In the most recent tutorial there is a question about converting a Riemann sum to a definite integral, and it seems to be tripping up quite a few students. I wanted to run through one of the calculations in detail so you can see how to answer such a question.

Let’s look at the example:

\[\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(9\left(4+(i-1)\frac{6}{n}\right)^2-8\left(4+(i-1)\frac{6}{n}\right)+7\right)\frac{13}{n}\]

There are many ways to tackle such a question but let’s take one particular path. Let’s start by the fact that when the limit is defined, the limit of a sum is the sum of the limits. We can split up our expression into 3, which looks like:

\[\lim_{n\rightarrow\infty}\sum_{i=1}^n9\left(4+(i-1)\frac{6}{n}\right)^2\frac{13}{n}-\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(8\left(4+(i-1)\frac{6}{n}\right)\right)\frac{13}{n}+\lim_{n\rightarrow\infty}\sum_{i=1}^n7\frac{13}{n}\]

Let’s tackle each of these separately. Let’s look at the first term:

\[\lim_{n\rightarrow\infty}\sum_{i=1}^n9\left(4+(i-1)\frac{6}{n}\right)^2\frac{13}{n}\]

Well, we can take the factor of 13 outside the front of the whole thing to start with, along with the factor of 9, and this will give

\[13\times 9\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(4+(i-1)\frac{6}{n}\right)^2\frac{1}{n}\]

We see here that we have a sum of terms, and a factor which looks like \(\frac{1}{n}\) in each term.…