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 we integrate is negative? Then is positive, and is the area between the horizontal axis and the graph of . But moving from to 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 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” with . 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 We can factor out all these signs to give an overall negative sign, along with a Riemann sum for the integral over . 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 . Then when we integrate over a signed interval, we take its sign into account. Notice that if we integrate over both and the two parts cancel each other out, and we get .