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We know the average of n numbers, a1, a2, . . . , an, is their sum divided by n. But what if we need to find the average temperature over a day’s time—there are too many possible temperatures to add them up. This is a job for the definite integral. The average value of a function f(x) on the interval [a, b] is given by $\frac{1}{b-a}\int_{a}^{b}f(x)dx$ The average value of a positive f has a nice geometric interpretation. Imagine that the area under f (figure 1a) is a liquid that can "leak" through the graph to form a rectangle with the same area (figure 1b). If the height of the rectangle is H, then the area of the rectangle is H(ba). We know the area of the rectangle is the same as the area under f so $H(b - a) = \int_{a}^{b}f(x)dx$. Then $H = \frac{1}{b-a}\int_{a}^{b}f(x)dx$, the average value of f on [a,b].
Figure 1
The average value of a positive f is the height H of the rectangle whose area is the same as the area under f.

## Example

During a 9 hour work day, the production rate at time t hours after the start of the shift was given by the function $r(t) = 5 + \sqrt t$ cars per hour. Find the average hourly production rate.

### Solution

The average hourly production is $\frac{1}{9-0}\int_{0}^{9}(5 + \sqrt{t})dt = 7$ cars per hour. A note about the units—remember that the definite integral has units (cars per hour) × (hours) = cars. But the $\frac{1}{(b-a)}$ in front has units 1/hours—the units of the average value are cars per hour, just what we expect an average rate to be. In general, the average value of a function will have the same units as the integrand. Function averages, involving means and more complicated averages, are used to "smooth" data so that underlying patterns are more obvious and to remove high frequency "noise" from signals. In these situations, the original function f is replaced by some "average of f." If f is rather jagged time data, then the ten year average of f is the integral $g(x) = \frac{1}{10} \int_{x-5}^{x+5} f(t)dt$, an average of f over 5 units on each side of x. For example, figure 2 shows the graphs of a Monthly Average (rather “noisy” data) of surface temperature data, an Annual Average (still rather “jagged), and a Five Year Average (a much smoother function). Typically the average function reveals the pattern much more clearly than the original data. This use of a “moving average” value of “noisy” data (weather information, stock prices) is a very common.
Figure 2

## Example

The graph in figure 3 shows the amount of water in a reservoir over a 12 hour period. Estimate the average amount of water in the reservoir over this period.
Figure 3

### Solution

If V(t) is the volume of the water (in millions of liters) after t hours, then the average amount is $\frac{1}{12}\int_{0}^{12}V(t)dt$. In order to find the definite integral, we’ll have to estimate. I’ll use 6 rectangles, and I’ll take the heights from their right edges.
Figure 4
My estimate of the integral is $\int_{0}^{12}V(t)dt \cong (18)(2) + (9.7)(2) + (8.2)(2) + (12)(2) + (19.9)(2) + (22)(2) = 179.6$. The units of this integral are millions of liters × feet. So my estimate of the average volume is millions of liters. Your estimate might be a little different. In figure 5, you can see the same graph with the line drawn in. The area under the curve and the area under the rectangle are (approximately) the same.
Figure 5
In fact, that would be a different way to estimate the average value. We could have estimated the placement of the horizontal line so that the area under the curve and under the line were equal.