1. **State the problem:** Find the area of the region bounded by the curve $y = 2x - x^2$ and the x-axis. 2. **Identify the points of intersection:** The area is bounded where the curve intersects the x-axis, i.e., where $y=0$. Solve $2x - x^2 = 0$: $$x(2 - x) = 0$$ So, $x=0$ or $x=2$. 3. **Set up the integral:** The area $A$ is the integral of the function from $x=0$ to $x=2$: $$A = \int_0^2 (2x - x^2) \, dx$$ 4. **Calculate the integral:** $$\int (2x - x^2) \, dx = \int 2x \, dx - \int x^2 \, dx = x^2 - \frac{x^3}{3} + C$$ Evaluate from 0 to 2: $$A = \left[x^2 - \frac{x^3}{3}\right]_0^2 = \left(2^2 - \frac{2^3}{3}\right) - (0 - 0) = 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}$$ 5. **Interpretation:** The area bounded by the curve and the x-axis is $\frac{4}{3}$. **Final answer:** $\boxed{\frac{4}{3}}$