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
Relate average value to the Mean Value Theorem for Integrals

The mean height of a curve

The average value of a continuous function f over [a, b] is the constant height that would give the same total accumulation (same area). You get it by integrating f over the interval and dividing by the interval’s length: (1/(b − a))·∫ₐᵇ f(x) dx. Intuitively, it is the height of the rectangle on base [a, b] whose area equals the area under f. This is the integral analog of averaging a list of numbers.

Average value of a function

f_avg = (1/(b − a)) · ∫ₐᵇ f(x) dx

Integrate, then divide by the length of the interval. Do not confuse with average rate of change, [f(b) − f(a)]/(b − a).

Average value vs. average rate of change

These two "averages" are different and easily mixed up. Average value of f uses an integral and answers "what is the typical y-value." Average rate of change uses the endpoints, [f(b) − f(a)]/(b − a), and answers "what is the typical slope" — it is the average value of the derivative f′. The Mean Value Theorem for Integrals ties the first to a point: some c in (a, b) has f(c) equal to the average value f_avg.

Worked example

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

1.Use f_avg = (1/(b − a))·∫ₐᵇ f dx with a = 0, b = 3.
2.Integrate: ∫₀³ x² dx = [x³/3]₀³ = 27/3 = 9.
3.Divide by the interval length: (b − a) = 3.
4.Average value: f_avg = 9 / 3 = 3.

Answer: The average value is 3. A rectangle of height 3 over [0, 3] has the same area (9) as the region under y = x².

Watch out

Do not forget to divide by (b − a). Computing the integral alone gives total accumulation, not the average. The division by interval length is what turns a total into a mean.

Checkpoint

What is the average value of f(x) = 6x over the interval [0, 2]?

Checkpoint

The Mean Value Theorem for Integrals guarantees that a continuous function on [a, b] attains its average value. This means there is a c in (a, b) with:

On the exam

Read the wording precisely: "average value of the velocity" means integrate v and divide by the time interval, while "average acceleration" means [v(b) − v(a)]/(b − a). One is an integral average, the other an endpoint average — pick the right formula.

Answer the 2 checkpoints as you read.