, the integrand is “upper curve minus lower curve”:

$$\text{Area}= \int_{-1}^{2}\left[\dfrac{x+2}{4}-\dfrac{x^{2}}{4}\right]dx =\dfrac14\int_{-1}^{2}\bigl(x+2-x^{2}\bigr)\,dx.$$

Expanding inside the integral gives

$$\dfrac14\int_{-1}^{2}\left(-x^{2}+x+2\right)dx.$$

We now integrate term by term. The antiderivative of $$-x^{2}$$ is $$-\dfrac{x^{3}}{3}$$; that of $$x$$ is $$\dfrac{x^{2}}{2}$$; and that of the constant $$2$$ is $$2x$$. Hence

$$\int\left(-x^{2}+x+2\right)dx = -\dfrac{x^{3}}{3}+\dfrac{x^{2}}{2}+2x.$$

Evaluating this from $$x=-1$$ to $$x=2$$ we get

At $$x=2$$: $$-\dfrac{(2)^{3}}{3}+\dfrac{(2)^{2}}{2}+2(2) =-\dfrac{8}{3}+2+4 =-\dfrac{8}{3}+6 =\dfrac{10}{3}.$$

At $$x=-1$$: $$-\dfrac{(-1)^{3}}{3}+\dfrac{(-1)^{2}}{2}+2(-1) =-\dfrac{-1}{3}+\dfrac{1}{2}-2 =\dfrac{1}{3}+\dfrac{1}{2}-2 =\dfrac{2+3-12}{6} =-\dfrac{7}{6}.$$

The definite integral value is therefore

$$\dfrac{10}{3}-\Bigl(-\dfrac{7}{6}\Bigr)=\dfrac{10}{3}+\dfrac{7}{6} =\dfrac{20}{6}+\dfrac{7}{6}=\dfrac{27}{6}=\dfrac{9}{2}.$$

Finally, multiplying by the factor $$\dfrac14$$ that we took outside earlier, the required area is

$$\dfrac14\times\dfrac{9}{2}=\dfrac{9}{8}\text{ square units}.$$

Hence, the correct answer is Option B.