1. **State the problem:** Find the area enclosed between the curve $y = x^2 - 2x - 3$ and the x-axis for $0 \leq x \leq 3$. 2. **Formula and rules:** The area between a curve and the x-axis from $a$ to $b$ is given by the definite integral $$\text{Area} = \int_a^b |f(x)| \, dx.$$ Since the curve lies below the x-axis in this interval, we take the negative of the function to get positive area. 3. **Set up the integral:** $$\text{Area} = \int_0^3 -(x^2 - 2x - 3) \, dx = \int_0^3 (-x^2 + 2x + 3) \, dx.$$ 4. **Integrate term-by-term:** $$\int_0^3 (-x^2 + 2x + 3) \, dx = \left[-\frac{x^3}{3} + x^2 + 3x \right]_0^3.$$ 5. **Evaluate at the bounds:** $$= \left(-\frac{3^3}{3} + 3^2 + 3 \times 3\right) - \left(-\frac{0^3}{3} + 0^2 + 3 \times 0\right) = \left(-\frac{27}{3} + 9 + 9\right) - 0 = (-9 + 9 + 9) = 9.$$ 6. **Final answer:** The area enclosed between the curve and the x-axis from $x=0$ to $x=3$ is $$\boxed{9}.$$