1. **State the problem:** Find the area in the first quadrant bounded by the curve $f(x) = 4x - x^2$ and the x-axis. 2. **Identify the region:** The first quadrant means $x \geq 0$ and $y \geq 0$. The curve intersects the x-axis where $f(x) = 0$. 3. **Find the x-intercepts:** Solve $4x - x^2 = 0$. $$x(4 - x) = 0 \implies x = 0 \text{ or } x = 4$$ 4. **Determine the interval for the area:** Since we want the area in the first quadrant, the region is between $x=0$ and $x=4$ where $f(x) \geq 0$. 5. **Set up the integral for the area:** The area $A$ is given by $$A = \int_0^4 (4x - x^2) \, dx$$ 6. **Calculate the integral:** $$\int_0^4 4x \, dx = \left[2x^2\right]_0^4 = 2(4)^2 - 0 = 32$$ $$\int_0^4 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^4 = \frac{64}{3} - 0 = \frac{64}{3}$$ 7. **Subtract to find the area:** $$A = 32 - \frac{64}{3} = \frac{96}{3} - \frac{64}{3} = \frac{32}{3}$$ 8. **Final answer:** The area bounded by the curve and the x-axis in the first quadrant is $$\boxed{\frac{32}{3}}$$