To express the integral as a limit of Riemann sums using the right endpoints, follow these steps:

The given integral is:

$\int_{6}^{9} \left(x^2 + \frac{1}{x}\right) dx$

1. Divide the interval

Let the interval $[6, 9]$ be divided into $n$ subintervals of equal width $\Delta x$, where:

$\Delta x = \frac{9 - 6}{n} = \frac{3}{n}.$

2. Choose the right endpoints

The right endpoint of the $i$-th subinterval is:

$x_i = 6 + i \cdot \Delta x = 6 + i \cdot \frac{3}{n}.$

3. Evaluate the function at the right endpoint

The integrand is $f(x) = x^2 + \frac{1}{x}$. At the right endpoint $x_i$, this becomes:

$f(x_i) = \left(6 + \frac{3i}{n}\right)^2 + \frac{1}{6 + \frac{3i}{n}}.$

4. Write the Riemann sum

The Riemann sum using right endpoints is:

$\text{Sum} = \sum_{i=1}^{n} f(x_i) \cdot \Delta x.$

Substitute $\Delta x = \frac{3}{n}$ and $f(x_i)$:

$\text{Sum} = \sum_{i=1}^{n} \left[\left(6 + \frac{3i}{n}\right)^2 + \frac{1}{6 + \frac{3i}{n}}\right] \cdot \frac{3}{n}.$

5. Take the limit as $n \to \infty$

The integral as a limit of Riemann sums is:

$\lim_{n \to \infty} \sum_{i=1}^{n} \left[\left(6 + \frac{3i}{n}\right)^2 + \frac{1}{6 + \frac{3i}{n}}\right] \cdot \frac{3}{n}.$

This is the required expression. The limit represents the definite integral without evaluating it.

Let me know if you'd like a step-by-step breakdown of any part! Here are related questions for practice:

1. How would the Riemann sum change if left endpoints were used instead of right endpoints?
2. Can you compute the definite integral directly to verify its value?
3. What happens to the expression for $\Delta x$ if the interval changes to $[a, b]$?
4. How does the summation formula for $\sum_{i=1}^n i^2$ help simplify the process?
5. What are the properties of the definite integral that make Riemann sums work?

Tip: Always check your substitution for $x_i$ carefully when setting up Riemann sums—it determines the accuracy of the sum!