The area function is a bit weird. Brace yourself. Say you’ve got any old function, *f*(*t*). Imagine that at some *t-*value, call it *s, *you draw a fixed vertical line. (Note that because this line is fixed, *s* is a constant, not a variable.) Check out the figure below.

Then, you add a moveable vertical line (the dotted line in the figure) at the *t*-value *x*. You start with the dotted line at *s** *(“*s*” is for *starting* point), and then drag it to the right. As you drag the line, you sweep out a larger and larger area under the curve between *s* and *x*. This area is a function of *x,* the position of the moving line.

In symbols, you write

The *dt* is a little increment along the *t-*axis — actually an infinitesimally small increment.

Here’s a simple example to make sure you’ve got a handle on how the area function works. By the way, don’t feel bad if you find this extremely hard to grasp — you’ve got lots of company. Say you’ve got the simple function *f*(*t*) = 10 — that’s a horizontal line at *y* = 10. If you sweep out area beginning at *s* = 3, you get the following area function:

You can see that the area swept out from 3 to 4 is 10 because, in dragging the line from 3 to 4, you sweep out a rectangle with a width of 1 and a height of 10, which has an area of 1 times 10, or 10. See the figure below.

Now, imagine that you drag the line across at a rate of one *unit per second*. You start at *x* = 3, and you hit 4 at 1 second, 5 at 2 seconds, 6 at 3 seconds, and so on. How much area are you sweeping out per second? Ten *square units per second* because each second you sweep out another 1-by-10 rectangle. Notice — this is huge — that because the width of each rectangle you sweep out is 1, the area of each rectangle — which is given by *height* times *width* — is the same as its height because anything times 1 equals itself. You’ll see why this is huge in a minute.

Okay, are you sitting down? You’ve reached one of the big *Ah ha!* moments in the history of mathematics. Recall that a derivative is a rate. So, because the rate at which the previous area function grows is 10 *square units per second*, you can say its derivative equals 10. Thus, you can write —

Now here’s the critical thing: Notice that this rate or derivative of 10 is the same as the height of the original function *f*(*t*) = 10 because as you go across 1 unit, you sweep out a rectangle that’s 1 by 10, which has an area of 10, the height of the function.

This works for any function, not just horizontal lines. The next figure shows the function *g*(*t*) and its area function

that sweeps out the area beginning at *s* = 2.

You can see that

equals roughly 20 because the area swept out between 2 and 3 has a width of 1 and the curved top of the “rectangle” has an average height of about 20. So, during this interval, the rate of growth of

is about 20 *square units per second*. Between 3 and 4, you sweep out about 15 square units of area because that’s roughly the average height of *g*(*t*) between 15 and 4. So, during the second number two — the interval from *x* = 3 to *x* = 4 — the rate of growth of

is about 15.

The *rate* of area being swept out under a curve by an area function at a given *x*-value is equal to the *height* of the curve at that *x*-value.

Although it’s a bit loose — in the discussion of the above figure — saying things like “roughly” this and “average” that, don’t worry; when you do the math, it all works out. The important thing to focus on is that the rate of area being swept out under a curve is the same as the height of the curve.