Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.
A note on notation
This page has no equation renderer, so integrals are written in a compact form: ∫ (a to b) f(x) dx means the definite integral of f(x) from x = a to x = b. On paper, write a at the bottom of the integral sign and b at the top. The average value of f on [a, b] is often written f_avg.
From averaging a list to averaging a function
To average five test scores, you add them and divide by 5. A function on an interval has infinitely many values, so you cannot list them all. But you can sample.
Take f(x) = x² on [0, 3]. Split [0, 3] into n equal pieces of width Δx = 3/n and take the value of f at the right end of each piece. The average of those n values is
(f(x₁) + f(x₂) + … + f(xₙ)) / n
Since n = 3/Δx, dividing by n is the same as multiplying by Δx/3. So the sample average is
(1/3) × [f(x₁)Δx + f(x₂)Δx + … + f(xₙ)Δx]
The bracket is a right Riemann sum. As n grows, it approaches ∫ (0 to 3) x² dx. Here is what happens to the sample average:
| Number of samples n | Sample average |
|---|---|
| 3 | 4.667 |
| 6 | 3.792 |
| 100 | 3.045 |
| 1000 | 3.005 |
The values settle down towards 3. And (1/3) ∫ (0 to 3) x² dx = (1/3)(9) = 3. The “3” in front is the length of the interval. That is the whole idea.
The definition
Average value. If f is continuous on [a, b], the average value of f on [a, b] is f_avg = (1/(b − a)) ∫ (a to b) f(x) dx
Read it as “total accumulated amount, shared out evenly over the interval”.
Units. The integral has units of f times units of x. Dividing by b − a (units of x) leaves the units of f. If f(t) is a temperature in °C and t is in hours, the average value is in °C, not °C·hours.
Signs. The integral is a signed area, so the average value can be negative or zero. A function that spends more time below the axis than above it has a negative average.
The picture: a rectangle with the same area
Rearrange the definition: f_avg × (b − a) = ∫ (a to b) f(x) dx. The left side is the area of a rectangle with base b − a and height f_avg. So:
The average value is the height of the rectangle on [a, b] that has the same signed area as the region under f.
Think of the region as water in a tank with a curved surface. If the water settles flat, its level is the average value. The “extra” water flows into the “gap”.
The function reaches its average. If f is continuous on [a, b], the average value lies between the smallest and largest values of f there. By the Intermediate Value Theorem, there is at least one c in [a, b] with f(c) = f_avg. (This result is sometimes called the Mean Value Theorem for integrals. You may use the idea, but questions will usually just ask you to find c.) In Figure 1, c = √3. Note that c is not the midpoint of the interval, and there may be several such c.
Not the middle of the range. In Figure 1, f runs from 0 to 9, but the average is 3, not 4.5. The curve spends most of its time low and shoots up near the end. A function can also have an average above the middle of its range, as in Worked example 1.
Average value or average rate of change?
These two are easy to mix up. Keep them apart:
| Average value of f | Average rate of change of f | |
|---|---|---|
| Formula | (1/(b − a)) ∫ (a to b) f(x) dx | (f(b) − f(a))/(b − a) |
| Uses | every value of f on [a, b] | only f(a) and f(b) |
| Units | units of f | units of f per unit of x |
| Picture | height of the equal-area rectangle | slope of the secant line |
The link. Apply the average value formula to a derivative f′ instead of f. By the Fundamental Theorem, ∫ (a to b) f′(x) dx = f(b) − f(a). So
average value of f′ on [a, b] = (f(b) − f(a))/(b − a) = average rate of change of f
For example, the average value of 3x² on [0, 2] is (1/2)(8) = 4, and the average rate of change of x³ on [0, 2] is (8 − 0)/2 = 4. Always ask: “the average of which function?”
Worked example 1: a formula, by hand
Question. Let f(x) = 3√x. (a) Find the average value of f on [1, 4]. (b) Find every c in [1, 4] with f(c) equal to that average value.
(a)
- Write the definition. f_avg = (1/(4 − 1)) ∫ (1 to 4) 3√x dx = (1/3) ∫ (1 to 4) 3x^(1/2) dx.
- Antiderivative. 3x^(1/2) has antiderivative 3 × x^(3/2)/(3/2) = 2x^(3/2).
- Evaluate. 2(4)^(3/2) − 2(1)^(3/2) = 2(8) − 2(1) = 14.
- Divide by the length. f_avg = 14/3 ≈ 4.667.
(b) Solve 3√c = 14/3. Then √c = 14/9, so c = 196/81 ≈ 2.420. This is inside [1, 4], so it counts.
Check. f(1) = 3 and f(4) = 6, and 14/3 lies between them, as it must. It is a little above the middle of the range, 4.5. The graph of 3√x is concave down: it climbs quickly and then flattens out near its top, so it spends more time at high values. Compare Figure 1, where the opposite happens.
Notice also that the average rate of change of f on [1, 4] is (6 − 3)/3 = 1. That is a completely different number, in different units.
Worked example 2: an average from a table
Question. The temperature T(t) inside a greenhouse, in °C, is recorded at selected times t, in hours after 6 a.m. (The data are invented.) T is continuous.
| t (hours) | 0 | 2 | 5 | 8 | 12 |
|---|---|---|---|---|---|
| T(t) (°C) | 18 | 22 | 27 | 25 | 19 |
Use a trapezoidal sum with the four subintervals in the table to estimate the average temperature over 0 ≤ t ≤ 12.
