Unit 8 · Topic 8.1 · about 30 minutes · 7 parts · 10 scored checks, 2 of them multiple choice, plus 2 from earlier lessons

AP Calculus AB 8.1: Finding the Average Value of a Function on an Interval

Q: Is the average value the same as the average rate of change?

No. The average value averages the outputs of f and uses the integral of f. The average rate of change, f(b) - f(a)b - a, says how fast f changed and uses only the endpoints. It equals the average value of f'.

Q: Why divide by b - a?

The integral adds up the values of f over the interval, so it grows with the interval. Dividing by the length turns that total into one typical height: the height of a rectangle on the same base with the same area.

Q: Is the average value of a velocity the same as the average velocity?

Yes. The integral of velocity over [a, b] is the displacement, so dividing it by b - a gives displacement over elapsed time.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 8: 10 to 15% of the multiple-choice section

What you'll be able to do

Find the average value of a function on an interval by dividing its definite integral by the length of the interval.

Key vocabulary

Term	What it means	The exam trap
Average value of a function	For $f$ continuous on $[a, b]$ , the number $\frac{1}{b - a} \int_{a}^{b} f (x) d x$ .	It is in the same units as the function, so check the units of your answer before you report it.
Definite integral	$\int_{a}^{b} f (x) d x$ , the limit of Riemann sums of $f$ on $[a, b]$ ; for $f \geq 0$ it is the area under the graph.	On its own it is a total, in units such as degree-hours, not an average.
Average rate of change	$\frac{f (b) - f (a)}{b - a}$ , the change in a function's output over the change in its input.	Its units are the function's units per unit of input, one quick way to tell it apart from an average value.

Graph of f(x) = (x - 1) squared + 1 for x from 0 to 4, falling from 2 to 1 at x = 1 and rising to 10 at x = 4. The region under the graph is shaded and has area about 13.33. A rectangle on the same base, 4 wide and about 3.33 tall, is outlined with a dashed top and has the same area.

Part 1 Think first What was the car's average speed?

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A car on a test track speeds up from rest. Its speed $t$ seconds later is $v (t) = t^{2}$ meters per second for $0 \leq t \leq 6$ , so it goes from 0 to 36 meters per second. What was its average speed over those 6 seconds?

In short. The average value of a continuous function $f$ on $[a, b]$ is $\frac{1}{b - a} \int_{a}^{b} f (x) d x$ : the definite integral spread evenly over the length of the interval.

The integral is a total, and dividing it by $b - a$ gives the height of a rectangle on $[a, b]$ with the same area.
The average value uses every value of $f$ on the interval; the average rate of change, $\frac{f (b) - f (a)}{b - a}$ , uses only the two endpoints.
With a table or a graph instead of a formula, find or approximate the integral first, then divide by the length of the interval the question names.

Part 2 Learn it Average value in three short ideas

8a. The integral, spread evenly over the interval

To average five test scores you add them and divide by 5. A continuous function has a value at every input in $[a, b]$ , far too many to add one at a time, so the definite integral does the adding. The average value of a continuous function $f$ on $[a, b]$ is

\frac{1}{b - a} \int_{a}^{b} f (x) d x .

The integral plays the part of the total, and $b - a$ , the length of the interval, plays the part of the count.

There is a picture for it. When $f \geq 0$ , the integral is the area under the graph. Divide that area by the base, $b - a$ , and you get the height of a rectangle on the same base with the same area. That height is the average value. In the graph at the top, the shaded region and the rectangle with the dashed top have equal areas, so the parts of the region above the dashed line would exactly fill the gaps below it.

Example: $f (x) = x^{3}$ on $[1, 3]$ . The integral is ${[\frac{x^{4}}{4}]}_{1}^{3} = \frac{81}{4} - \frac{1}{4} = 20$ , and the interval is $3 - 1 = 2$ long, so the average value is $\frac{20}{2} = 10$ .

Quick check · Calculate

No calculator. What is the average value of $f (x) = x^{2} + 2$ on the interval $[1, 4]$ ?

