Average value of a function | Applications of integrals

What you’ll learn:

What it means to find the average value of a continuous function and how to calculate it
How to interpret it geometrically and in real-world contexts

To find the average of a set of numbers, we add them and divide by how many numbers there are. The average value of a continuous function on an interval works the same way, but instead of summing discrete values, we integrate over the interval for the total accumulated value and divide that “sum” by the length of the interval.

Average value of a function

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

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

The units are the same as those of $f (x)$ .

Geometric interpretation

Geometrically, the average value of a function $f (x)$ on the interval $[a, b]$ is the height of a rectangle with base length $b - a$ that has the same area as the region under $f (x)$ from $a$ to $b$ .

If you compute the area under $f (x)$ :

$Area = \int_{a}^{b} f (x) d x$

And the average value is

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

then the rectangle of height $f_{avg}$ and width $b - a$ has area:

$width (b - a) \times height f_{avg} = \int_{a}^{b} f (x) d x$

This helps us visualize $f_{avg}$ as the “flattened” version of the function that would accumulate the same total value.

Examples

1. Find the average value of $f (x) = x^{2}$ on the interval $[1, 3]$ .

$f_{avg} = \frac{1}{3 - 1} \int_{1}^{3} x^{2} d x$

$= \frac{1}{2} \int_{1}^{3} x^{2} d x$

$= \frac{1}{2} \cdot \frac{x ^{3}}{3}_{1}^{3}$

$= \frac{3 ^{3}}{6} - \frac{1 ^{3}}{6}$

$\approx 4.333$

2. A room’s temperature, in degrees Fahrenheit, is given by $T (t) = t^{3} - 5 t^{2} + 60$ , where time $t = 0$ hours is midnight. Find the average temperature between $3$ AM and $7$ AM and when that value occurs.

Solution

(spoiler)

The average temperature over interval $3 \leq t \leq 7$ is

$T_{avg} = \frac{1}{7 - 3} \int_{3}^{7} (t^{3} - 5 t^{2} + 60) d t$

$= \frac{1}{4} \int_{3}^{7} (t^{3} - 5 t^{2} + 60) d t$

$\approx 73.333$

The average temperature over those 4 hours is $73.333° F$ .

The function reaches this value when

$73.333 = t^{3} - 5 t^{2} + 60$

Solving with a calculator,

$t = 5.449$

which is roughly around 5:27 am.

3. Water flows into a tank at a rate of $r (t)$ liters per minute for $0 \leq t \leq 30$ , where

$r (t) = 10 + 2 cos (\frac{π t}{3})$

a) Find the total volume of water that flows into the tank over the $30$ -minute interval.

b) Find the average flow rate over that time period.

c) Let $V (t) = \int_{0}^{t} r (x) d x$ represent the total amount of water (in liters) that has flowed into the tank from time $t = 0$ to time $t$ . Is $V (t)$ concave up, concave down, or neither on the interval $4 < t < 6$ ?

Solutions

a) Total volume of water over $0 \leq t \leq 30$

(spoiler)

The total/accumulated volume is

$\int_{0}^{30} r (t) d t = \int_{0}^{30} (10 + 2 cos (\frac{π t}{3})) d t$

$= 300 liters$

b) Average flow rate

(spoiler)

The average flow rate is the average value of $r (t)$ over $[0, 30]$ :

$r_{avg} = \frac{1}{30 - 0} \int_{0}^{30} (10 + 2 cos (\frac{π t}{3})) d t$

$= \frac{1}{30} (300)$

$= 10 liters/min$

c) Concavity of $V (t)$

(spoiler)

$V (t)$ is defined as an accumulation function

$V (t) = \int_{0}^{t} r (x) d x$

By the Fundamental theorem of calculus,

$V^{'} (t) = r (t)$

and

$V^{''} (t) = r^{'} (t)$

If:

$V^{''} (t) > 0$ , then $V (t)$ is concave up.
$V^{''} (t) < 0$ , then $V (t)$ is concave down.

So we need to analyze the sign of $r^{'} (t)$ . Differentiating $r (t)$ ,

$r^{'} (t) = - \frac{2 π}{3} sin (\frac{π t}{3})$

To find the sign of $r^{'} (t)$ , or whether the graph is above or below the $x$ -axis, let’s find where it equals $0$ .

