Average Value of a Function

Averaging infinitely many values

The average of a list of numbers is their sum divided by how many there are. A function on $[a, b]$ takes infinitely many values, so start with a sample. Split $[a, b]$ into $n$ equal pieces of width $\Delta x = \tfrac{b - a}{n}$ and take one value $f(x_i)$ from each piece. Since $\tfrac{1}{n} = \tfrac{\Delta x}{b - a}$, the average of the sample is

$$\frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n} = \frac{1}{b - a}\sum_{i=1}^{n} f(x_i)\,\Delta x$$

The sum on the right is a Riemann sum. As $n$ grows, the sample fills the interval, and the sum becomes an integral.

Definition

The average value of $f$ on $[a, b]$ is

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

Why the average value is a height

Multiply both sides of the definition by $b - a$:

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

The left side is the area of a rectangle with base $[a, b]$ and height $f_{\text{avg}}$. The right side is the area under the curve. So the average value is the height of the rectangle whose area matches the region’s.

Think of the region as water in a tank, with a wavy surface. Let the water settle: the crests flow into the troughs, the amount stays the same, and the level surface ends at height $f_{\text{avg}}$. The average value is the level the region settles to when its area is spread evenly across the interval.

• y = x²

For $y = x^2$ on $[0, 3]$ the area is $9$ and the width is $3$, so the average value is $3$. The piece of the region sticking out above the rectangle exactly fills the part of the rectangle the curve leaves empty; each has area $2\sqrt{3}$.

The Mean Value Theorem for integrals

Mean Value Theorem for integrals

If $f$ is continuous on $[a, b]$, then there is a number $c$ in $[a, b]$ with

$$f(c) = \frac{1}{b - a}\int_a^b f(x)\,dx$$

A continuous function has a smallest value $m$ and a largest value $M$ on $[a, b]$, so the region lies between the rectangles of heights $m$ and $M$:

$$m(b - a) \le \int_a^b f(x)\,dx \le M(b - a) \quad\Longrightarrow\quad m \le f_{\text{avg}} \le M$$

By the Intermediate Value Theorem, $f$ takes every value from $m$ to $M$, so it takes $f_{\text{avg}}$ somewhere. In the figure, that happens at $c = \sqrt{3}$, where $x^2 = 3$.

The average value of a rate

When the function is a rate $f'$, the Fundamental Theorem turns its average into something familiar:

$$\frac{1}{b - a}\int_a^b f'(x)\,dx = \frac{f(b) - f(a)}{b - a}$$

The average value of a rate is the average rate of change. Average velocity, for instance, is displacement divided by time. Applied to a continuous rate $f'$, the Mean Value Theorem for integrals becomes the Mean Value Theorem for derivatives: at some instant, the rate equals its average.

Worked examples

Common mistakes

Practice problems