Average Value of a Function — AP® Calculus AB Notes

You know how to average a list: add the numbers and divide by how many there are. A function on an interval has infinitely many values, so "add them up" becomes an integral and "how many" becomes the length of the interval. The result is the average value of a function, and it answers questions like "what was the average temperature this afternoon?" or "what was the average flow through this pipe during the storm?"

From a list to an integral

Split $[a, b]$ into $n$ equal subintervals of width $\Delta x = \frac{b - a}{n}$ and sample $f$ once in each. The average of the samples is

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

because $\frac{1}{n} = \frac{\Delta x}{b - a}$ . As $n$ grows, the sum becomes a Riemann sum for the definite integral, so the average of more and more samples settles on

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

The integrand is $f$ itself. Not $f'$ , not $|f|$ : just $f$ .

The picture: an equal-area rectangle

Multiply both sides by $b - a$ : $(b - a) \cdot f_{\text{avg}}$ $= \int_a^b f(x)\,dx$ . The left side is the area of a rectangle with base $b - a$ and height $f_{\text{avg}}$ . So the average value is the height of the rectangle that has the same (signed) area as the region under $f$ .

The parabola f(x) = 6x − x² on 0 ≤ x ≤ 6 with the area under it shaded in crimson, 36 square units. A dashed blue rectangle of width 6 and height 6 has the same area, so the average value of f on [0, 6] is 6. The curve meets the line y = 6 at x = 3 − √3 and x = 3 + √3, marked with blue points. — **Figure 1.** For $f(x) = 6x - x^2$ on $[0, 6]$ , the area under the curve is 36 and the rectangle of height 6 has the same area: $f_{\text{avg}} = 6$ . The curve crosses height 6 at $x = 3 \pm \sqrt{3}$ . LevlPrep original diagram.

If $f$ is continuous on $[a, b]$ , it reaches its average value somewhere: the average lies between the minimum and maximum of $f$ (the Extreme Value Theorem gives them), and the Intermediate Value Theorem says a continuous $f$ takes every value in between. In the figure, that happens twice.

Worked example 1. Find the average value of $f(x) = 6x - x^2$ on $[0, 6]$ , and every $c$ where $f(c)$ equals it.

Integrate: $\int_0^6 (6x - x^2)\,dx$ $= \left[3x^2 - \frac{x^3}{3}\right]_0^6$ $= 108 - 72$ $= 36$ .

Divide by the length: $f_{\text{avg}} = \frac{36}{6 - 0} = 6$ .

Solve $6c - c^2 = 6$ : $c^2 - 6c + 6 = 0$ , so $c = 3 \pm \sqrt{3}$ , both inside $[0, 6]$ .

Average value or average rate of change?

These two sound alike and are different numbers with different units. Keep them apart:

	Average value of $f$	Average rate of change of $f$
Formula	$\frac{1}{b - a}\int_a^b f(x)\,dx$	$\frac{f(b) - f(a)}{b - a}$
Picture	height of the equal-area rectangle	slope of the secant line
Uses	every value of $f$ on the interval	two values of $f$
Units (temperature in °C, time in hours)	°C	°C per hour

There is a neat link between them. The average value of $f'$ is $\frac{1}{b - a}\int_a^b f'(x)\,dx$ $= \frac{f(b) - f(a)}{b - a}$ by the Fundamental Theorem. So the average rate of change of $f$ is the average value of its derivative. If a question says "average value of $f'$ ", that is the formula to use.

Averages from graphs, tables and calculators

Worked example 2 (graph). $f$ is made of line segments through $(0, 2)$ , $(2, 6)$ , $(4, 6)$ , $(6, 0)$ and $(8, -2)$ . Find its average value on $[0, 8]$ .

Use geometry: a trapezoid of area 8 on $[0, 2]$ , a rectangle of area 12 on $[2, 4]$ , a triangle of area 6 on $[4, 6]$ , and a triangle of area 2 below the axis on $[6, 8]$ , which counts as $-2$ .

$\int_0^8 f(x)\,dx$ $= 8$ $+ 12$ $+ 6 - 2$ $= 24$ , so $f_{\text{avg}} = \frac{24}{8} = 3$ .

From a table, estimate the integral with a Riemann or trapezoidal sum using the table's own widths, then divide by the length of the interval. With a formula that has no easy antiderivative, use the calculator's numerical integral and write the setup: " $\frac{1}{8}\int_6^{14} T(t)\,dt$ $\approx 19.297$ °C". Interpret the answer with the quantity, its units and the interval: "From 6 a.m. to 2 p.m. the average temperature was about 19.297 °C."

Common slips

Averaging two endpoint values, $\frac{f(a) + f(b)}{2}$ , instead of integrating.
Forgetting to divide by $b - a$ , or dividing by $b$ when $a \ne 0$ .
Counting area below the axis as positive.
Integrating $f'$ when the question asks for the average value of $f$ .