- Set up. Average temperature = (1/12) ∫ (0 to 12) T(t) dt. You do not have a formula, so estimate the integral.
- Trapezoids. The widths are unequal (2, 3, 3 and 4 hours), so work out each one:
- [0, 2]: 2 × (18 + 22)/2 = 40
- [2, 5]: 3 × (22 + 27)/2 = 73.5
- [5, 8]: 3 × (27 + 25)/2 = 78
- [8, 12]: 4 × (25 + 19)/2 = 88
- Add. ∫ (0 to 12) T(t) dt ≈ 40 + 73.5 + 78 + 88 = 279.5 °C·hours.
- Divide by the length. Average ≈ 279.5/12 ≈ 23.29 °C.
Interpretation. Over the 12 hours from 6 a.m., the greenhouse temperature averaged about 23.3 °C. The units are °C, not °C·hours, because you divided by hours.
Why not average the five readings? (18 + 22 + 27 + 25 + 19)/5 = 22.2. That treats each reading as if it covered the same length of time. It does not: the reading of 19 at t = 12 is the end of a 4-hour gap, while 18 at t = 0 starts a 2-hour gap. The trapezoidal sum weights each value by the time it represents.
Read the interval carefully. If a question asked for the average over 0 ≤ t ≤ 8 instead, you would use only the first three trapezoids and divide by 8, even though the table goes further.
Worked example 3: an average from a graph
Question. The graph of g on [0, 8] is made of straight segments joining (0, 1), (2, 5), (4, 5), (6, −1) and (8, −1), as in Figure 2. (a) Find the average value of g on [0, 8]. (b) Find the average rate of change of g on [0, 8]. (c) Find every x in [0, 8] where g(x) equals its average value.
(a) Find ∫ (0 to 8) g(x) dx from signed areas. For a straight segment, the signed area is width × (average of the two end heights), even if the segment crosses the axis.
- [0, 2]: trapezoid, 2 × (1 + 5)/2 = 6
- [2, 4]: rectangle, 2 × 5 = 10
- [4, 6]: 2 × (5 + (−1))/2 = 4. (Check: the segment crosses the axis at x = 17/3. The triangle above has area 25/6 and the one below has area 1/6, and 25/6 − 1/6 = 4.)
- [6, 8]: rectangle below the axis, 2 × (−1) = −2
Total: 6 + 10 + 4 − 2 = 18. So g_avg = 18/8 = 9/4.
(b) (g(8) − g(0))/(8 − 0) = (−1 − 1)/8 = −1/4.
(c) The line y = 9/4 meets the graph twice.
- On [0, 2], g(x) = 1 + 2x. Solve 1 + 2x = 9/4: x = 5/8.
- On [4, 6], g(x) = 5 − 3(x − 4). Solve 5 − 3(x − 4) = 9/4: x − 4 = 11/12, so x = 59/12 ≈ 4.917.
Interpretation. The average value is positive because most of the area lies above the axis. The average rate of change is negative because g ends lower than it starts. They answer different questions.
With technology
Some average values have no antiderivative you can write down. For example, the average value of sin(x²) on [0, 2] is (1/2) ∫ (0 to 2) sin(x²) dx. On a graphing calculator, the numerical integral is about 0.80478, so the average value is about 0.402. Write the full expression first, then the decimal. In calculator-active questions, keep full accuracy until the last step and give answers correct to three decimal places.
Common misconceptions
- Forgetting to divide by b − a. ∫ (a to b) f(x) dx is the total, not the average.
- Mixing up average value and average rate of change. Ask which function is being averaged. The average value of f′ is the average rate of change of f, not the average value of f.
- “The average value is (f(a) + f(b))/2.” That shortcut is guaranteed only for linear functions; for other functions it can match only by coincidence. Worked example 1 and Figure 1 both show it failing.
- Averaging table entries. Adding the listed values and dividing by how many there are ignores unequal gaps. Use a sum that weights each value by its width.
- Wrong units. The average value has the units of f, not units of f times units of x.
- Using total (unsigned) area. Regions below the axis count as negative, so the average value can be negative.
- Assuming c is the midpoint of [a, b]. The value c with f(c) = f_avg can be anywhere in the interval, and there may be more than one.
- Using the wrong interval. If a table runs past the interval in the question, stop at the interval’s end and divide by its length.
Where this leads
The average value turns an accumulation into one representative number. In Topic 8.2, Connecting Position, Velocity, and Acceleration of Functions Using Integrals, the same idea gives average velocity: the average value of v(t) over a time interval, which equals displacement divided by time. Later topics in Unit 8 use integrals for net change in context, then for areas and volumes. If you need to revisit what came just before, see Topic 7.8. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.
Try the practice questions now, then use the revision notes and the checklist to consolidate.
Quick questions
Is this page for Calculus AB or Calculus BC?
Both. Topic 8.1 is common content, so the same page serves AB and BC students.
Why not just add up some values of f and divide by how many there are?
That only uses a few points, and it treats unequal gaps as if they were equal. The integral uses every value of f on the interval, weighted by how long f spends near it. A list average is only an estimate, and only when the points are equally spaced.
Does the function actually reach its average value?
If f is continuous on [a, b], yes: there is at least one c in [a, b] with f(c) equal to the average value. This follows from the Intermediate Value Theorem. There can be more than one such c.