To set up the Riemann sum for the integral $\int_0^4 (5x - x^2) \, dx$, let's follow these steps:

1. Identify $f(x)$: The function $f(x)$ in this case is $f(x) = 5x - x^2$.

2. Set the Interval and Partition: We are integrating over the interval $[0, 4]$. Divide this interval into $n$ subintervals, each with width $\Delta x = \frac{4 - 0}{n} = \frac{4}{n}$.

3. Determine $\overline{x}_k$: Let $\overline{x}_k$ be the midpoint or right-endpoint of the $k$-th interval, which can be represented as $\overline{x}_k = 0 + k \Delta x = \frac{4k}{n}$.

4. Set up the Riemann Sum: Substitute $f(\overline{x}_k) = 5\left(\frac{4k}{n}\right) - \left(\frac{4k}{n}\right)^2$ and $\Delta x = \frac{4}{n}$ into the Riemann sum:

$\int_0^4 (5x - x^2) \, dx = \lim_{n \to \infty} \sum_{k=1}^n \left(5\frac{4k}{n} - \left(\frac{4k}{n}\right)^2\right) \cdot \frac{4}{n}$

Thus, the answer in Riemann sum form is:

$\int_0^4 (5x - x^2) \, dx = \lim_{n \to \infty} \sum_{k=1}^n \left(20 \frac{k}{n} - 16 \frac{k^2}{n^2}\right) \cdot \frac{4}{n}$

Would you like more details on the steps, or do you have any specific questions?

Related Questions:

1. What is the purpose of a Riemann sum in calculus?
2. How would you choose between left, right, and midpoint Riemann sums?
3. What happens to the Riemann sum as $n$ approaches infinity?
4. How would you evaluate this Riemann sum explicitly for a finite $n$?
5. How can this Riemann sum be used to approximate definite integrals?

Tip:

When setting up Riemann sums, remember that $\Delta x$ is the width of each subinterval, and $\overline{x}_k$ represents the specific point in each subinterval where the function is evaluated.