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Average Value of a Function — Formula & Examples

Average Value of a Function

The average height of the graph of a function. For y = f(x) over the domain [a, b], the formula for average value is given below.

 

Graph of y=f(x) over [a,b] showing shaded area beneath curve, with average value formula: (1/(b-a))∫f(x)dx shown as horizontal...

 

See also

Mean value theorem for integrals

Key Formula

favg=1b−a∫abf(x) dxf_{\text{avg}} = \frac{1}{b - a} \int_a^b f(x)\, dx
Where:
  • favgf_{\text{avg}} = The average value of the function over the interval
  • f(x)f(x) = The function being averaged
  • aa = The left endpoint of the interval
  • bb = The right endpoint of the interval
  • b−ab - a = The length of the interval

Worked Example

Problem: Find the average value of f(x) = x² on the interval [0, 3].
Step 1: Write the average value formula with the given function and interval.
favg=13−0∫03x2 dxf_{\text{avg}} = \frac{1}{3 - 0} \int_0^3 x^2\, dx
Step 2: Find the antiderivative of x².
∫x2 dx=x33\int x^2\, dx = \frac{x^3}{3}
Step 3: Evaluate the definite integral from 0 to 3.
x33∣03=273−03=9\frac{x^3}{3}\Bigg|_0^3 = \frac{27}{3} - \frac{0}{3} = 9
Step 4: Divide by the length of the interval to get the average value.
favg=13⋅9=3f_{\text{avg}} = \frac{1}{3} \cdot 9 = 3
Answer: The average value of f(x) = x² on [0, 3] is 3.

Another Example

This example uses a trigonometric function and produces an irrational answer, showing that the average value does not have to be a clean number. It also illustrates that the average value of a function that reaches a maximum of 1 can be noticeably less than that maximum.

Problem: Find the average value of f(x) = sin(x) on the interval [0, π].
Step 1: Set up the average value formula.
favg=1π−0∫0πsin⁡(x) dxf_{\text{avg}} = \frac{1}{\pi - 0} \int_0^{\pi} \sin(x)\, dx
Step 2: Find the antiderivative of sin(x).
∫sin⁡(x) dx=−cos⁡(x)\int \sin(x)\, dx = -\cos(x)
Step 3: Evaluate the definite integral from 0 to π.
−cos⁡(x)∣0π=−cos⁡(π)−(−cos⁡(0))=−(−1)+1=2-\cos(x)\Big|_0^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) + 1 = 2
Step 4: Divide by the interval length π.
favg=1π⋅2=2π≈0.637f_{\text{avg}} = \frac{1}{\pi} \cdot 2 = \frac{2}{\pi} \approx 0.637
Answer: The average value of sin(x) on [0, π] is 2/π ≈ 0.637.

Frequently Asked Questions

What is the difference between the average value of a function and the average rate of change?
The average value of a function measures the mean output (y-value) over an interval using integration: (1/(b−a))∫f(x)dx. The average rate of change measures how much the output changes per unit of input and equals (f(b)−f(a))/(b−a), which is the slope of the secant line. One averages the function's heights; the other averages how fast the function is changing.
How does the average value of a function relate to the Mean Value Theorem for Integrals?
The Mean Value Theorem for Integrals guarantees that if f is continuous on [a, b], there exists at least one point c in (a, b) where f(c) equals the average value. In other words, the function actually attains its average value at some point in the interval. This gives a geometric meaning: there is a rectangle with height f(c) and width (b − a) whose area equals the area under the curve.
Can the average value of a function be negative?
Yes. If the function spends enough of the interval below the x-axis, the definite integral will be negative, making the average value negative. For example, f(x) = −1 on [0, 2] has an average value of −1. The average value simply reflects the net signed area divided by the interval length.

Average Value of a Function vs. Average Rate of Change

Average Value of a FunctionAverage Rate of Change
What it measuresThe mean output (y-value) of a function over an intervalHow fast the output changes per unit of input over an interval
Formula(1/(b−a)) ∫ₐᵇ f(x) dx(f(b) − f(a)) / (b − a)
Requires calculus?Yes — requires integrationNo — only arithmetic with function values
Related theoremMean Value Theorem for IntegralsMean Value Theorem (for derivatives)
UnitsSame units as f(x)Units of f(x) per unit of x

Why It Matters

The average value of a function appears frequently in AP Calculus (both AB and BC) and is a standard topic in college calculus courses. It has practical applications: the average temperature over a day, the average velocity over a trip, or the average concentration of a substance over time are all computed this way. Understanding it also deepens your grasp of the connection between integrals and the concept of a mean.

Common Mistakes

Mistake: Forgetting to divide by (b − a) and just computing the definite integral.
Correction: The definite integral alone gives the total net area under the curve, not the average height. You must multiply by 1/(b − a) to convert total area into average height.
Mistake: Averaging the endpoint values, f(a) and f(b), instead of integrating.
Correction: Taking (f(a) + f(b))/2 only gives the average of two specific outputs and ignores the function's behavior between those points. The correct method requires integrating f(x) across the entire interval.

Related Terms