Your answer

8b. Average value or average rate of change?

Both have "average" in the name and both divide by $b - a$ , so they are easy to mix up. They answer different questions.

The average value of $f$ on $[a, b]$ is $\frac{1}{b - a} \int_{a}^{b} f (x) d x$ . It averages the outputs of $f$ , and it is in the units of $f$ .
The average rate of change of $f$ on $[a, b]$ is $\frac{f (b) - f (a)}{b - a}$ . It uses only the two endpoints and says how fast $f$ changed, in units of $f$ per unit of input.

For $f (x) = x^{3}$ on $[1, 3]$ the average value is 10, from the integral, while the average rate of change is $\frac{27 - 1}{2} = 13$ .

The two meet in one place. When $f^{'}$ is continuous, the Fundamental Theorem of Calculus gives $\int_{a}^{b} f^{'} (x) d x = f (b) - f (a)$ , so

\frac{1}{b - a} \int_{a}^{b} f^{'} (x) d x = \frac{f (b) - f (a)}{b - a} .

The average rate of change of $f$ is the average value of its derivative. For $x^{3}$ , the average value of $f^{'} (x) = 3 x^{2}$ on $[1, 3]$ is $\frac{1}{2} {[x^{3}]}_{1}^{3} = 13$ . So before you start, ask which function you are averaging: $f$ itself, or its rate of change $f^{'}$ .

Quick check · Sort it

No calculator. The function $f$ has a continuous derivative, and $F$ is an antiderivative of $f$ . Sort each expression by what it equals for every such $f$ . Tap a card, then a bin.

The average value of $f$ on $[0, 5]$

The average rate of change of $f$ on $[0, 5]$

Neither, in general

8c. Tables, graphs and the size of the answer

Often $f$ comes as a graph or a table instead of a formula. The method does not change: find or approximate the integral, then divide by the length of the interval.

From a graph made of segments and arcs, find the integral from areas, counting area below the axis as negative. From a table, approximate the integral with a Riemann sum or a trapezoidal sum on the subintervals the table gives, even when their widths differ. Then divide by the length of the interval the question names. A table may run past it: if the readings go from 0 to 16 but the question asks about $[0, 12]$ , use only the readings from 0 to 12 and divide by 12.

Then check the size of your answer. The average value is not automatically halfway between the least and greatest values of $f$ . It depends on how long $f$ stays at each height. A line spends as long below its middle height as above it, so its average value is halfway between its endpoint values. A function that stays low for most of the interval and climbs late, like the car in Think first, has an average value below that halfway mark. One that rises early and stays high has an average value above it.

Quick check · Decide and justify

No calculator. The function $f$ is increasing and its graph is concave up on $[0, 10]$ , with $f (0) = 2$ and $f (10) = 12$ . A classmate says the average value of $f$ on $[0, 10]$ is 7, halfway between 2 and 12. Is the classmate right?

Which reason supports your answer?

Common mistakes

The mistake	Why it is wrong	Do this instead
Reporting $\int_{a}^{b} f (x) d x$ as the average value.	The integral is a total, such as meters or degree-hours. It grows with the length of the interval, and an average should not.	Multiply the integral by $\frac{1}{b - a}$ , with $b - a$ the length of the interval the question names.
Using $\frac{f (b) - f (a)}{b - a}$ when a question asks for the average value of $f$ .	That quotient is the average rate of change of $f$ . It measures how fast $f$ changed, not how large $f$ was.	Integrate $f$ over $[a, b]$ and divide by $b - a$ . Use the difference quotient only when the question asks how fast $f$ changed.
Averaging the two endpoint values, or the least and greatest values, of $f$ .	Those averages ignore how long $f$ spends at each height. A line is the one case where they always match the average value.	Find the integral, which accounts for every value of $f$ , and then divide by the length of the interval.

Part 3 Watch it done An average temperature from a formula

Watch each step, then do the last one yourself.

