The Unapologetic Mathematician

Mathematics for the interested outsider

Integration gives signed areas

I haven’t gotten much time to work on the promised deconstruction, so I’ll punt to a math post I wrote up earlier.

Okay, 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 what happens as we extend the formalism of integration to handle more general situations?

What if the function f we integrate is negative? Then -f is positive, and \int_a^b-f(x)dx is the area between the horizontal axis and the graph of -f. But moving from f to -f is just a reflection through the horizontal axis. The horizontal axis stays in the same place, and it seems the area should be the same. But by the basic rules of integration we spun off at the end of yesterday’s post, we see that


That is, we don’t get the same answer; we get its negative. So, integration counts areas below the horizontal axis as negative. We could also see this from the Riemann sums, where we replace all the function evaluations with their negatives, and factor out a -1 from the whole sum.

How else could we extend the formalism of integration? What if we ran it “backwards”, from the right endpoint of our interval to the left? That is, let’s take an “interval” \left[b,a\right] with a<b. Then when we partition the interval we should get a string of partition points decreasing as we go along. Then when we set up the Riemann sum we’ll get negative values for each x_i-x_{i-1} We can factor out all these signs to give an overall negative sign, along with a Riemann sum for the integral over \left[a,b\right]. The upshot is that we can integrate over an interval from right to left at the cost of introducing an overall negative sign.

We can handle this by attaching a sign to an interval, just like we did to points yesterday. We write \left[b,a\right]^-=\left[a,b\right]. Then when we integrate over a signed interval, we take its sign into account. Notice that if we integrate over both \left[a,b\right] and \left[a,b\right]^- the two parts cancel each other out, and we get {0}.


February 19, 2008 - Posted by | Analysis, Calculus


  1. I am not sure how relevant you may find this but you might be interested in adding some “identities” (on definite integrals) that I have posted here and some more here.

    Comment by Vishal | February 20, 2008 | Reply

  2. They’re interesting, Vishal, and I’ll recommend interested readers go to your weblog to read about them, but they feel to me a lot more useful to find actual answers, and less useful for the theory.

    Of course, if it turns out to be useful for my exposition, I’ll gladly come back to them.

    Comment by John Armstrong | February 20, 2008 | Reply

  3. Hi

    Thank you for this analysis. I see too many lecture notes that contain whoppers like “to find the area under the graph, you find the integral”. This dooms students to incorrect answers later…

    But it seems to me that what you have is more complicated than it needs to be, especially for the ‘generally interested lay audience’. You say:

    So, integration counts areas below the horizontal axis as negative.

    No, the integration of a function that sits below the horizontal axis is negative. There is a semantic problem here, where “area” can be taken to mean “a space enclosed by a boundary” or the mathematical quantity measured in meters squared, or similar.

    To find the area of that space, you simply need to find the absolute value of the integral.

    In your previous post, you have:

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

    So if the rectangles are below the horizontal axis, the areas of the bunch of rectangles will each be positive – we just refer to the function values as positive values, since an area can only be positive.

    Thanks again for the stimulus.

    Comment by Zac | February 20, 2008 | Reply

  4. Zac: That’s another way to look at it, yes. To find the “actual” area, you have to take the absolute value of the area of each part. That much is exactly right.

    But my point here is that integration as a technique actually doesn’t give us area, except in the case of the graph of a positive function integrated over an interval from left to right. What integration alone tells us is signed area.

    Now, there’s a really great way to put both area and signed area on an equal footing, but that will come later, once we get the basic tools set up.

    Comment by John Armstrong | February 20, 2008 | Reply

  5. John, a comment to the theme rather than the actual post. I hope you’ll keep up the series. I’m too busy with school to get a chance to read them carefully right now, but you can bet I’ll be poring over them this summer. All that I’ve seen so far is enormously enticing to this second year undergrad who should’ve taken honors calc and is trying to make up for it 🙂

    Comment by rick | February 20, 2008 | Reply

  6. Oh I’ll be keeping on going, rick, but I’ll warn you to pay more attention to your class than to my own take, since I’m doing a really high-level view on things and emphasizing a lot more of an abstract point of view, especially as I move forwards from here.

    Comment by John Armstrong | February 20, 2008 | Reply

  7. […] from the definition we can see the same “additivity” (using signed intervals) in the region of integration that the Riemann integral […]

    Pingback by The Riemann-Stieltjes Integral I « The Unapologetic Mathematician | March 15, 2008 | Reply

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