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Average Value of a Function Calculator

Find the average value of f(x) on [a, b] with f_avg = (1/(b − a))∫ f(x) dx: the definite integral, the average height, the point c where f(c) = f_avg, and the minimum and maximum.

f(x)

Lower limit a

Upper limit b

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How it works

f_avg = (1 / (b − a)) · ∫ₐᵇ f(x) dx · Mean Value Theorem for integrals: there is a c in [a, b] with f(c) = f_avg (f continuous) · integral by Simpson’s rule with 2,000 panels

The average value of a function is the height of the rectangle on [a, b] that has the same area as the region under the curve: the definite integral divided by the interval length. For a continuous function the Mean Value Theorem for integrals guarantees a point c where the function actually takes that average value (OpenStax Calculus Vol. 1 §5.2). The integral is computed with Simpson’s rule on 2,000 panels, which is exact for polynomials up to degree 3 and accurate to many digits for smooth functions.

Worked example

f(x) = x² on [0, 3]: ∫₀³ x² dx = 3³/3 = 9, and the interval length is 3, so f_avg = 9 / 3 = 3. Solving x² = 3 gives c = √3 = 1.732051, which lies in [0, 3]. On this interval f ranges from 0 to 9.

Sources: OpenStax Calculus Vol. 1 — 5.2 The Definite Integral (average value) · Paul’s Online Math Notes — Average Function Value

Frequently asked questions

What is the formula for the average value of a function?

f_avg = (1/(b − a)) ∫ from a to b of f(x) dx: integrate over the interval, then divide by its length.

Is the average value the same as the average of f(a) and f(b)?

Not in general. [f(a) + f(b)]/2 uses only the endpoints; the average value accounts for every point in between. For x² on [0, 3] the endpoint average is 4.5 but the true average value is 3.

What does the c value mean?

It is a point where the function equals its own average value — guaranteed to exist for continuous functions by the Mean Value Theorem for integrals.

Can the average value be negative?

Yes. If more signed area lies below the x-axis than above it, the integral and therefore the average value are negative.

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