Average Function Value

Main idea

To average a function, add up all its values with an integral and divide by the length of the interval, just as you'd add up a list of numbers and divide by how many there are. Picture it as the height of a flat rectangle with the same area as the region under the curve.

{\htmlData{sym=0}{f_{\text{avg}}}} = {\htmlData{sym=1}{\frac{1}{b - a}}}\int_a^b f(x)\,dx

What the symbols mean

$f_{\text{avg}}$
Average value
The constant height with the same total area over the interval.
$\frac{1}{b - a}$
Divide by the length
Like dividing a sum by the number of items, for a continuous quantity.

Mean Value Theorem for integrals

f(c) = f_{\text{avg}} \text{ for some } c \in [a, b] \quad (f \text{ continuous})

Explanation

The average value of a function is simply its typical height over an interval: the average temperature over a day, say, instead of the temperature at one moment. All we're really doing is the ordinary average, add everything up and divide by how many, adapted to a function that has infinitely many values. The formula has two pieces: $\int_a^b f(x)\,dx$ (adds up all the values of $f$ across the interval) and $\frac{1}{b - a}$ (divides by the length of the interval). Add up, divide. That's all an average ever is.

Think about why dividing by $b - a$ is the right move. To average $n$ numbers you add them and divide by $n$ . Now sample $f$ at $n$ evenly spaced points, each $\Delta x = \frac{b - a}{n}$ apart. Their average is $\frac1n\sum f(x_i) = \frac{1}{b - a}\sum f(x_i)\,\Delta x$ , and as $n$ grows, that sum becomes the integral. So the integral does the adding, and $b - a$ plays the role of $n$ .

Geometrically, $f_{\text{avg}}\cdot(b - a) = \int_a^b f(x)\,dx$ : a rectangle of height $f_{\text{avg}}$ over $[a, b]$ has exactly the area under the curve. The parts of the curve above the rectangle fill the gaps below it.

A continuous function actually hits its average value somewhere in the interval. It makes sense: it can't stay above its average the whole time, or the average would be higher, and it can't stay below either. That's the Mean Value Theorem for integrals.

See it

This is f(x) = 3 + 2 sin x and its average value on [0, b], the height of a rectangle with the same area as the shaded region. Stretch b and notice the average wobbling less and less as it settles toward 3, the line the wave swings around. Each full wave adds as much above 3 as below, so in the long run they cancel.

Worked examples

Example 1

The temperature over a day is $T(t) = 50 + 14\sin\frac{\pi t}{12}$ degrees, $t$ hours after 6 a.m. Find the average temperature from 6 a.m. to 6 p.m.

1
From 6 a.m. to 6 p.m. is $t = 0$ to $t = 12$ , so divide the integral by 12: $\displaystyle\frac{1}{12}\int_0^{12}\left(50 + 14\sin\frac{\pi t}{12}\right)dt = \frac{1}{12}\left[50t - \frac{168}{\pi}\cos\frac{\pi t}{12}\right]_0^{12}$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find the average value of $f(x) = 2x$ on $[0, 4]$ .

Problem 2+5 XP

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

Problem 3+5 XP

The average of $x^2$ on $[0, 3]$ is 3. Where does $x^2$ equal its average?

Problem 4+5 XP

Find the average value of $e^x$ on $[0, 1]$ .

Problem 5+5 XP

Find the average temperature in the example to two decimals.

Problem 6+5 XP

US household voltage is $V(t) = 170\sin(120\pi t)$ volts. Its RMS value is the square root of the average of $V^2$ over one cycle, $0 \le t \le \frac{1}{60}$ s. Find it.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

For which $b > 0$ is the average value of $f(x) = 2x - 3x^2$ on $[0, b]$ equal to $-1$ ?

Where it's used

Average power in an AC circuit, a drug's average concentration over a dosing interval, and average daily temperature are all integral averages.

Why 120 V AC Peaks at 170 VCalc I · Electrical Engineering

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