4.08e - Mean value of a function

A constant height with the same integral

If a quantity follows $y = x^2$ from $x = 0$ to $x = 3$, what single height should represent its average: $3$, $\frac{9}{2}$, or something else?

The endpoint values are $0$ and $9$, but averaging those two numbers ignores every value between them. A curve can spend much of an interval below that endpoint average. Instead, replace the curve by a horizontal line with the same signed area over the whole interval.

A constant function with height $m$ has integral $m(b-a)$. To give it the same integral as $f$, require $m(b-a) = \int_a^b f(x)\,dx$, then divide by the interval's width.

For the opening curve, the integral is $\int_0^3 x^2\,dx = 9$. The rectangle must therefore have height $9/3 = 3$, even though the average of the two endpoint heights is $9/2$.

[DIAGRAM: asset_name: A mean height replaces a curve; asset_slug: h245_mean_equal_area; caption: On the interval from 0 to 3, the curve y = x squared and the rectangle of height 3 both have area 9; description: Two side-by-side coordinate plots of y equals x squared on zero to three, with the curved region shaded blue on the left and a rectangle of width three and height three shaded clay on the right. Both plots show the horizontal line y equals three, and each shaded area is labelled nine.]

The same definition works when the graph crosses the axis. For $f(x) = x$ on $[-1,1]$, the two triangular regions cancel, so the mean is $0$. Averaging $|x|$ would give $1/2$: that is the mean magnitude, a different question. A function can have mean zero while being non-zero almost everywhere.

This is the continuous counterpart of adding many equally spaced heights and dividing by their number. If $N$ strips have width $\Delta x = (b-a)/N$, then their average height is $\frac{1}{N}\sum f(x_i) = \frac{1}{b-a}\sum f(x_i)\Delta x$; the sum of the strip areas approaches the integral.

Test that distinction on a deliberately flawed calculation.

Evaluating a mean value

Write the averaging factor before integrating, so that the final answer is a height rather than an area.

The definite integral has units of function value multiplied by independent variable. Dividing by $b-a$ cancels the independent variable's units. For example, integrating power in kilowatts over hours gives energy in kilowatt-hours; dividing by the number of hours gives mean power in kilowatts.

The second example changes only the lower endpoint. Removing the lowest part of an increasing curve should raise its mean, and the averaging factor must change too.