Average Value of a Function

The central idea

If $f$ is integrable on $[a,b]$ with $a<b$, its average value is $f_{\text{avg}}=\frac1{b-a}\int_a^b f(x)\,dx$. Continuity is a common sufficient hypothesis. The factor $1/(b-a)$ divides accumulated output by the interval length, just as an arithmetic mean divides a sum by the number of terms.

Definitions, hypotheses, and notation

Average value is sensitive to how long the function spends at each height, not merely to its largest and smallest values. For a time-dependent temperature, the integral weights every instant equally in time. If a different variable or probability density supplies the weight, the appropriate mean changes to a weighted integral. The standard formula therefore assumes uniform weighting with respect to the integration variable.

For continuous $f$, the bound $m\le f_{\text{avg}}\le M$ follows by integrating $m\le f(x)\le M$ and dividing by the positive length $b-a$. The integral mean-value theorem then says the horizontal average line meets the graph. Neither result says the meeting point is unique: a periodic or nonmonotone function may equal its average many times.

Conceptual meaning

The average value is the height of a rectangle of width $b-a$ having the same signed area as the region under $f$. For continuous $f$, the Mean Value Theorem for Integrals guarantees a point $c$ with $f(c)=f_{\text{avg}}$.

A dependable method and decision rule

1. Identify the full interval and compute its length $b-a$.
2. Evaluate the definite integral of the function over that interval.
3. Divide the integral by $b-a$, not by $b$ or by the number of algebraic terms.
4. Keep units: the average has the same units as the function values.
5. If asked, solve $f(c)=f_{\text{avg}}$ for points attaining the average.

Fully worked example

Common mistakes and why they fail