1. **State the problem:** Find the area bounded by the curve $y = n - \frac{n^2}{4}$, the vertical line $x = \frac{n}{2}$, and the x-axis. 2. **Rewrite the function:** The function is given as $y = n - \frac{n^2}{4}$. Here, $y$ is expressed in terms of $n$, but since $x$ is also related to $n$ by $x = \frac{n}{2}$, we can express $n$ in terms of $x$ as $n = 2x$. 3. **Express $y$ in terms of $x$:** Substitute $n = 2x$ into $y$: $$y = 2x - \frac{(2x)^2}{4} = 2x - \frac{4x^2}{4} = 2x - x^2$$ 4. **Set the limits of integration:** The area is bounded between $x=0$ (since the curve meets the x-axis at $y=0$) and $x=\frac{n}{2} = x$ (upper limit). Since $n=2x$, the upper limit is $x$ itself, so the limits are from $0$ to $x$. 5. **Find the points where the curve meets the x-axis:** Set $y=0$: $$0 = 2x - x^2 \implies x^2 = 2x \implies x(x-2) = 0$$ So, $x=0$ or $x=2$. 6. **Determine the upper limit:** Since $x=\frac{n}{2}$ and $n=2x$, the upper limit is $x=2$. 7. **Calculate the area:** The area $A$ is given by the integral of $y$ from $0$ to $2$: $$A = \int_0^2 (2x - x^2) \, dx$$ 8. **Integrate:** $$\int_0^2 2x \, dx = \left[x^2\right]_0^2 = 4$$ $$\int_0^2 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}$$ 9. **Subtract the integrals:** $$A = 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3}$$ 10. **Final answer:** The area bounded by the curve, the line $x=\frac{n}{2}$, and the x-axis is $\boxed{\frac{4}{3}}$ square units. This corresponds to option (c).