Concept explainerCalculus IFoundational7 min read

Average value of a function on an interval

Total accumulated output divided by interval length—the continuous counterpart of an arithmetic mean.

f_{\mathrm{avg}}=\frac1{b-a}\int_a^bf(x)\,dx

Concept explainer

What the idea means, why its conditions matter, and where it connects.

Reviewed July 11, 2026

FormulaStart here

\boxed{f_{\mathrm{avg}}=\frac1{b-a}\int_a^bf(x)\,dx}

The integral supplies total signed accumulation; dividing by interval length converts it into a representative function value.

Why divide by interval length

A longer interval naturally accumulates more area even when the function’s typical height is unchanged. Dividing by b − a removes that duration or width effect.

Units confirm the formula: output-units times input-units, divided by input-units, returns output-units.

Geometric interpretation

The average value is the height of a rectangle with the same signed area as the region under the function over the interval.

If the function is continuous, the Integral Mean Value Theorem guarantees at least one point where the function actually equals its average.

Signed versus physical averages

For velocity, the average value gives average velocity, not average speed. If direction changes and speed is requested, average the absolute value of velocity instead.

Worked exampleAverage a quadratic

Find the average value of x² on [0, 3].

f_{\mathrm{avg}}=\frac1{3-0}\int_0^3x^2\,dx

Divide by interval length.

=\frac13\left[\frac{x^3}{3}\right]_0^3

Evaluate the accumulation.

=\frac13\cdot9

Normalize the total area.

Result

\boxed{3}

Watch for

Common mistakes

Forgetting the factor 1/(b−a).
Using endpoint average instead of function average.
Confusing average velocity with average speed.

Keep

Three takeaways

Average equals accumulation divided by interval length.
The result has the same units as the function.
Interpret signs according to the modeled quantity.