Average Value of a Function

You know how to average a list of numbers: add them up and divide by how many there are. But what is the average temperature over a whole day, when the temperature changes every instant? A definite integral does the “adding up” for infinitely many values, and dividing by the length of the interval finishes the job. That’s the average value of a function.

Key ideas

The formula

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

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

Compare it with averaging a list: the integral plays the role of the sum, and the length $b - a$ plays the role of “how many.”

The picture: a rectangle with the same area

Multiply both sides by $b - a$:

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

So $f_{\text{avg}}$ is the height of the rectangle on $[a, b]$ that has exactly the same (signed) area as the region under the curve. The parts of the curve above the rectangle “fill in” the parts below it.

For $y = x^2$ on $[0, 3]$, the area is $9$, so the average value is $\dfrac{9}{3} = 3$. The curve reaches that height at $x = \sqrt{3}$.

Where the function equals its average

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

$$f(c) = f_{\text{avg}}$$

This is sometimes called the Mean Value Theorem for integrals. It makes sense from the picture: a continuous curve can’t stay entirely above or entirely below the rectangle, so it has to cross the line $y = f_{\text{avg}}$ somewhere. To find $c$, set $f(c)$ equal to the average and solve.

Units

The average value has the same units as $f$, not the units of the integral. If $T(t)$ is a temperature in °C and $t$ is in hours, then $\int T\, dt$ is in °C·hours, and dividing by the hours leaves °C.

Average value vs average rate of change

These sound alike but answer different questions:

	Formula	Meaning
Average value of $f$	$\dfrac{1}{b - a}\displaystyle\int_a^b f(x)\, dx$	the typical height of $f$
Average rate of change of $f$	$\dfrac{f(b) - f(a)}{b - a}$	the slope of the secant line

They connect nicely: the average value of $f'$ on $[a, b]$ is $\dfrac{1}{b - a}\displaystyle\int_a^b f'(x)\, dx = \dfrac{f(b) - f(a)}{b - a}$, the average rate of change of $f$.

On the AP exam, average value shows up in both calculator and no-calculator questions. When a calculator is allowed, write the setup (the integral with its $\frac{1}{b - a}$) before giving the number; graders award a point for the setup.

Worked examples

Example 1: A polynomial

Find the average value of $f(x) = x^2$ on $[0, 3]$, and find every $c$ in the interval where $f(c)$ equals that average.

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 $c = \sqrt{3} \approx 1.732$ is in $[0, 3]$.

Example 2: A trig function

Find the average value of $f(x) = \sin x$ on $[0, \pi]$. (Radians, as always in calculus.)

Solution.

$$f_{\text{avg}} = \frac{1}{\pi - 0} \int_0^{\pi} \sin x\, dx = \frac{1}{\pi} \Big[ -\cos x \Big]_0^{\pi} = \frac{1}{\pi} \big( 1 - (-1) \big) = \frac{2}{\pi}$$

That’s about $0.637$, a little above $\tfrac{1}{2}$. Sensible: the arch is wide near its top, so it spends more of the interval high than low.

Example 3: Average temperature

On a spring day, the temperature $t$ hours after 6 a.m. is modelled by $T(t) = 15 + 6\sin\!\left(\dfrac{\pi t}{12}\right)$ degrees Celsius. Find the average temperature from 6 a.m. to 6 p.m.

Solution. The interval is $0 \le t \le 12$.

$$\begin{aligned} T_{\text{avg}} &= \frac{1}{12} \int_0^{12} \left( 15 + 6\sin\frac{\pi t}{12} \right) dt \\ &= \frac{1}{12} \left[ 15t - \frac{72}{\pi} \cos\frac{\pi t}{12} \right]_0^{12} \\ &= \frac{1}{12} \left[ \left( 180 + \frac{72}{\pi} \right) - \left( 0 - \frac{72}{\pi} \right) \right] \\ &= \frac{1}{12} \left( 180 + \frac{144}{\pi} \right) = 15 + \frac{12}{\pi} \end{aligned}$$

The average temperature is $15 + \dfrac{12}{\pi} \approx 18.820$ °C. (The units are °C, the same as $T$.)

