The Mean Value of a Function

A height, not an area

The region under $y=x^2$y = x² from x = 0 to x = 4 has area ${\int }_0^4{x}^{2}dx={\left[\frac{x^3}{3}\right]}_0^4=\frac{64}{3}\approx 21.33$∫₀⁴ x² dx = [x³/3]₀⁴ = 64/3 ≈ 21.33. The curve starts at height 0 and ends at height 16, so its height changes all the way across.

A rectangle on the same base, 4 wide, needs some fixed height to hold the same area. That height is the mean value of $x^2$x² on the interval from 0 to 4. It is measured in the units of y, like any height, and it is a single number standing in for every height the curve takes.

Divide the integral by the width

A rectangle of height H on a base from a to b has area H(b − a). Setting that equal to the area under the curve and dividing by the width gives the mean value: $\bar{f}=\left(\frac{1}{b-a}\right){\int }_a^{b}f\left(x\right)dx$f̄ = (1/(b − a)) ∫ₐᵇ f(x) dx.

Where the curve is above the line y = f̄ it holds area the rectangle lacks, and where it is below it lacks area the rectangle has. The mean height is the one at which those two pieces are equal.

The gold curve $y=x^2$y = x² and the dashed line $y=\frac{16}{3}\approx 5.33$y = 16/3 ≈ 5.33, with the rectangle under the line shaded from x = 0 to x = 4. Its area is $4\times \frac{16}{3}=\frac{64}{3}$4 × 16/3 = 64/3, the same as the area under the curve. The curve crosses the line at the dot, $x=\frac{4}{\sqrt{3}}\approx 2.31$x = 4/√3 ≈ 2.31. Left of the dot the rectangle holds 8.21 square units more than the region under the curve; right of it the curve holds 8.21 more than the rectangle.

Work it out

For $x^2$x² on the interval from 0 to 4 the area is $\frac{64}{3}$64/3 and the width is 4, so the mean value is $\left(\frac{64}{3}\right)÷4=\frac{16}{3}\approx 5.33$(64/3) ÷ 4 = 16/3 ≈ 5.33.

It lies between the lowest value of $x^2$x² there, 0, and the highest, 16, as a mean must. It is not 8, the value halfway between them: the curve spends most of its width low down and climbs steeply only near x = 4, so its mean sits nearer the bottom.

The curve reaches its mean height at $x^2=\frac{16}{3}$x² = 16/3, that is at $x=\frac{4}{\sqrt{3}}\approx 2.31$x = 4/√3 ≈ 2.31. A continuous function always takes its mean value somewhere on the interval, because its graph runs from below that height to above it.

An average of many readings

Read $x^2$x² at the middles of four equal strips, at x = 0.5, 1.5, 2.5 and 3.5. The readings are 0.25, 2.25, 6.25 and 12.25, and their average is 21 ÷ 4 = 5.25. At the middles of eight strips the average is 5.3125, and at sixteen it is 5.328.

Those averages close in on $\frac{16}{3}\approx 5.333$16/3 ≈ 5.333. The mean value of a function is that limit: the average of readings spread evenly across the width, taken with more and more readings. Each reading stands for an equal share of the width, which is why the integral is divided by the width and not by a count.

More functions

On the interval from 0 to $\pi$π, sin x has area ${\int }_0^{\pi }\sin xdx={\left[-\cos x\right]}_0^{\pi }=1+1=2$∫₀^π sin x dx = [−cos x]₀^π = 1 + 1 = 2. The width is $\pi$π, so the mean value is $\frac{2}{\pi }\approx 0.637$2/π ≈ 0.637. The curve reaches that height twice, at $x\approx 0.690$x ≈ 0.690 and at $x\approx 2.451$x ≈ 2.451.

On the interval from 0 to $2\pi$2π the area above the axis and the area below it cancel, so the integral is 0 and the mean value of sin x is 0. A mean value can be 0 or negative, because it is an integral divided by a width, and an integral counts area below the axis as negative.

For a straight line the mean is the value in the middle of the interval. On the interval from 0 to 4, y = 2x + 1 has area ${\left[x^2+x\right]}_0^4=20$[x² + x]₀⁴ = 20, and 20 ÷ 4 = 5, which is its height at x = 2. Halving the two end heights, (1 + 9) ÷ 2 = 5, works for a line and for nothing that bends.

The gold curve y = sin x from 0 to $\pi$π, and the dashed line at its mean height $\frac{2}{\pi }\approx 0.637$2/π ≈ 0.637. The shaded rectangle is $\pi$π wide, so its area is $\pi \times \frac{2}{\pi }=2$π × 2/π = 2, the area under the arch. The dots, at $x\approx 0.690$x ≈ 0.690 and $x\approx 2.451$x ≈ 2.451, are where the curve is at its mean height.

The usual mistakes

Giving the area as the mean. For $x^2$x² on the interval from 0 to 4 the area is $\frac{64}{3}$64/3; the mean value is that area divided by the width 4, which is $\frac{16}{3}$16/3.

Dividing by b instead of b − a. On the interval from 1 to 4, ${\int }_1^4{x}^{2}dx=\frac{64}{3}-\frac{1}{3}=21$∫₁⁴ x² dx = 64/3 − 1/3 = 21, and the width is 3, so the mean value is 7, not $\frac{21}{4}$21/4.

Halving the two end heights. That gives (0 + 16) ÷ 2 = 8 for $x^2$x² on the interval from 0 to 4, but the true mean is $\frac{16}{3}$16/3, because the curve bends.

A roof over a grain store

In the application below, the mean height of a curved roof comes from its integral divided by the width of the floor. Five readings averaged plainly give less, because they count the two zero readings at the walls as fully as the one at the crown.