Accumulation Functions (Calculus)

I'm so excited. I think I found a more effective way (for me) to explain a certain part of Accumulation Functions in calculus. These are functions defined by

f(x) = integral (from some number to x) of r(t) dt

where r(t) is a graph. The graph can be (say) from -4 to 8 and the lower bound of f(x) could be 1, so: f(x) = integral (from 1 to x) r(t) dt.

Anyway, if r(t) is ABOVE the x-axis to the right of 1, then f(x) is accumulating "things" and getting larger, and if r(t) is BELOW the x-axis to the right of 1, then f(x) is losing "things" and getting smaller.

Well, everything is "reversed" in this example if you pick an x value to the left of 1. Say, f(-2) = integral (from 1 to -2) r(t) dt. Then if the graph is ABOVE the x-axis, between -2 and 1, then this defined f(x) is getting smaller.

This always confused the kids, and I hadn't a effective way to explain it. This year I tried: Suppose you took a movie of how f(x) is changing from start to finish on the SHOWN graph (regardless of the lower bound of your integral), so in this case, the movie would run from -4 to 8.

Now if f(x) = integral (from 1 to x) r(t) dt, you start this movie at "1" and show it either forward (for x>1) or backward (for x<1) and you see what is happening to f(x). This seemed to make sense to them, since ABOVE the graph r(t) means you're accumulating, and so if you show the movie "backwards", then you're doing the opposite.

Anyway, it looks kind of confusing written out here, but it was a small joy of my day to see their looks of comprehension.