Check: the antiderivative of $\sin\dfrac{\pi t}{12}$ is $-\dfrac{12}{\pi}\cos\dfrac{\pi t}{12}$, and $6 \cdot \dfrac{12}{\pi} = \dfrac{72}{\pi}$.

Example 4: Working backwards

The average value of $f(x) = 6x^2$ on $[0, k]$ is $8$. Find $k \gt 0$.

Solution.

$$\frac{1}{k} \int_0^k 6x^2\, dx = \frac{1}{k} \Big[ 2x^3 \Big]_0^k = \frac{2k^3}{k} = 2k^2$$

Set $2k^2 = 8$, so $k^2 = 4$ and $k = 2$ (since $k \gt 0$).

Common mistakes

Forgetting to divide by b − a. The integral alone is the area (or total), not the average. In Example 1, $\int_0^3 x^2\, dx = 9$, but the average value is $3$.

Dividing by the wrong length. It’s $b - a$, the length of the interval, not $b$. On $[2, 6]$ you divide by $4$, not $6$.

Mixing up average value and average rate of change. “Average value of $f$” uses an integral of $f$. “Average rate of change of $f$” uses $\dfrac{f(b) - f(a)}{b - a}$. Read the question carefully.

Giving the wrong units. The average value has the units of $f$. The average of a velocity in m/s is in m/s, not metres.

Averaging just the endpoints. $\dfrac{f(a) + f(b)}{2}$ is the average of two numbers, not the average value of the function. For $x^2$ on $[0, 3]$ it gives $4.5$, not $3$.

Keeping a c outside the interval. When you solve $f(c) = f_{\text{avg}}$, throw away solutions outside $[a, b]$.

Practice

1. (Warm-up) Find the average value of $f(x) = 4x$ on $[0, 5]$.

Solution

$$\frac{1}{5} \int_0^5 4x\, dx = \frac{1}{5} \Big[ 2x^2 \Big]_0^5 = \frac{1}{5}(50) = 10$$

Check: $f$ is linear, so its average is the average of its endpoint values, $\dfrac{0 + 20}{2} = 10$. (This shortcut works only for linear functions.)

2. (Warm-up) You are told that $\displaystyle\int_1^5 g(x)\, dx = 12$. What is the average value of $g$ on $[1, 5]$?

Solution

$$g_{\text{avg}} = \frac{1}{5 - 1}(12) = 3$$

3. (Warm-up) Find the average value of $f(x) = x^3$ on $[0, 2]$.

Solution

$$\frac{1}{2} \int_0^2 x^3\, dx = \frac{1}{2} \left[ \frac{x^4}{4} \right]_0^2 = \frac{1}{2}(4) = 2$$

4. (Core) Find the average value of $f(x) = \dfrac{1}{x}$ on $[1, e]$. Give the exact value and a decimal to 3 places.

Solution

$$\frac{1}{e - 1} \int_1^e \frac{1}{x}\, dx = \frac{1}{e - 1} \Big[ \ln x \Big]_1^e = \frac{1}{e - 1}(1 - 0) = \frac{1}{e - 1} \approx 0.582$$

5. (Core) Find the average value of $f(x) = e^{2x}$ on $[0, \ln 3]$.

Solution

$$\int_0^{\ln 3} e^{2x}\, dx = \left[ \frac{e^{2x}}{2} \right]_0^{\ln 3} = \frac{e^{2\ln 3} - 1}{2} = \frac{9 - 1}{2} = 4$$

(Remember $e^{2\ln 3} = \left(e^{\ln 3}\right)^2 = 9$.) So

$$f_{\text{avg}} = \frac{4}{\ln 3} \approx 3.641$$

6. (Core) Find the average value of $f(x) = \sec^2 x$ on $\left[0, \dfrac{\pi}{4}\right]$ (radians).

