8.1 Finding the Average Value of a Function on an Interval

Average value formula

The average value of a continuous function f on [a, b] is given by $$f_{avg} = \frac{1}{b-a}\int_a^b f(x)\,dx$$. This generalizes the idea of averaging a finite list of numbers to averaging infinitely many outputs of a function over an interval.

Why divide by (b - a)?

The integral $$\int_a^b f(x)\,dx$$ gives the net area under the curve. Dividing by the interval's length (b - a) rescales that total area into a single representative height — the average value — as if the region were a rectangle with the same area and base (b − a).

Mean Value Theorem for Integrals

This theorem guarantees that for a continuous f on [a, b], there exists at least one c in [a, b] such that f(c) equals the average value. Geometrically, a horizontal line at height f(c) creates a rectangle with the same area as the region under the curve.

Key terms

Average value formula

f_avg = (1/(b-a)) ∫ from a to b of f(x) dx

Why divide by (b - a)?

It rescales the total signed area under f into a single height representing the function's average over the interval.

Mean Value Theorem for Integrals

Guarantees some c in [a, b] where f(c) equals the average value of f on [a, b], for continuous f.

Geometric meaning of average value

The height of a rectangle with base (b-a) that has the same area as the region under the curve of f.

Average value of a constant function

Equals the constant itself, since the area is just the constant times (b - a).

Try an AP-style question

What is the average value of f(x) = x² on [0, 3]?

1. 9
2. 3
3. 4.5
4. 9/2