## Example Riemann Sums

1. Introduction. There are different ways to do the Riemann Sum. What I mean is, recall the definition for the Riemann sum looked like
$\displaystyle\sum^{n}_{k=1}f(c_{k})\Delta x_{k}$
for some $x_{k-1}\leq c_{k}\leq x_{k}$. There are many different choices for $c_{k}$, but three are most popular:
1. The midpoint $c_{k}=(x_{k-1}+x_{k})/2$
2. The lower sum, i.e., choosing $c_{k}$ such that $f(c_{k})\leq f(x)$ for any $x_{k-}\leq x\leq x_{k}$
3. The upper sum, choosing the $c_{k}$ such that $f(c_{k})\geq f(x)$ for any $x_{k-}\leq x\leq x_{k}$
This is all rather abstract, so how about we doodle some pictures and see what happens as the number of points increases?

2. Specification. We will specifically be considering the example
(1)$\displaystyle f(x)=x^{2}+1$
and integrate it on the interval $[0,4]$.

3. Lower Sum. We consider the partition with 8 points equally distanced from each other. Lets draw out what this would look like:

So we see that $c_{k}=x_{k-1}$. We can write out the Riemann sum for this as
(2)$\displaystyle \sum^{8}_{k=1}(x_{k-1}^{2}+1)\cdot\Delta x_{k}$
We see that $\Delta x_{k}=1/2$ since the partition points are equidistant. Also we can see that
(3)$\displaystyle x_{k}=\frac{k}{8}(4-0)=\frac{k}{2}$
We can rewrite our sum (2) as
(4)$\displaystyle \frac{1}{2}\sum^{8}_{k=1}\frac{(k-1)^{2}+4}{4}=\frac{1}{8}\sum^{7}_{k=1}k^{2}+\frac{1}{2}\sum^{8}_{k=1}1$
WARNING: we juggled the dummy index a little. Since $(k-1)$ ranges over $k=1$ until $8$, this is the same as $k$ ranging from $0$ until $7$.

The astute reader will realize that the right hand side of (4) has two sums.

The second one is just the sum for the function $g(x)=1$. This is no mistake: recall the property “linearity of the integrand” for definite integrals!

Invoking exercise 1 of finite series, which states
(5)$\displaystyle\sum^{n}_{k=1}k^{2}=\frac{1}{6}n(n+1)(2n+1)$
we can rewrite our sum as
(6)$\displaystyle \frac{1}{8}\sum^{8}_{k=1}\frac{k^{2}+4}{4}=\frac{1}{8}\left(\frac{7(8)(15)}{6}\right)+\frac{1}{2}(8)=\frac{35}{2}+4=21.5$
This gives us a lower bound on the area.

3. Upper Sum. So we just seen the lower sum, lets try using the same partition for an upper sum. We doodle it out again:

The upper sum picks $c_{k}=x_{k}$. So plugging this into the Riemann sum gives us:
(6)$\displaystyle \sum^{8}_{k=1}f(x_{k})\Delta x_{k}=\sum^{8}_{k=1}\left[\left(\frac{k}{2}\right)^{2}+1\right]\cdot\frac{1}{2}$
We see this sum becomes
(7)$\displaystyle \sum^{8}_{k=1}\left[\left(\frac{k}{2}\right)^{2}+1\right]\cdot\frac{1}{2}=\frac{1}{8}\sum^{8}_{k=1}k^{2}+\frac{1}{2}\sum^{8}_{k=1}1$
Again, using our Equation (5) this simplifies to
(8)$\displaystyle \frac{1}{8}\sum^{8}_{k=1}k^{2}+\frac{1}{2}\sum^{8}_{k=1}1=\frac{1}{8}\left(\frac{8(9)(17)}{6}\right)+\frac{1}{2}(8)=\frac{51}{2}+4=25.5+4=29.5$
Basic arithmetic simplifies this to
(8′)$\displaystyle\frac{51}{2}+4=25.5+4=29.5$
This gives us an upper bound for the area.

Exercise 1. So far we have been observing, but the reader has not been foisted into any work. So lets consider the same partition, and use the midpoint rule. The picture for this looks like:

What is the area described by this Riemann sum?

Exercise 2. Suppose we refine the lower sum to work with a partition taking $x_{k}=4k/80$, so there are 80 partition points. The doodle for this looks like:

What is its Riemann sum?

Exercise 3. Using the partition from exercise 2, consider the upper sum. The graph for this partitioning scheme is:

What is its Riemann sum?

Exercise 4. Remember that the function we are working with is $f(x)=x^{2}+1$. What is its definite integral
$\displaystyle\int^{4}_{0}f(x)\,\mathrm{d}x$
using antiderivatives?

Acknowledgements: I borrowed these images, produced by sage, from the online example notebook page on Riemann Sums.