1. **State the problem:** We need to find the area of the region bounded by the curve $y = x^2 - 4$, the x-axis ($y=0$), and the vertical lines $x=0$ and $x=2$. 2. **Understand the curve and boundaries:** The curve is a parabola opening upwards shifted down by 4 units. The x-axis is $y=0$. The vertical lines $x=0$ and $x=2$ limit the region horizontally. 3. **Find points of intersection with the x-axis:** Solve $x^2 - 4 = 0$. $$x^2 = 4$$ $$x = \pm 2$$ So the curve crosses the x-axis at $x=-2$ and $x=2$. 4. **Determine the region between $x=0$ and $x=2$:** On this interval, $y = x^2 - 4$ is negative because $x^2$ is at most 4, so $x^2 - 4 \leq 0$. The curve lies below the x-axis. 5. **Set up the integral for the area:** Since the curve is below the x-axis, the area between the curve and the x-axis from $x=0$ to $x=2$ is $$\text{Area} = \int_0^2 |x^2 - 4| \, dx = \int_0^2 (4 - x^2) \, dx$$ 6. **Calculate the integral:** $$\int_0^2 (4 - x^2) \, dx = \left[4x - \frac{x^3}{3}\right]_0^2 = \left(4 \times 2 - \frac{2^3}{3}\right) - (0) = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3}$$ 7. **Final answer:** The area of the region bounded by the curve, the x-axis, and the lines $x=0$ and $x=2$ is $$\boxed{\frac{16}{3}}$$