Solution

$$\frac{1}{\pi/4} \int_0^{\pi/4} \sec^2 x\, dx = \frac{4}{\pi} \Big[ \tan x \Big]_0^{\pi/4} = \frac{4}{\pi}(1 - 0) = \frac{4}{\pi} \approx 1.273$$

7. (Core) A cyclist’s velocity is $v(t) = t^2 + 2$ metres per second for $0 \le t \le 4$ seconds. Find her average velocity over these $4$ seconds, with units.

Solution

$$v_{\text{avg}} = \frac{1}{4} \int_0^4 (t^2 + 2)\, dt = \frac{1}{4} \left[ \frac{t^3}{3} + 2t \right]_0^4 = \frac{1}{4} \left( \frac{64}{3} + 8 \right) = \frac{1}{4} \cdot \frac{88}{3} = \frac{22}{3}$$

The average velocity is $\dfrac{22}{3} \approx 7.333$ m/s.

8. (Challenge) Let $f(x) = x^2 - 4x + 5$. Find the average value of $f$ on $[0, 3]$, then find every $c$ in $[0, 3]$ where $f(c)$ equals that average.

Solution

$$\int_0^3 (x^2 - 4x + 5)\, dx = \left[ \frac{x^3}{3} - 2x^2 + 5x \right]_0^3 = 9 - 18 + 15 = 6$$

So $f_{\text{avg}} = \dfrac{6}{3} = 2$.

Solve $c^2 - 4c + 5 = 2$: $c^2 - 4c + 3 = 0$, so $(c - 1)(c - 3) = 0$ and $c = 1$ or $c = 3$. Both are in $[0, 3]$, so both count.

9. (Challenge) The average value of a continuous function $f$ on $[0, 6]$ is $5$, and its average value on $[0, 2]$ is $8$. Find the average value of $f$ on $[2, 6]$.

Solution

Turn each average back into an integral by multiplying by the length:

$$\int_0^6 f(x)\, dx = 6(5) = 30, \qquad \int_0^2 f(x)\, dx = 2(8) = 16$$

So $\displaystyle\int_2^6 f(x)\, dx = 30 - 16 = 14$, and the average on $[2, 6]$ is $\dfrac{14}{6 - 2} = 3.5$.

Notice you can’t just “subtract the averages”: the intervals have different lengths.

Homework

Try these independently after the practice questions. Show your setup and explain your reasoning. Use exact answers where possible, and check that your answers fit the question. Homework solutions are not provided on the published site. H1. (Foundations) If $\int_2^8 f(x)\,dx=21$, find the average value of $f$. Explain why you divide by 6 rather than 8.

H2. (Foundations) Find the average value of $f(x)=3x-2$ for $1\le x\le5$, using an integral and then an endpoint check.

H3. (Core) Find the average value of $f(x)=x^2+1$ for $-2\le x\le2$ and every input in that interval where it is attained.

H4. (Core) Find the average value of $e^x$ for $0\le x\le\ln4$. Show why averaging only the endpoint heights gives a different answer.

H5. (Core) Find the average value of $\cos x$ for $-\pi/3\le x\le\pi/3$. All trigonometric inputs in this homework are in radians.

H6. (Core) A room’s temperature is modelled by $T(t)=18+4\sin(\pi t/8)$ °C for $0\le t\le8$ hours. Find its average temperature exactly. Explain the units of the integral and the average.

H7. (Challenge) The averages of a continuous function on $1\le x\le9$ and $1\le x\le4$ are 6 and 10, respectively. Find its average on $4\le x\le9$. Explain why simply subtracting 10 from 6 fails.

H8. (Challenge) Find every $b\gt1$ such that the average value of $3x^2$ on $1\le x\le b$ is 13. Check your answer and reject any inadmissible algebraic roots.

H9. (Challenge) For $f(x)=x^3-3x$ on $-2\le x\le2$, find its average value, average rate of change, and every input where it reaches its average value. Explain why the first two quantities differ.

H10. (Challenge) For $v(t)=2t-4$ m/s on $0\le t\le5$ seconds, find the average velocity and average speed. Find every time when each equals the corresponding instantaneous velocity or speed. Explain how direction changes affect the two averages.