Average Value of a Function

You’ll be able to

Compute the average value of a function over an interval
Distinguish the average value of f from the average rate of change
Interpret the average value geometrically and in context

Averaging over a continuum

To average finitely many numbers you add and divide by the count. For a continuous function there are infinitely many values, so averaging uses an integral instead: the average value of f on [a, b] is (1/(b − a))·∫[a to b] f(x) dx. The integral totals up the function’s values, and dividing by the interval length b − a spreads that total evenly — exactly the continuous analog of "sum over count".

A geometric reading

The average value is the height of the rectangle on [a, b] whose area equals the area under the curve. In other words, f_avg is the constant level that would give the same accumulated total. This is closely tied to the Mean Value Theorem for Integrals, which guarantees some c in [a, b] where f(c) actually equals this average value.

Average value vs. average rate of change

Do not confuse two different "averages". The average value of f integrates f and divides by the interval length. The average rate of change of f is [f(b) − f(a)]/(b − a), the secant slope — which is the average value of the derivative f′. When a problem gives you a rate and asks for the average rate, it usually wants the average value of that rate function via the integral.

Average value of a function

f_avg = (1/(b − a)) ∫[a to b] f(x) dx

The integral of f divided by the interval length. Different from the average rate of change [f(b) − f(a)]/(b − a).

Worked example

Find the average value of f(x) = x² on the interval [0, 3].

1.Apply the formula: f_avg = (1/(3 − 0))·∫[0 to 3] x² dx.
2.Integrate x²: the antiderivative is x³/3, evaluated from 0 to 3 gives 27/3 − 0 = 9.
3.Divide by the interval length: f_avg = (1/3)·9.
4.That gives 3.

Answer: The average value is 3. A horizontal line at y = 3 encloses the same area over [0, 3] as the curve y = x².

Tip

Remember to divide by the interval length b − a — that is the step people skip. Computing the integral alone gives the accumulated total, not the average; the division is what turns a total into an average.

Checkpoint

The average value of f on [a, b] is given by:

Checkpoint

If ∫[2 to 6] f(x) dx = 20, what is the average value of f on [2, 6]?

On the exam

When a free-response question asks for the "average value" of a quantity, use the integral formula and divide by the interval length. If it asks for the "average rate of change", use the endpoint difference over the length instead — the two prompts want different computations.

Answer the 2 checkpoints as you read.