1. **State the problem:** Find the area of the region bounded by the curve $y=4-x^2$, the x-axis, and the vertical lines $x=0$ and $x=2$. 2. **Formula and rules:** The area under a curve $y=f(x)$ from $x=a$ to $x=b$ is given by the definite integral $$\text{Area} = \int_a^b f(x)\,dx.$$ Since the region is bounded by the curve and the x-axis, and $y=4-x^2$ is positive between $x=0$ and $x=2$, we can directly integrate. 3. **Set up the integral:** $$\text{Area} = \int_0^2 (4 - x^2)\,dx.$$ 4. **Calculate the integral:** $$\int_0^2 4\,dx = 4x \Big|_0^2 = 4(2) - 4(0) = 8,$$ $$\int_0^2 x^2\,dx = \frac{x^3}{3} \Big|_0^2 = \frac{8}{3} - 0 = \frac{8}{3}.$$ 5. **Combine results:** $$\text{Area} = 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3}.$$ 6. **Interpretation:** The area of the region bounded by the curve, x-axis, and lines $x=0$ and $x=2$ is $$\frac{16}{3}.$$