$- \frac{2 π}{3} sin (\frac{π t}{3}) = 0$

$\frac{π t}{3} = 0, π, 2 π ...$

$t = 0, 3, 6, ...$

This means that $r^{'} (t)$ is either above or below the $x$ -axis on the entire interval $4 < t < 6$ . Plug in a value to check:

$r^{'} (4) = - \frac{2 π}{3} sin (\frac{4 π}{3})$

$= - \frac{2 π}{3} (- \frac{3}{2})$

$= \frac{2 3}{6}$

Since $r^{'} (t)$ is positive on the interval $4 < t < 6$ , $R (t)$ is concave up.

4. The function

$R (t) = 20 + 5 sin (\frac{π t}{6})$

models the rate, in customers per hour, at which people enter a bookstore, where $t$ is the number of hours after opening. The store is open from 9 AM to 9 PM.

a) How many customers enter the store between 12 PM and 6 PM? Include units.

b) At what rate were customers entering the store, on average, between 12 PM and 6 PM? Include units.

c) Between 12 PM and 6 PM, how did the rate of customer flow change per hour, on average?

Solutions

a) Total customers

(spoiler)

Since $t$ is the number of hours after the opening time of 9 AM, the time interval from 12 PM to 6 PM corresponds to $t = 3$ to $t = 9$ . Integrating the rate function over this time interval, gives the accumulated value, or the total number of customers.

$\int_{3}^{9} R (t) d t$

$= \int_{3}^{9} (20 + 5 sin (\frac{π t}{6})) d t$

$= 120 customers$

b) Average value

(spoiler)

This question asks for the average rate of customers entering, or the average value of the rate function over the 6 hours from $t = 3$ to $t = 9$ , which is

$R_{avg} = \frac{1}{9 - 3} \int_{3}^{9} R (t) d t$

$= \frac{1}{6} \cdot 120$

$= 20 customers per hour$

c) Average rate of change

(spoiler)

The difference in wording is subtle, but this question asks for how the inflow rate itself changed, or the average rate of change of the rate function over the 6 hours. The previous part was about typical inflow of customers, while this question is about how the inflow rate trends over time.

The average rate of change of the rate is

$\frac{R ( 9 ) - R ( 3 )}{9 - 3}$

$= - 1.667$

This value is negative, which means that overall the inflow rate at $t = 9$ was lower than that at $t = 3$ , with a net decrease of about $1.67$ customers per hour per hour on average.

What you’ll learn:

What it means to find the average value of a continuous function and how to calculate it
How to interpret it geometrically and in real-world contexts

Average value of a function

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

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

The units are the same as those of $f (x)$ .

Geometric interpretation

If you compute the area under $f (x)$ :

$Area = \int_{a}^{b} f (x) d x$

And the average value is

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

then the rectangle of height $f_{avg}$ and width $b - a$ has area:

$width (b - a) \times height f_{avg} = \int_{a}^{b} f (x) d x$

This helps us visualize $f_{avg}$ as the “flattened” version of the function that would accumulate the same total value.

Examples

1. Find the average value of $f (x) = x^{2}$ on the interval $[1, 3]$ .

$f_{avg} = \frac{1}{3 - 1} \int_{1}^{3} x^{2} d x$

$= \frac{1}{2} \int_{1}^{3} x^{2} d x$

$= \frac{1}{2} \cdot \frac{x ^{3}}{3}_{1}^{3}$

$= \frac{3 ^{3}}{6} - \frac{1 ^{3}}{6}$

$\approx 4.333$

2. A room’s temperature, in degrees Fahrenheit, is given by $T (t) = t^{3} - 5 t^{2} + 60$ , where time $t = 0$ hours is midnight. Find the average temperature between $3$ AM and $7$ AM and when that value occurs.

Solution

(spoiler)

The average temperature over interval $3 \leq t \leq 7$ is

$T_{avg} = \frac{1}{7 - 3} \int_{3}^{7} (t^{3} - 5 t^{2} + 60) d t$

$= \frac{1}{4} \int_{3}^{7} (t^{3} - 5 t^{2} + 60) d t$

$\approx 73.333$

The average temperature over those 4 hours is $73.333° F$ .

