Math Core

You already know how to average a list of numbers: add them up and divide by how many there are. But a function like a temperature over a day takes infinitely many values. The definite integral is exactly the tool that "adds up" a continuous quantity, so it gives you a way to find the average of a function over an interval. This is the first application in the unit, and it shows up on almost every AP exam.

From a finite average to an integral

Suppose you want the average value of a continuous function $f$ on $[a, b]$ . A natural first attempt is to sample it. Split $[a, b]$ into $n$ equal pieces of width $\Delta x = \dfrac{b - a}{n}$ , pick a point $x_i$ in each piece, and average the samples:

\frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n}.

Since $n = \dfrac{b - a}{\Delta x}$ , you can rewrite this average as

\frac{1}{b - a}\sum_{i=1}^{n} f(x_i)\,\Delta x.

The sum is a Riemann sum. As you take more and more samples ( $n \to \infty$ ), it approaches $\displaystyle\int_a^b f(x)\,dx$ . That limit is the definition of the average value.

Definition

Average value of a function

If $f$ is continuous on $[a, b]$ , the average value of $f$ on $[a, b]$ is

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

In words: take the total accumulated amount (the integral) and divide by the length of the interval.

A picture of the average

Think of the region between the graph of a positive function and the $x$ -axis as water in a tank seen from the side. If the water sloshed flat, its level would be the average value. The rectangle of height $f_{\text{avg}}$ and width $b - a$ has exactly the same area as the region under the curve:

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

The area under y = x² on [0, 3] is 9, the same as a rectangle of height 3 and width 3. The curve reaches the average height at x = √3.Open in grapher →

This equation is also handy in reverse: if you know the average value and the interval, you know the integral.

The Mean Value Theorem for integrals

Look at the graph above again. The curve starts below the average height and ends above it, so somewhere in between it must cross that height. That is always true for a continuous function.

Mean Value Theorem for integrals

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

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

A continuous function actually takes on its average value somewhere in the interval.

This is the Mean Value Theorem from differentiation in disguise. If $F$ is an antiderivative of $f$ , the ordinary MVT says there is a $c$ with $F'(c) = \dfrac{F(b) - F(a)}{b - a}$ , and $F'(c) = f(c)$ while $F(b) - F(a) = \int_a^b f(x)\,dx$ .

Average value versus average rate of change

These two phrases sound alike and are tested side by side on the AP exam.

You are asked for	Formula	Uses
Average value of $f$ on $[a, b]$	$\dfrac{1}{b - a}\displaystyle\int_a^b f(x)\,dx$	an integral of $f$
Average rate of change of $f$ on $[a, b]$	$\dfrac{f(b) - f(a)}{b - a}$	two values of $f$

Here is how they connect. If $v(t)$ is a velocity, the average value of $v$ on $[a, b]$ is $\dfrac{1}{b - a}\int_a^b v(t)\,dt$ , which equals $\dfrac{s(b) - s(a)}{b - a}$ where $s$ is position. So the average value of a rate is the average rate of change of the amount.

Common mistake

Don't divide by the wrong thing, and don't forget to divide at all. The factor in front is $\dfrac{1}{b - a}$ , the reciprocal of the interval's length. Students often compute only $\int_a^b f(x)\,dx$ , or divide by $b$ instead of $b - a$ when the interval doesn't start at 0.

Worked examples

Worked example: A polynomial

Find the average value of $f(x) = x^2$ on $[0, 3]$ , and find every $c$ in $[0, 3]$ guaranteed by the Mean Value Theorem for integrals.

Solution.

f_{\text{avg}} = \frac{1}{3 - 0}\int_0^3 x^2\,dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3}(9) = 3.

Now solve $f(c) = 3$ : $c^2 = 3$ , so $c = \pm\sqrt{3}$ . Only $c = \sqrt{3} \approx 1.732$ lies in $[0, 3]$ .

Worked example: A trigonometric function

Find the average value of $g(x) = \sin x$ on $[0, \pi]$ .

Solution.

g_{\text{avg}} = \frac{1}{\pi}\int_0^\pi \sin x\,dx = \frac{1}{\pi}\Big[-\cos x\Big]_0^\pi = \frac{1}{\pi}\big(1 - (-1)\big) = \frac{2}{\pi} \approx 0.637.

The maximum of $\sin x$ is 1, and the average is a bit less than two thirds of that, which matches the shape of one arch.

Worked example: Working backward

The average value of a continuous function $h$ on $[2, 8]$ is $5$ , and $\displaystyle\int_2^4 h(x)\,dx = 12$ . Find $\displaystyle\int_4^8 h(x)\,dx$ .

Solution. The total integral is (average) $\times$ (length):

\int_2^8 h(x)\,dx = 5 \cdot (8 - 2) = 30.

Split the interval: $\displaystyle\int_4^8 h(x)\,dx = 30 - 12 = 18$ .

Tip

Sanity check every average value: it must lie between the minimum and maximum of $f$ on the interval. If $f$ ranges from 0 to 9 and you get 12, something went wrong.