8.1 Average value of a function

Achievable AP Calculus AB

8. Applications of integrals

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Average value of a function

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What you’ll learn

Average value: Calculate the average value of a continuous function over a closed interval using a definite integral.
Interpretation: Analyze its meaning in real-world contexts and distinguish it from average rate of change.

The average value of a continuous function on an interval follows the same logic as a standard arithmetic average. Instead of averaging finitely many values, a definite integral accumulates function values continuously across an interval.

If $f$ is continuous on the closed interval $[a, b]$ , its average value $f_{avg}$ is:

$f_{avg} = \frac{1}{b - a} \int_{a}^{b} f (x) d x$

The units of $f_{avg}$ are always identical to the units of $f (x)$ .

Example: Finding average temperature

The temperature of a metal rod, in degrees Celsius ( $^{\circ} C$ ), $t$ minutes after heating begins is modeled by the function $T (t) = 20 + 6 t^{2}$ . Find the average temperature of the rod during the first $2$ minutes.

Solution

(spoiler)

Identify the components: the interval is $[0, 2]$ , so $a = 0$ and $b = 2$ .

Apply the formula:

$T_{avg} = \frac{1}{2 - 0} \int_{0}^{2} (20 + 6 t^{2}) d t = \frac{1}{2} [20 t + 2 t^{3}]_{0}^{2} = \frac{1}{2} ([20 (2) + 2 (2)^{3}] - [0]) = 2 8^{\circ} C$

The average temperature of the rod during the first $2$ minutes is $2 8^{\circ} C$ .

Average value vs. Average rate of change

Watch out for one of the most common semantic traps on the exam: confusing average value with average rate of change (ARC).

When an AP question uses the word “average,” look directly at the units to identify the correct formula.

Context 1: Water accumulation & flow

1. Average value

Given: A rate $R (t)$ in gallons per hour
Asked for: “Average rate at which water enters”
Units check: Answer matches the given function ( $gal/hr$ ). Find average value (integral formula):

$R_{avg} = \frac{1}{b - a} \int_{a}^{b} R (t) d t$

2. Average rate of change

Given: A total volume $V (t)$ in gallons
Asked for: “Average rate at which the volume of water changes”
Units check: Answer requires a rate, so you must introduce a time component ( $gal/hr$ ). Find average rate of change (slope formula):

$ARC = \frac{V ( b ) - V ( a )}{b - a}$

Context 2: Temperature dynamics

1. Average value

Given: Temperature $T (t)$ in $^{\circ} C$
Asked for: Average temperature
Units check: Answer match the given function ( $^{\circ} C$ ). Find the average value:

$T_{avg} = \frac{1}{b - a} \int_{a}^{b} T (t) d t$

2. Average rate of change

Given: Temperature $T (t)$ in $^{\circ} C$
Asked for: “Average rate at which the temperature changes”
Units check: Answer must add a time unit ( $° C / hr$ )

$ARC = \frac{T ( b ) - T ( a )}{b - a}$

Sidenote

The calculus connection

Mathematically, these two concepts merge when you link a rate to its total accumulation. Since integrating a rate gives net change in amount:

$\int_{a}^{b} R (t) d t = V (b) - V (a)$

Substituting this into the average value formula highlights the link:

$R_{avg} = \frac{\int _{a}^{b} R ( t ) d t}{b - a} = \frac{V ( b ) - V ( a )}{b - a}$

So your formula choice depends entirely on which “level” of data the problem provides.

AP practice FRQ

(Calculator-active question)

Water is pumped into and out of an underground storage tank. The net rate at which the volume of water in the tank changes is modeled by the function

$R (t) = 12 + 4 e^{0.2 t} sin (\frac{π t}{6})$

where $R (t)$ is measured in gallons per hour and $t$ is measured in hours for $0 \leq t \leq 12$ . At time $t = 0$ , the tank contains $150$ gallons of water.

a) Find the average rate at which the volume of water in the tank is changing over the 12-hour period. Include units.

b) Find the total amount of water in the tank at time $t = 12$ .

c) Is the rate at which the volume of water changes increasing or decreasing at $t = 5$ ? Give a reason for your answer.

d) Is the amount of water increasing or decreasing at time $t = 9$ ? Justify your answer.

Answers

(spoiler)

a) $6.432$ gallons per hour
b) $227.178$ gallons c) Decreasing ( $R^{'} (5) < 0$ )
d) Decreasing ( $R (9) < 0$ )

Solutions

a) Find the average rate at which the volume of water in the tank is changing over the 12-hour period. Include units.

(spoiler)

Keyword translation: The function given is already a rate, so the “average rate at which the volume is changing” is the average value of the rate function.

