The Unapologetic Mathematician

Mathematics for the interested outsider

Riemann Integration

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

An area to be integrated

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:

Approximating the integral

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 ith sample point, and \Delta_i is the width of the ith 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<x_1<...<x_{n-1}<x_n=b. 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 John Armstrong | Analysis, Calculus | | 11 Comments

11 Comments »

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

    Comment by Vishal | January 30, 2008

  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

  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

  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

  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

  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

  7. Very super your explanation

    Comment by lalitha | February 27, 2008

  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

  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

  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

  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

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