Unit 8: Applications of Integration

Average Value of a Function

Using a definite integral to find the average value of a function.

Advanced25 min lesson3 min readUpdated August 12, 2026Author not yet attributed

The Average of Infinitely Many Values

Averaging a finite list of numbers means summing them and dividing by how many there are. Before reading on: for a function with infinitely many output values across an interval, what would 'summing them all and dividing' even mean — and could a definite integral be exactly that idea, taken to its limit?

Definition — Average Value of a Function

The average value of f on [a, b] is (1/(b − a)) ∫ₐᵇ f(x) dx — the definite integral (a continuous sum) divided by the interval's length (how many 'terms' were summed, in the continuous sense). If f is continuous on [a, b], the Mean Value Theorem for Integrals guarantees some c in [a, b] where f(c) exactly equals this average — the same 'the average is actually achieved somewhere' guarantee the Mean Value Theorem gives for derivatives, applied here to integrals instead.

f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx

Worked Example — Finding an Average Value and the Point That Achieves It

Find the average value of f(x) = x² on [0, 3]. Average = (1/3)∫₀³ x² dx = (1/3)[x³/3]₀³ = (1/3)(27/3) = (1/3)(9) = 3. Find c in [0, 3] with f(c) = 3, as the Mean Value Theorem for Integrals guarantees: c² = 3, so c = √3 ≈ 1.732 — which is indeed in [0, 3], confirming the guarantee concretely.

Worked Example — Average Velocity from a Velocity Function

A particle's velocity is v(t) = 6t − t² (m/s) for t in [0, 4]. Find its average velocity. Average = (1/4)∫₀⁴(6t − t²)dt = (1/4)[3t² − t³/3]₀⁴ = (1/4)[3(16) − 64/3] = (1/4)[48 − 21.333] = (1/4)(26.667) = 6.667 m/s. In fraction form: (1/4)(80/3) = 20/3 m/s.

Tip

Average value is not the same as averaging just the endpoint values, f(a) and f(b) — it accounts for every value f takes across the whole interval, weighted by the definite integral, not just the two boundary values.

Common Mistakes

Computing average value as [f(a) + f(b)]/2, treating it like averaging two numbers.
That formula only applies to a linear function — average value in general requires the full definite integral divided by the interval length, since it accounts for every value f takes, not just its two endpoints.
Forgetting to divide by (b − a) after computing the definite integral.
∫ₐᵇf(x)dx alone is the total accumulated area, not the average — dividing by the interval's length converts that total into a genuine per-unit average.

Key Takeaways

Average value of f on [a, b] is (1/(b−a))∫ₐᵇf(x)dx — a definite integral divided by the interval's length.
The Mean Value Theorem for Integrals guarantees a continuous function actually achieves its average value at some point in the interval.
Average value accounts for the function's entire behavior across the interval, not just its endpoint values.

Summary

Average value uses a single definite integral. The next lesson uses a definite integral to measure something genuinely two-dimensional — the area enclosed between two curves.

Track your progress

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