In Desmos, define the rate function:

$R (t) = 12 + 4 e^{0.2 t} sin (\frac{π t}{6})$

Then on the next line, setup the integral for the average value, $R_{avg}$ :

$\frac{1}{12} \int_{0}^{12} R (t) d t$

The answer is $6.432$ and $R_{avg}$ has the same units as $R (t)$ of gallons per hour. Keep $R (t)$ defined for the remaining problems.

b) Find the total amount of water in the tank at time $t = 12$ .

(spoiler)

Let $V (t)$ be the amount of water in the tank at any time $t$ .

Initially, the tank contains $150$ gallons of water, so $V (0) = 150$ .

To find $V (12)$ , use the Fundamental theorem of calculus. Add the initial amount to the net change in volume (found by integrating the rate):

$V (12) = V (0) + \int_{0}^{12} R (t) d t$

In Desmos, entering

$150 + \int_{0}^{12} R (t) d t$

yields $227.178$ gallons as the amount of water at $t = 12$ hours.

c) Is the rate at which the volume of water changes increasing or decreasing at $t = 5$ ?

(spoiler)

Keyword translation: If the rate $R (t)$ is increasing, then $R^{'} (t) > 0$ (and if decreasing, $R^{'} (t) < 0$ ). Evaluate the derivative at $t = 5$ by entering:

$R^{'} (5)$

which gives $- 3.843$ .

Justification: The rate at which the volume changes is decreasing at $t = 5$ because $R^{'} (5) < 0$ .

d) Is the amount of water increasing or decreasing at time $t = 9$ ?

(spoiler)

Keyword translation: If amount of water $V (t)$ is increasing, then $V^{'} (t) > 0$ (and if decreasing, its derivative is negative).

Since $V^{'} (t) = R (t)$ , we find $V^{'} (9)$ by entering:

$R (9)$

in Desmos, which gives $- 12.199$ gallons per hour as the rate of change of the volume at $t = 9$ .

Since $V^{'} (9) = R (9) < 0$ , the rate of change is negative, meaning the amount of water in the tank is decreasing at $t = 9$ .

Average value of a function

Formula: $f_{avg} = \frac{1}{b - a} \int_{a}^{b} f (x) d x$
Units of $f_{avg}$ always match units of $f (x)$

Average value vs. average rate of change

Key strategy: check units to identify which formula to use
Average value (integral formula): given data already in target units — integrate and divide by interval length
Average rate of change (slope formula): answer requires introducing a new unit (e.g., per hour) — use $\frac{f ( b ) - f ( a )}{b - a}$

Calculus connection between the two

Integrating a rate gives net change: $\int_{a}^{b} R (t) d t = V (b) - V (a)$
Average value of a rate = ARC of its accumulation function: $R_{avg} = \frac{V ( b ) - V ( a )}{b - a}$
Formula choice depends on which “level” (rate or total) the problem provides

FRQ problem-solving strategies

“Average rate of change of volume” with rate given → average value of $R (t)$
Total amount at time $t$ : $V (t) = V (0) + \int_{0}^{t} R (s) d s$
Rate increasing/decreasing at a point → evaluate $R^{'} (t)$ ; sign determines answer
Amount increasing/decreasing → evaluate $R (t)$ directly (since $V^{'} (t) = R (t)$ )

Average value of a function

What you’ll learn

Average value: Calculate the average value of a continuous function over a closed interval using a definite integral.
Interpretation: Analyze its meaning in real-world contexts and distinguish it from average rate of change.

If $f$ is continuous on the closed interval $[a, b]$ , its average value $f_{avg}$ is:

$f_{avg} = \frac{1}{b - a} \int_{a}^{b} f (x) d x$

The units of $f_{avg}$ are always identical to the units of $f (x)$ .

Example: Finding average temperature

The temperature of a metal rod, in degrees Celsius ( $^{\circ} C$ ), $t$ minutes after heating begins is modeled by the function $T (t) = 20 + 6 t^{2}$ . Find the average temperature of the rod during the first $2$ minutes.

Solution

(spoiler)

Identify the components: the interval is $[0, 2]$ , so $a = 0$ and $b = 2$ .

Apply the formula:

$T_{avg} = \frac{1}{2 - 0} \int_{0}^{2} (20 + 6 t^{2}) d t = \frac{1}{2} [20 t + 2 t^{3}]_{0}^{2} = \frac{1}{2} ([20 (2) + 2 (2)^{3}] - [0]) = 2 8^{\circ} C$

The average temperature of the rod during the first $2$ minutes is $2 8^{\circ} C$ .

Average value vs. Average rate of change

Watch out for one of the most common semantic traps on the exam: confusing average value with average rate of change (ARC).

When an AP question uses the word “average,” look directly at the units to identify the correct formula.

