The average of a continuous function

Averaging a few numbers is something we know how to do: add them up and divide by how many there are. But how do you average a function, which takes infinitely many values between a and b? We can no longer add or count. The integral solves exactly this problem.

The idea: replace the sum of the values by their integral (the continuous sum), and the number of values by the length of the interval. The average value of f on [a, b] is:

   μ = 1/(b − a) · ∫ (from a to b) f(x) dx

The resemblance with the discrete average is striking:

   average of numbers:      (x₁ + x₂ + ... + xₙ) / n        (sum / count)
   average of a function:   (∫ f) / (b − a)                 (integral / length)

The integral plays the role of the sum, and the length (b − a) that of the count. It is the natural generalization of the average to the continuous case.

An example: the average value of f(x) = x² on [0, 1].

   μ = 1/(1−0) · ∫ from 0 to 1 x² dx = 1 · (1/3) = 1/3

On average, the squaring function equals 1/3 on [0, 1] — well below its maximum value 1, because it spends a lot of time near 0. Remember the formula and above all its meaning: the average value is the integral "smoothed out" over the length of the interval.