# The Unapologetic Mathematician

## Riemann Integration

Before continuing with methods of antidifferentiation, let’s consider another geometric problem: integration. Here’s an example:

We’ve got a function whose graph is drawn in red, and we want to find the area contained between the graph, the $x$-axis, and the two blue lines at $x=3$ and $x=7$. We’ll approximate this by cutting up this interval into $n$ pieces and choosing a sample point $t_i$ in each piece, like so:

Now we’ve just got a bunch of rectangles, and we can add up their areas to get

$\displaystyle\sum\limits_{i=1}^nf(t_i)\Delta_i$

where $f(x_i)$ is the value of the function at the $i$th sample point, and $\Delta_i$ is the width of the $i$th strip. Now as we cut the strips thinner and thinner, our stairstep-like approximation to the function should get closer and closer to the real function, and our approximation to the area we’re interested in should get better and better.

So how can we formalize this process? First, let’s take an interval $\left[a,b\right]$ and think about how to cut it up the strips. We do this by picking a collection of points $a=x_0. We get a bunch of smaller intervals $\left[x_{i-1},x_i\right]$, and in each one we pick some $t_i$. This structure we call a “tagged partition” of the interval $\left[a,b\right]$. We define the “mesh” of a partition to be its thickest subinterval, $\max\limits_{1\leq i\leq n}(x_i-x_{i-1})$, and we’ll want to somehow take this down to zero.

We can now see that the collection of all the tagged partitions of an interval form a directed set! We say that a tagged partition $y=((y_0,...,y_m),(s_1,...,s_m))$ is a “refinement” of a tagged partition $x=((x_0,...,x_n),(t_1,...,t_n))$ if every partition point $x_i$ is one of the $y_j$, and every tag $t_i$ is one of the $s_j$. That is, we get from $x$ to $y$ by splitting up some of the slices of $x$ and adding new tags to the new slices. Then we define $x\preceq y$ if $y$ is a refinement of $x$. This makes the collection of tagged partitions into a partially-ordered set.

To show that this is a directed set, consider any two tagged partitions $x=((x_0,...,x_n),(t_1,...,t_n))$ and $y=((y_0,...,y_m),(s_1,...,s_m))$, and make a new partition by using all the partition points from each one. Now look at each slice in the new partition. It can’t have more than one $t$ tag or $s$ tag, so it has either zero, one, or two distinct tags. If it has no tags, add one. If it has one tag, do nothing. If it has two distinct tags, split it between them (notice how we’re using the topology of $\mathbb{R}$ to say we can make this split). At the end, we’ve got a new partition that refines both of $x$ and $y$. And thus we have a directed set.

Now if we have a function $f$ on $\left[a,b\right]$, we can get a net on this directed set. Given any tagged partition $x=((x_0,...,x_n),(t_1,...,t_n))$, we define the “Riemann sum”

$\displaystyle f_x=\sum\limits_{i=1}^nf(t_i)(x_i-x_{i-1})$

Finally, we say that the function $f$ is “Riemann integrable” if this net converges to a limit $s$, and in this case we define the “Riemann integral” of $f$:

$\displaystyle\int\limits_a^b f(x)dx=s$

which is, at last, the area under the curve as we set out to find.

January 29, 2008 - Posted by | Analysis, Calculus

1. What software do you use for plotting graphs? I presume it is open source.

Comment by Vishal | January 30, 2008 | Reply

2. Actually, no. I have an old version of Maple banging around which gets more and more brain-damaged with the passing years. I doubt anyone but me could make it work anymore, since I happen to know its quirks. For example, the above images are in GIF format because the JPEG exporter is broken.

If I had a readily available open-source (or even freeware) program to plot graphs I’d use it. As it stands, Maple is around and I use the tools at hand. That might change when I get my new computer soon.

Comment by John Armstrong | January 30, 2008 | Reply

3. […] Okay, defining the integral as the limit of a net of Riemann sums is all well and good, but it’s a huge net, and it seems impossible to calculate with. We need […]

Pingback by Darboux Integration « The Unapologetic Mathematician | January 30, 2008 | Reply

4. […] Okay, we know what it means for a function to be integrable (in either of the equivalent Riemann or Darboux senses), but we don’t yet know any functions to actually be integrable. I […]

Pingback by Some integrable functions « The Unapologetic Mathematician | February 1, 2008 | Reply

5. […] we have a function and an interval , and we need to find . We’ve got these big, messy Riemann sums (or Darboux sums), and there’s a lot of work to compute the integral by hand. But notice that […]

Pingback by How to Use the FToC « The Unapologetic Mathematician | February 18, 2008 | Reply

6. […] let’s look back and see what integration is really calculating. We started in on integration by trying to find the area between the horizontal axis and the graph of a positive function. But […]

Pingback by Integration gives signed areas « The Unapologetic Mathematician | February 19, 2008 | Reply

Comment by lalitha | February 27, 2008 | Reply

8. […] Riemann-Stieltjes Integral I Today I want to give a modification of the Riemann integral which helps give insight into the change of variables […]

Pingback by The Riemann-Stieltjes Integral I « The Unapologetic Mathematician | February 28, 2008 | Reply

9. […] Integrals I We’ve dealt with Riemann integrals and their extensions to Riemann-Stieltjes integrals. But these are both defined to integrate a […]

Pingback by Improper Integrals I « The Unapologetic Mathematician | April 18, 2008 | Reply

10. riemann integral is realy good..but u can also lay your hands on other integrals like lebesgue integrals,henstock integrals and darboux integrals.they will realy help you.. UNAABite

Comment by obi kenneth-CEO ,KOBIA-TECHNOLOGY | July 15, 2008 | Reply

11. Yes, I’m aware of other integrals. I’m trying to cover things in some semblance of order, though, and I’m not nearly there yet.

Comment by John Armstrong | July 15, 2008 | Reply

12. […] notice that this is just a Riemann sum for the function . Since is continuous, we know that it’s Riemann integrable, and so as […]

Pingback by Step Function Integrators « The Unapologetic Mathematician | March 14, 2009 | Reply

13. […] What’s Next? Last May we started talking about linear algebra, with a little aside into complex numbers and another into power series along the way. Before that all, long long ago, we were talking about single variable calculus. Specifically, we were studying functions which took a real number in and gave a real number back out, and the two main aspects to this study: differentiation and integration. […]

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14. […] Our coverage of multiple integrals will actually parallel our earlier coverage of Riemann integrals pretty closely. Only now we have to change our notion of “interval” to a […]

Pingback by Higher-Dimensional Riemann Integrals « The Unapologetic Mathematician | December 1, 2009 | Reply

15. […] measures the volume contained between the graph of the function and the – plane within the rectangle. If we fix some constant between and we can calculate the single integral […]

Pingback by Iterated Integrals I « The Unapologetic Mathematician | December 16, 2009 | Reply

16. […] : the measure ! In fact, if you go back to the “area under the curve” definition of the Riemann integral, this makes sense: the graph of is a “rectangle” (possibly in many pieces) with one […]

Pingback by Integrating Simple Functions « The Unapologetic Mathematician | May 24, 2010 | Reply

17. […] as the “area under the graph”, which was the motivation behind our definition of the Riemann integral, way back when we introduced […]

Pingback by The Measures of Ordinate Sets « The Unapologetic Mathematician | July 26, 2010 | Reply