The area of the region bounded by the curves $y = x^2 + 2$ and $x$-axis, between $x = 0$ and $x = 3$ in the first quadrant is:

The problem asks for the area of the region bounded by $y = x^2 + 2$, the x-axis, between $x = 0$ and $x = 3$ in the first quadrant.

Since $y = x^2 + 2$ is always positive (minimum value is 2 when $x = 0$), the curve is always above the x-axis.

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The area under the curve is given by:

$\text{Area} = \int_0^3 (x^2 + 2) \, dx$

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Using basic integration rules:

$\int_0^3 (x^2 + 2) \, dx = \left[\frac{x^3}{3} + 2x\right]_0^3$

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At $x = 3$:

$\frac{3^3}{3} + 2(3) = \frac{27}{3} + 6$

$= 9 + 6$

$= 15$

At $x = 0$:

$\frac{0^3}{3} + 2(0) = 0$

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$\text{Area} = 15 - 0 = 15$

Therefore, the area of the region is $15$ square units.