To solve this problem, we need to set up the Riemann sum for the integral

$\int_{0}^{6}(5x - x^2)dx$

Step-by-Step Solution:

1. Define the Components:

   • $a = 0$, $b = 6$
   • The function $f(x) = 5x - x^2$
   • Divide the interval $[0,6]$ into $n$ subintervals, each of width: $\Delta x = \frac{b-a}{n} = \frac{6-0}{n} = \frac{6}{n}$
   • The $k$-th sample point in the interval is: $x_k = a + k\Delta x = 0 + k\frac{6}{n} = \frac{6k}{n}$

2. Riemann Sum Form: Using the definition of a Riemann sum: $\int_{0}^{6}(5x - x^2)dx = \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k)\Delta x$ Substitute $f(x) = 5x - x^2$ and the sample point $x_k = \frac{6k}{n}$: $f(x_k) = 5\left(\frac{6k}{n}\right) - \left(\frac{6k}{n}\right)^2$

3. Simplify $f(x_k)$: $f(x_k) = 5\frac{6k}{n} - \left(\frac{6k}{n}\right)^2$ $f(x_k) = \frac{30k}{n} - \frac{36k^2}{n^2}$

4. Substitute into the Riemann Sum: $\int_{0}^{6}(5x - x^2)dx = \lim_{n \to \infty} \sum_{k=1}^{n} \left(\frac{30k}{n} - \frac{36k^2}{n^2}\right)\frac{6}{n}$ Expand the terms: $\int_{0}^{6}(5x - x^2)dx = \lim_{n \to \infty} \sum_{k=1}^{n} \left(\frac{180k}{n^2} - \frac{216k^2}{n^3}\right)$

This is the setup for the Riemann sum. Would you like further assistance in simplifying this sum or evaluating it? Let me know!

Related Questions:

1. How do you compute the actual value of this definite integral using limits?
2. What is the role of the sample point $x_k$ in defining the Riemann sum?
3. What happens if you change the sampling method (e.g., midpoint or right-endpoint)?
4. Can this integral be evaluated directly using the Fundamental Theorem of Calculus?
5. How does the width of each subinterval ($\Delta x$) affect the accuracy of the approximation for a finite $n$?

Tip:

Always double-check your substitution for $x_k$ and $\Delta x$ in a Riemann sum setup to avoid computational errors!