The area (in sq. units) of the region $$\{(x, y) : x \geq 0, x + y \leq 3, x^{2} \leq 4y \text{ and } y \leq 1 + \sqrt{x}\}$$ is

We have to find the area enclosed by the set $$\{(x,y):x\ge 0,\;x+y\le 3,\;x^{2}\le 4y,\;y\le 1+\sqrt{x}\}.$$

From $$x+y\le 3$$ we rewrite the line as $$y\le 3-x.$$ From $$x^{2}\le4y$$ we rewrite the parabola as $$y\ge\dfrac{x^{2}}{4}.$$ The curve $$y=1+\sqrt{x}$$ already gives $$y\le 1+\sqrt{x}.$$

So, for each non-negative $$x$$ the ordinate $$y$$ must lie between the lower curve $$y=\dfrac{x^{2}}{4}$$ and the smaller of the two upper curves $$y=3-x$$ and $$y=1+\sqrt{x}.$$ We therefore compare the two upper curves:

Set $$3-x=1+\sqrt{x}\;.$$ Let $$t=\sqrt{x}$$ (so $$t\ge 0,\;x=t^{2}$$). Substituting,

$$3-t^{2}=1+t\;\Longrightarrow\;t^{2}+t-2=0\;\Longrightarrow\;t=\frac{-1+\sqrt{9}}{2}=1.$$ Hence $$x=t^{2}=1.$$ Thus, $$\begin{cases} y\le 1+\sqrt{x}, & 0\le x\le 1,\\[4pt] y\le 3-x, & 1\le x\le 2\;, \end{cases}$$ because for $$x<1$$ the quantity $$1+\sqrt{x}$$ is smaller, while for $$x>1$$ the quantity $$3-x$$ is smaller.

Now we check where the region actually exists by requiring that the lower bound does not exceed the chosen upper bound.

For $$0\le x\le 1$$ we need $$\dfrac{x^{2}}{4}\le 1+\sqrt{x}.$$ Writing $$x=t^{2}$$ again, $$\dfrac{t^{4}}{4}\le 1+t.$$ Multiplying by 4 gives $$t^{4}\le 4+4t,$$ which is clearly true for $$0\le t\le 1.$$ Hence the whole interval $$[0,1]$$ is admissible.

For $$x\ge 1$$ we need $$\dfrac{x^{2}}{4}\le 3-x.$$ Multiplying by 4,

$$x^{2}+4x-12\le 0\;\Longrightarrow\;(x+6)(x-2)\le 0.$$ Thus $$-6\le x\le 2.$$ Because we already have $$x\ge 1,$$ the effective interval is $$1\le x\le 2.$$

Consequently, the region splits into two vertical strips:

$$\displaystyle \text{Area}= \int_{0}^{1}\!\Bigl[(1+\sqrt{x})-\frac{x^{2}}{4}\Bigr]\,dx\;+\;\int_{1}^{2}\!\Bigl[(3-x)-\frac{x^{2}}{4}\Bigr]\,dx.$$

We evaluate the first integral term by term.

Using $$\int x^{n}\,dx=\dfrac{x^{n+1}}{n+1},$$

$$\int_{0}^{1}1\,dx =\Bigl[x\Bigr]_{0}^{1}=1,$$ $$\int_{0}^{1}x^{1/2}\,dx =\Bigl[\dfrac{2}{3}x^{3/2}\Bigr]_{0}^{1}=\dfrac{2}{3},$$ $$\int_{0}^{1}\dfrac{x^{2}}{4}\,dx =\dfrac{1}{4}\Bigl[\dfrac{x^{3}}{3}\Bigr]_{0}^{1}= \dfrac{1}{12}.$$

Adding those contributions,

$$\text{First part}=1+\dfrac{2}{3}-\dfrac{1}{12}=\dfrac{12}{12}+\dfrac{8}{12}-\dfrac{1}{12}=\dfrac{19}{12}.$$

Next we evaluate the second integral.

$$\int_{1}^{2}(3-x)\,dx =\Bigl[3x-\dfrac{x^{2}}{2}\Bigr]_{1}^{2} =\Bigl(6-2\Bigr)-\Bigl(3-\dfrac12\Bigr)=4-\dfrac52=\dfrac{3}{2},$$

and

$$\int_{1}^{2}\dfrac{x^{2}}{4}\,dx =\dfrac14\Bigl[\dfrac{x^{3}}{3}\Bigr]_{1}^{2} =\dfrac14\Bigl(\dfrac{8}{3}-\dfrac{1}{3}\Bigr)=\dfrac14\cdot\dfrac{7}{3}=\dfrac{7}{12}.$$

Subtracting the latter from the former gives

$$\text{Second part}=\dfrac{3}{2}-\dfrac{7}{12} =\dfrac{18}{12}-\dfrac{7}{12}=\dfrac{11}{12}.$$

Finally, we add the two parts:

$$\text{Total area}= \dfrac{19}{12}+\dfrac{11}{12}= \dfrac{30}{12}= \dfrac{5}{2}\text{ sq. units}.$$

Hence, the correct answer is Option D.