On a spring day the temperature in a greenhouse $t$ hours after 6 a.m. is $H (t) = 50 + 12 t - t^{2}$ degrees Fahrenheit, for $0 \leq t \leq 12$ . Find the average temperature in the greenhouse from 6 a.m. to 6 p.m.

Step 1 · Write the average value
The interval is $0 \leq t \leq 12$ , so the average temperature is $\frac{1}{12 - 0} \int_{0}^{12} (50 + 12 t - t^{2}) d t$ .
Why this step?
Why is the integral divided by 12?
The integral adds up temperature times time over the whole day, in degree-hours. Dividing by the 12 hours gives the one steady temperature that would add up to the same total: the height of a rectangle 12 hours wide with the same area.
Step 2 · Find an antiderivative
An antiderivative of $50 + 12 t - t^{2}$ is $50 t + 6 t^{2} - \frac{t^{3}}{3}$ . In a definite integral the constant $C$ cancels, so it can be left off here.
Your step · Calculate

No calculator. Finish it. Using the antiderivative $50 t + 6 t^{2} - \frac{t^{3}}{3}$ , what is the average value of $H (t) = 50 + 12 t - t^{2}$ on $0 \leq t \leq 12$ , in degrees Fahrenheit?

Your answer

Part 4 Try it A calculator integral and a cosine

Try it · Calculate

Graphing calculator required. Find the average value of $f (x) = \sqrt{1 + x^{3}}$ on the interval $[0, 2]$ . Give your answer to three decimal places.

Your answer

Try it · Multiple choice

No calculator. What is the average value of $\cos x$ on the interval $[0, \frac{π}{2}]$ ?

Part 5 Explain it Write it the way the exam asks

Free response is handwritten on the exam, so write your answer on paper first. Nothing you type here leaves this page. Then open the model answer and score yourself on each point.

Free response is 50% of the exam score. This one is shaped like Question 3, from free-response Part B (calculator not permitted).

Free-response practice · Question 3 style

No calculator. During a storm, rainwater flows into a drain at a rate of $R (t) = 24 t - 3 t^{2}$ cubic feet per hour, where $t$ is measured in hours and $0 \leq t \leq 8$ .

(a) Find the average value of $R (t)$ over the interval $0 \leq t \leq 8$ . Show the setup. Using correct units, explain the meaning of this value in the context of the problem.

(b) Find the average rate of change of $R (t)$ over the interval $0 \leq t \leq 6$ . Indicate units of measure.

(c) The greatest value of $R (t)$ on $0 \leq t \leq 8$ is 48 and the least is 0. A student says the average value of $R (t)$ on $0 \leq t \leq 8$ must be 24, halfway between them. Explain why the student's reasoning is not correct.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) $\frac{1}{8 - 0} \int_{0}^{8} (24 t - 3 t^{2}) d t = \frac{1}{8} {[12 t^{2} - t^{3}]}_{0}^{8} = \frac{1}{8} (768 - 512) = \frac{256}{8} = 32$ cubic feet per hour. Over the 8 hours, rainwater flowed into the drain at an average rate of 32 cubic feet per hour: a steady 32 cubic feet per hour for 8 hours would bring in the same 256 cubic feet.

(b) $\frac{R (6) - R (0)}{6 - 0} = \frac{36 - 0}{6} = 6$ cubic feet per hour per hour.

(c) The average value is the integral of $R$ divided by the length of the interval, which part (a) shows is 32, not 24. It depends on how long $R$ stays at each height, not only on its greatest and least values. Here $R$ stays near its peak for much of the storm, so the integral, 256, is more than $8 \cdot 24 = 192$ .

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Writes $\frac{1}{8} \int_{0}^{8} (24 t - 3 t^{2}) d t$ with the antiderivative $12 t^{2} - t^{3}$ , and gets 32.
Explains 32 cubic feet per hour as the average rate at which water flowed into the drain over the 8 hours, with units.
Writes $\frac{R (6) - R (0)}{6 - 0}$ and gets 6 cubic feet per hour per hour.
Explains that the average value comes from the integral of $R$ over the interval, not from its greatest and least values.