Context 1: Water accumulation & flow

1. Average value

Given: A rate $R (t)$ in gallons per hour
Asked for: “Average rate at which water enters”
Units check: Answer matches the given function ( $gal/hr$ ). Find average value (integral formula):

$R_{avg} = \frac{1}{b - a} \int_{a}^{b} R (t) d t$

2. Average rate of change

Given: A total volume $V (t)$ in gallons
Asked for: “Average rate at which the volume of water changes”
Units check: Answer requires a rate, so you must introduce a time component ( $gal/hr$ ). Find average rate of change (slope formula):

$ARC = \frac{V ( b ) - V ( a )}{b - a}$

Context 2: Temperature dynamics

1. Average value

Given: Temperature $T (t)$ in $^{\circ} C$
Asked for: Average temperature
Units check: Answer match the given function ( $^{\circ} C$ ). Find the average value:

$T_{avg} = \frac{1}{b - a} \int_{a}^{b} T (t) d t$

2. Average rate of change

Given: Temperature $T (t)$ in $^{\circ} C$
Asked for: “Average rate at which the temperature changes”
Units check: Answer must add a time unit ( $° C / hr$ )

$ARC = \frac{T ( b ) - T ( a )}{b - a}$

Sidenote

The calculus connection

Mathematically, these two concepts merge when you link a rate to its total accumulation. Since integrating a rate gives net change in amount:

$\int_{a}^{b} R (t) d t = V (b) - V (a)$

Substituting this into the average value formula highlights the link:

$R_{avg} = \frac{\int _{a}^{b} R ( t ) d t}{b - a} = \frac{V ( b ) - V ( a )}{b - a}$

So your formula choice depends entirely on which “level” of data the problem provides.

AP practice FRQ

(Calculator-active question)

Water is pumped into and out of an underground storage tank. The net rate at which the volume of water in the tank changes is modeled by the function

$R (t) = 12 + 4 e^{0.2 t} sin (\frac{π t}{6})$

where $R (t)$ is measured in gallons per hour and $t$ is measured in hours for $0 \leq t \leq 12$ . At time $t = 0$ , the tank contains $150$ gallons of water.

a) Find the average rate at which the volume of water in the tank is changing over the 12-hour period. Include units.

b) Find the total amount of water in the tank at time $t = 12$ .

c) Is the rate at which the volume of water changes increasing or decreasing at $t = 5$ ? Give a reason for your answer.

d) Is the amount of water increasing or decreasing at time $t = 9$ ? Justify your answer.

Answers

(spoiler)

a) $6.432$ gallons per hour
b) $227.178$ gallons c) Decreasing ( $R^{'} (5) < 0$ )
d) Decreasing ( $R (9) < 0$ )

Solutions

a) Find the average rate at which the volume of water in the tank is changing over the 12-hour period. Include units.

(spoiler)

Keyword translation: The function given is already a rate, so the “average rate at which the volume is changing” is the average value of the rate function.

In Desmos, define the rate function:

$R (t) = 12 + 4 e^{0.2 t} sin (\frac{π t}{6})$

Then on the next line, setup the integral for the average value, $R_{avg}$ :

$\frac{1}{12} \int_{0}^{12} R (t) d t$

The answer is $6.432$ and $R_{avg}$ has the same units as $R (t)$ of gallons per hour. Keep $R (t)$ defined for the remaining problems.

b) Find the total amount of water in the tank at time $t = 12$ .

(spoiler)

Let $V (t)$ be the amount of water in the tank at any time $t$ .

Initially, the tank contains $150$ gallons of water, so $V (0) = 150$ .

To find $V (12)$ , use the Fundamental theorem of calculus. Add the initial amount to the net change in volume (found by integrating the rate):

$V (12) = V (0) + \int_{0}^{12} R (t) d t$

In Desmos, entering

$150 + \int_{0}^{12} R (t) d t$

yields $227.178$ gallons as the amount of water at $t = 12$ hours.

c) Is the rate at which the volume of water changes increasing or decreasing at $t = 5$ ?

(spoiler)

Keyword translation: If the rate $R (t)$ is increasing, then $R^{'} (t) > 0$ (and if decreasing, $R^{'} (t) < 0$ ). Evaluate the derivative at $t = 5$ by entering:

$R^{'} (5)$

which gives $- 3.843$ .

Justification: The rate at which the volume changes is decreasing at $t = 5$ because $R^{'} (5) < 0$ .

d) Is the amount of water increasing or decreasing at time $t = 9$ ?

(spoiler)

Keyword translation: If amount of water $V (t)$ is increasing, then $V^{'} (t) > 0$ (and if decreasing, its derivative is negative).

Since $V^{'} (t) = R (t)$ , we find $V^{'} (9)$ by entering:

$R (9)$

in Desmos, which gives $- 12.199$ gallons per hour as the rate of change of the volume at $t = 9$ .

Since $V^{'} (9) = R (9) < 0$ , the rate of change is negative, meaning the amount of water in the tank is decreasing at $t = 9$ .

Achievable AP Calculus AB

8. Applications of integrals

Our AP Calculus AB course is now in "early access" - get 50% off for a limited time.