Average Value of a Function

We define the average value of finitely many numbers as the sum of the numbers divided by number of numbers. How would we define the average value of an arbitrary function f(x) over a closed interval [a, b]? As there are infinitely many numbers to consider, adding them and then dividing by infinity is not an option.

In order to compute the average value of a function f(x) over a closed interval [a, b], divide the interval [a, b] into 'n' equal sub-intervals, each with length Δx = (b – a)/n. Then we choose sample points c_{1}, c_{2}, ....., c_{n} in successive sub-intervals and calculate the average of the numbers f(c_{1}), f(c_{2}), ....., f(c_{n}).

The average value of the numbers f(c_{1}), f(c_{2}), ....., f(c_{n}) is: . Since Δx = (b – a)/n, we can write n = (b – a)/Δx and the average value becomes:

If we let 'n' increase, we would be computing the average value of a large number of closely spaced values. The limiting value is: by the definition of a definite integral. Therefore, we define the average value of f(x) on a closed interval [a, b] as:

- Ex: Find the average value of the function f(x) = x
^{3} on the interval [1, 3].

Sol:

**Mean value theorem for definite integrals: **If the function f(x) is continuous on a closed interval [a, b], then there exists a number 'c' in [a, b] such that: **or**

Therefore, mean value theorem for definite integrals states that a continuous function on a closed interval always assumes its average value at least once in the interval.

- Ex: If f(x) is continuous and = 15, show that f(x) takes on the value 5 at least once on the interval [1, 4].

Sol: By the mean value theorem, there exists a number 'c' in [1, 4] such that f(c) = f_{avg} = .