You can use an area function to measure the area under a curve, even as the area changes. For example, say you’ve got any old function, f (t). Imagine that at some t-value, call it s, you draw a fixed vertical line, as shown here.

Then you take a moveable vertical line, starting at the same point, s (“s” is for starting point), and drag it to the right. As you drag the line, you sweep out a larger and larger area under the curve. This area is a function of x, the position of the moving line. In symbols, you write

Note that t is the input variable in f (t) instead of x because x is already taken — it’s the input variable in A_{f} (x). The subscript f in A_{f} indicates that A_{f} (x) is the area function for the particular curve f or f (t). 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 an 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, as shown here.

So, A_{f}_{ }(4), the area swept out as you hit 4, equals 10. A_{f}_{ }(5) equals 20 because when you drag the line to 5, you’ve swept out a rectangle with a width of 2 and height of 10, which has an area of 2 times 10, or 20. A_{f}_{ }(6) equals 30, and so on.

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 see why this is huge in a minute. (By the way, the real rate you care about here is not area swept out per second, but, rather, area swept out per unit change on the x-axis. This example explains it in terms of per second because it’s easier to think about a sweeping-out-area rate this way. And since you’re dragging the line across at one x-axis unit per one second, both rates are the same. Take your pick.)

The derivative of an area function equals the rate of area being swept out. Okay, are you sitting down? You’ve reached another 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

Again, this just tells you that with each 1 unit increase in x, A_{f} (the area function) goes up 10. 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.

And the rate works out to 10 regardless of the width of the rectangle. Imagine that you drag the vertical line fromx = 4 to x = 4.001. At a rate of one unit per second, that’ll take you 1/1000^{th} second, and you’ll sweep out a skinny rectangle with a width of 1/1000, a height of 10, and thus an area of 10 times 1/1000, or 1/100 square units. The rate of area being swept out would be, therefore,

which equals 10 square units per second. So you see that with every small increment along the x-axis, the rate of area being swept out equals the function’s height.

This works for any function, not just horizontal lines. Look at the function g (t) and its area function A_{g }(x) that sweeps out area beginning at s = 2 in the next figure.

Between x = 3.6 and x = 3.7, A_{g }(x) grows by the area of that skinny, dark shaded “rectangle” with a width of 0.1 and a height of about 15. (As you can see, it’s not really a rectangle; it’s closer to a trapezoid, but it’s not that either because its tiny top is curving slightly. But, in the limit, as the width gets smaller and smaller, the skinny “rectangle” behaves precisely like a real rectangle.) So, to repeat, A_{g }(x) grows by the area of that dark “rectangle” which has an area extremely close to 0.1 times 15, or 1.5. That area is swept out in 0.1 seconds, so the rate of area being swept out is

or 15 square units per second, the height of the function. This idea is so important that it deserves to be repeated:

The sweeping out area rate equals the height. 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.

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