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. We notice however that inside the bracket there is a \frac{6}{n}, so we try to make the term on the outside look the same as this. Multiply and divide by 6 to give:

 

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

 

Now, the \frac{6}{n} is a constant (as we sum over the i’s), and will correspond to the width of the rectangles in the area which is ultimately what a Riemann sum tells us about. Remember in a Riemann sum, we want a function value at a particular point, times a width. We will call the width of each rectangle \Delta x=\frac{6}{n}. So we can write:

 

\frac{117}{6}\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(4+(i-1)\frac{6}{n}\right)^2\Delta x

 

Now, in a Riemann sum, we have a function evaluated at various points, which we generally call x_i. As we increase i by 1, we move to the next rectangle in our approximation. Here we see a factor of (i-1)\frac{6}{n} (pulling out the factor of 6). Remembering that \frac{1}{n} is the width of our rectangles, when i=1, this is going to give the point 0, and when i=n, this is going to give the point (n-1)\frac{6}{n}. That’s not very interesting, but if we were to go to the next point (i=n+1), we would have n\frac{6}{n} which would be the point 6. So, running through the values from i=1 to i=n, the quantity (i-1)\frac{6}{n} goes from the value 0 to the value just below 6, where as we take n to infinity, we get closer and closer to 6 for that last point. We can treat (i-1)\frac{6}{n} as our x_{i-1}, so long as we know that we are going from x_0=0 to x_{n-1}=\frac{6(n-1)}{n}. So, we can write:

 

\frac{117}{6}\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(4+x_{i-1}\right)^2\Delta x

with x_0=0 to x_{n-1}=\frac{6(n-1)}{n}

 

This can be converted directly into the definite integral:

 

\frac{117}{6}\int_0^6\left(4+x\right)^2dx

 

Exactly the same procedure works with the other expressions in the sum, and we end up with:

 

\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}

 

\frac{117}{6}\int_0^6\left(4+x\right)^2dx-\frac{104}{6}\int_0^6(4+x)dx+91\int_0^1 dx

 

where the 117 comes from 13 x 9, the 104 comes from 13 x 8 and the 91 comes from 13 x 7.

 

The last integral we can write as:

 

91\int_0^1 dx=\frac{91}{6}\int_0^6 dx

 

(Make sure that you can see that this is true).

 

And so we can actually write the whole thing as a single integral:

 

\frac{1}{6}\int_0^6117\left(4+x\right)^2-104(4+x)+91 dx

Which can be calculated from the known antiderivatives and the fundamental theorem of calculus part 2.

How clear is this post?