The function reaches this value when

$73.333 = t^{3} - 5 t^{2} + 60$

Solving with a calculator,

$t = 5.449$

which is roughly around 5:27 am.

3. Water flows into a tank at a rate of $r (t)$ liters per minute for $0 \leq t \leq 30$ , where

$r (t) = 10 + 2 cos (\frac{π t}{3})$

a) Find the total volume of water that flows into the tank over the $30$ -minute interval.

b) Find the average flow rate over that time period.

c) Let $V (t) = \int_{0}^{t} r (x) d x$ represent the total amount of water (in liters) that has flowed into the tank from time $t = 0$ to time $t$ . Is $V (t)$ concave up, concave down, or neither on the interval $4 < t < 6$ ?

Solutions

a) Total volume of water over $0 \leq t \leq 30$

(spoiler)

The total/accumulated volume is

$\int_{0}^{30} r (t) d t = \int_{0}^{30} (10 + 2 cos (\frac{π t}{3})) d t$

$= 300 liters$

b) Average flow rate

(spoiler)

The average flow rate is the average value of $r (t)$ over $[0, 30]$ :

$r_{avg} = \frac{1}{30 - 0} \int_{0}^{30} (10 + 2 cos (\frac{π t}{3})) d t$

$= \frac{1}{30} (300)$

$= 10 liters/min$

c) Concavity of $V (t)$

(spoiler)

$V (t)$ is defined as an accumulation function

$V (t) = \int_{0}^{t} r (x) d x$

By the Fundamental theorem of calculus,

$V^{'} (t) = r (t)$

and

$V^{''} (t) = r^{'} (t)$

If:

$V^{''} (t) > 0$ , then $V (t)$ is concave up.
$V^{''} (t) < 0$ , then $V (t)$ is concave down.

So we need to analyze the sign of $r^{'} (t)$ . Differentiating $r (t)$ ,

$r^{'} (t) = - \frac{2 π}{3} sin (\frac{π t}{3})$

To find the sign of $r^{'} (t)$ , or whether the graph is above or below the $x$ -axis, let’s find where it equals $0$ .

$- \frac{2 π}{3} sin (\frac{π t}{3}) = 0$

$\frac{π t}{3} = 0, π, 2 π ...$

$t = 0, 3, 6, ...$

This means that $r^{'} (t)$ is either above or below the $x$ -axis on the entire interval $4 < t < 6$ . Plug in a value to check:

$r^{'} (4) = - \frac{2 π}{3} sin (\frac{4 π}{3})$

$= - \frac{2 π}{3} (- \frac{3}{2})$

$= \frac{2 3}{6}$

Since $r^{'} (t)$ is positive on the interval $4 < t < 6$ , $R (t)$ is concave up.

4. The function

$R (t) = 20 + 5 sin (\frac{π t}{6})$

models the rate, in customers per hour, at which people enter a bookstore, where $t$ is the number of hours after opening. The store is open from 9 AM to 9 PM.

a) How many customers enter the store between 12 PM and 6 PM? Include units.

b) At what rate were customers entering the store, on average, between 12 PM and 6 PM? Include units.

c) Between 12 PM and 6 PM, how did the rate of customer flow change per hour, on average?

Solutions

a) Total customers

(spoiler)

$\int_{3}^{9} R (t) d t$

$= \int_{3}^{9} (20 + 5 sin (\frac{π t}{6})) d t$

$= 120 customers$

b) Average value

(spoiler)

This question asks for the average rate of customers entering, or the average value of the rate function over the 6 hours from $t = 3$ to $t = 9$ , which is

$R_{avg} = \frac{1}{9 - 3} \int_{3}^{9} R (t) d t$

$= \frac{1}{6} \cdot 120$

$= 20 customers per hour$

c) Average rate of change

(spoiler)

The average rate of change of the rate is

$\frac{R ( 9 ) - R ( 3 )}{9 - 3}$

$= - 1.667$

This value is negative, which means that overall the inflow rate at $t = 9$ was lower than that at $t = 3$ , with a net decrease of about $1.67$ customers per hour per hour on average.