The area (in sq. units) of the region bounded by the parabola, $$y = x^2 + 2$$ and the lines, $$y = x + 1$$, $$x = 0$$ and $$x = 3$$, is

First, let us note the two relevant curves:

$$y = x^2 + 2$$ is a parabola opening upward, while $$y = x + 1$$ is a straight line with positive slope.

To decide which curve lies above the other in the interval $$0 \le x \le 3,$$ we compare their ordinates.

Set the two expressions equal to look for any intersection points:

$$x^2 + 2 = x + 1 \quad\Longrightarrow\quad x^2 - x + 1 = 0.$$

The discriminant is $$\Delta = (-1)^2 - 4 \!\cdot\! 1 \!\cdot\! 1 = 1 - 4 = -3 < 0,$$ so the quadratic has no real roots. Hence the two graphs do not meet.

Now test a sample value, say $$x = 0:$$ the parabola gives $$y = 0^2 + 2 = 2,$$ and the line gives $$y = 0 + 1 = 1.$$ Because $$2 > 1,$$ the parabola is above the line, and with no intersection points this order remains valid for every $$x$$ in $$[0,3].$$

Therefore, throughout the strip $$0 \le x \le 3,$$

upper curve  $$= y_{\text{upper}} = x^2 + 2,$$

lower curve  $$= y_{\text{lower}} = x + 1.$$

We now apply the standard formula for area between two curves over a closed interval:

$$\text{Area} = \int_{a}^{b} \bigl[y_{\text{upper}} - y_{\text{lower}}\bigr] \, dx.$$

Here $$a = 0$$ and $$b = 3,$$ so

$$\text{Area} = \int_{0}^{3} \bigl[(x^2 + 2) - (x + 1)\bigr] \, dx.$$

Simplify the integrand algebraically:

$$(x^2 + 2) - (x + 1) \;=\; x^2 - x + 1.$$

Hence

$$\text{Area} = \int_{0}^{3} \bigl(x^2 - x + 1\bigr) \, dx.$$

Now we integrate term by term, using the power‐rule formula $$\int x^n dx = \dfrac{x^{\,n+1}}{n+1} + C.$$

$$\int x^2 \,dx = \dfrac{x^3}{3}, \quad \int (-x)\,dx = -\dfrac{x^2}{2}, \quad \int 1\,dx = x.$$

Combining these, the antiderivative is

$$F(x) = \dfrac{x^3}{3} \;-\; \dfrac{x^2}{2} \;+\; x.$$

Evaluate $$F(x)$$ from $$x = 0$$ to $$x = 3:$$

$$F(3) = \dfrac{3^3}{3} - \dfrac{3^2}{2} + 3 = \dfrac{27}{3} - \dfrac{9}{2} + 3 = 9 - \dfrac{9}{2} + 3.$$

Convert to a common denominator (2):

$$9 = \dfrac{18}{2},\quad 3 = \dfrac{6}{2}.$$

So

$$F(3) = \dfrac{18}{2} - \dfrac{9}{2} + \dfrac{6}{2} = \dfrac{18 - 9 + 6}{2} = \dfrac{15}{2}.$$

At $$x = 0,$$ clearly $$F(0) = 0.$$

Therefore, by the Fundamental Theorem of Calculus,

$$\text{Area} = F(3) - F(0) = \dfrac{15}{2} - 0 = \dfrac{15}{2}.$$

Thus the required area is $$\dfrac{15}{2}$$ square units.

Hence, the correct answer is Option C.