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

Graphical or geometric meaning

The area under $x^2$ from $0$ to $3$ is $9$. A rectangle with the same width $3$ must have height $3$. The horizontal line $y=3$ intersects the curve at $x=\sqrt3$, visually realizing the integral mean-value theorem.

Common mistakes and why they fail

Verification and reasonableness checks

• Confirm the average lies between the minimum and maximum for continuous $f$.
• Multiply the average by $b-a$ and recover the definite integral.
• For a constant function, verify the formula returns that constant.

Interpreting the average as a balance height

The value $f_{\mathrm{avg}}=\frac1{b-a}\int_a^b f(x)\,dx$ is the height of a rectangle with the same signed area as the graph. This explains the divisor and the units: integration adds a factor of input units, then division by interval length removes it. For continuous $f$, some $c\in[a,b]$ satisfies $f(c)=f_{\mathrm{avg}}$, though it need not be unique. Before computing, bound the answer between the minimum and maximum of $f$. An average outside that range signals an error in the antiderivative or interval length. Symmetry also helps: an odd function has average zero on $[-a,a]$.

Practice

1. Find the average of $f(x)=2x$ on $[1,3]$.
2. Find the average of $f(x)=5$ on $[-2,4]$.
3. Where does $f(x)=x$ attain its average on $[0,6]$?

Connections and next steps

• Definite Integral
• Fundamental Theorem of Calculus
• Mean Value Theorem