MISSION · Calculus
The Average Value of a Function
Averaging infinitely many values, by integrating and dividing
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The idea in one sentence
To average a list you add and divide by how many; to average a function you integrate and divide by the length of the interval.
Building it from the ordinary average
The mean of \(n\) numbers is \(\frac{1}{n}\sum x_i\). A function on \([a,b]\) has infinitely many values, so there is no \(n\) to divide by — but the limit tells you what to do.
Sample \(f\) at \(n\) evenly spaced points and average those:
\[\frac{1}{n}\sum_{i=1}^{n} f(x_i)\]Since the spacing is \(\Delta x = \frac{b-a}{n}\), we have \(\frac{1}{n} = \frac{\Delta x}{b-a}\), so the sum becomes \(\frac{1}{b-a}\sum f(x_i)\Delta x\) — and that sum is a Riemann sum. Letting \(n \to \infty\):
\[f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx\]The \(n\) that had nowhere to go turned into the interval’s length.
The picture that makes it obvious
The integral is the area under the curve. Dividing by the width gives a height.
So the average value is the height of the rectangle with the same width and the same area as the region under the curve. Draw that rectangle and it slices through the curve, with the bits poking above exactly filling the gaps below.
That picture is also the check. If your answer is higher than the curve’s maximum or lower than its minimum, it is wrong, and you can see it immediately.
A worked one
\(f(x) = x^2\) on \([0, 3]\).
\[f_{\text{avg}} = \frac{1}{3-0}\int_0^3 x^2\,dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3}(9) = 3\]Sanity check: \(x^2\) runs from 0 to 9 on this interval, so an average of 3 is plausible. It is below the midpoint of 0 and 9 because the curve spends more of its time near the bottom — which is the honest answer, and is why averaging the endpoints would have given the wrong 4.5.
Where the average is attained
The Mean Value Theorem for Integrals says a continuous \(f\) actually hits its average value somewhere inside:
\[\text{there is } c \in (a,b) \text{ with } f(c) = \frac{1}{b-a}\int_a^b f(x)\,dx\]For our example, \(c^2 = 3\) gives \(c = \sqrt 3 \approx 1.73\) — the point where the rectangle’s top crosses the curve.
It is the integral’s counterpart to the Mean Value Theorem for derivatives, and it follows from the Intermediate Value Theorem: the average sits between the function’s minimum and maximum, so a continuous function must pass through it.
Continuity is required. A function that jumps can straddle its average without ever equalling it.
Why it comes up
- Average velocity. \(\frac{1}{b-a}\int_a^b v(t)\,dt\), which works out to total displacement over total time — the definition you already had, now derived.
- RMS voltage. Mains electricity is quoted at 230 V or 120 V, which is the average of the square of a sine wave, square-rooted. Averaging the wave itself gives zero.
- Average temperature over a day, average concentration over a reaction, average cost over a production run.
- Probability. The expected value of a continuous random variable is an average value weighted by its density, and it is the same construction.
Common mix-ups
- Forgetting to divide. The bare integral is an area, not a height.
- Dividing by \(b\), or by 2. The divisor is the interval’s width, \(b - a\).
- Averaging the endpoints. \(\frac{f(a) + f(b)}{2}\) is only right for a straight line.
- Expecting a positive answer. A function mostly below the axis has a negative average.
- Confusing it with the average rate of change, which is \(\frac{f(b)-f(a)}{b-a}\) — the average of the derivative, not of the function.
Check yourself
What do you divide the integral by, what is the average value of \(x^2\) on \([0,3]\), and what is the average of a constant function?
Divide by \(b - a\), the width. For \(x^2\): \(\frac13 \cdot 9 = 3\). And a constant function averages to that constant — \(\frac{1}{b-a}\cdot 7(b-a) = 7\).
LEVEL 3 · REFLECT
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