Part 6 Check yourself A tide pool

The two case questions use the same table. Answer them from it.

A scientist records the depth of the water in a tide pool, $D (t)$ centimeters, $t$ hours after she arrives. The function $D$ is differentiable, and $D^{'}$ is continuous. She stays 10 hours, and the table gives her readings.

Depth of the water in the tide pool
$t$ (hours)	$D (t)$ (centimeters)
0	30
2	36
3	40
6	44
8	38
10	32

Case question 1 · Calculate

No calculator. Use a trapezoidal sum with the four subintervals the table gives from $t = 0$ to $t = 8$ to approximate the average depth of the water over $0 \leq t \leq 8$ , in centimeters.

Your answer

Case question 2 · Multiple choice

No calculator. What is the average value of $D^{'} (t)$ over $0 \leq t \leq 8$ ?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds the average value of $g (x) = 3 x^{2} - 2 x$ on the interval $[- 1, 2]$ . Tap the first line with a mistake.

Check yourself · Match

No calculator. Tap a function and its interval, then its average value on that interval.

Part 7 Keep it Bring back earlier ideas

An average value is a definite integral divided by a length, and it is easy to confuse with an average rate of change. Both come from earlier lessons.

From 6.7 The Fundamental Theorem of Calculus and Definite Integrals · Calculate

No calculator. Evaluate $\int_{1}^{3} (3 x^{2} + 2 x) d x$ .

Your answer

From 2.1 Defining Average and Instantaneous Rates of Change at a Point · Calculate

No calculator. Let $f (x) = x^{3} - 2 x$ . What is the average rate of change of $f$ over the interval $[1, 3]$ ?

Your answer

In three lines

The average value of a continuous $f$ on $[a, b]$ is $\frac{1}{b - a} \int_{a}^{b} f (x) d x$ , the height of a rectangle with the same area.
The average rate of change, $\frac{f (b) - f (a)}{b - a}$ , is a different quantity: it is the average value of $f^{'}$ , not of $f$ .
From a table or a graph, find or approximate the integral, then divide by the length of the interval the question names.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

When a question asks for an average value, write $\frac{1}{b - a} \int_{a}^{b} f (x) d x$ with the actual limits before you integrate, and give units that match the function. When it asks for an average rate of change, write the difference quotient instead. Read which one the question names before you start.

Questions students ask

Is the average value the same as the average rate of change?

No. The average value averages the outputs of $f$ and uses the integral of $f$ . The average rate of change, $\frac{f (b) - f (a)}{b - a}$ , says how fast $f$ changed and uses only the endpoints. It equals the average value of $f^{'}$ .

Why divide by

b - a

The integral adds up the values of $f$ over the interval, so it grows with the interval. Dividing by the length turns that total into one typical height: the height of a rectangle on the same base with the same area.

Can an average value be negative?

Yes. Area below the axis counts as negative in a definite integral, so if $f$ is negative over enough of the interval, the integral and the average value are negative.

What if the function is given only by a table?

Approximate the integral with a Riemann sum or a trapezoidal sum on the subintervals the table gives, then divide by the length of the interval the question asks about, which may be shorter than the table.

Is the average value of a velocity the same as the average velocity?

Yes. The integral of velocity over $[a, b]$ is the displacement, so dividing it by $b - a$ gives displacement over elapsed time.

Course alignment, for teachers

AP Calculus AB topic 8.1, Unit 8: Applications of Integration.

Skill 1.E: Apply appropriate mathematical rules or procedures, with and without technology.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.

What you'll be able to do

Key vocabulary

Part 1 Think first What was the car's average speed?

Part 2 Learn it Average value in three short ideas

8a. The integral, spread evenly over the interval

8b. Average value or average rate of change?

8c. Tables, graphs and the size of the answer

Part 3 Watch it done An average temperature from a formula

Part 4 Try it A calculator integral and a cosine

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself A tide pool

Part 7 Keep it Bring back earlier ideas

Exam strategy for this topic

Questions students ask

Past exam questions on this topic