8.1 Average value of a function

Achievable AP Calculus AB

8. Applications of integrals

Our AP Calculus AB course is currently in development and is a work-in-progress.

Average value of a function

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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, you add them and divide by how many numbers there are. The average value of a continuous function on an interval works the same way. The difference is that, instead of adding up a list of separate values, you use an integral to accumulate values continuously across the interval, then divide by the interval length.

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 $b$ .

The area under $f (x)$ from $a$ to $b$ is:

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

The average value is:

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

So a rectangle with width $b - a$ and height $f_{avg}$ has area:

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

This is why you can think of $f_{avg}$ as a “flattened” height that would produce the same total accumulated area over the interval.

Examples

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

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

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

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

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

$\approx 4.333$

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 the interval $3 \leq t \leq 7$ is

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

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

$\approx 73.333$

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

To find when the temperature equals this average value, set $T (t)$ equal to $T_{avg}$ :

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

Solving with a calculator gives

$t = 5.449$

That is about 5:27 AM.

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 an accumulation function:

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

By the Fundamental theorem of calculus,

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

and

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

Use the second derivative test for concavity:

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

So we need the sign of $r^{'} (t)$ . Differentiate $r (t)$ :

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

To see where the sign could change, find where $r^{'} (t) = 0$ :

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

$sin (\frac{π t}{3}) = 0$

$\frac{π t}{3} = 0, π, 2 π ...$

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

On the interval $4 < t < 6$ , there are no zeros of $r^{'} (t)$ , so the sign stays consistent throughout that interval. Check a convenient point, such as $t = 4$ :

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

$= - \frac{2 π}{3} (- \frac{3}{2})$

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

Since $r^{'} (t) > 0$ on $4 < t < 6$ , we have $V^{''} (t) > 0$ there, so $V (t)$ is concave up on that interval.

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 $t$ measures hours after opening at 9 AM, the interval from 12 PM to 6 PM corresponds to $t = 3$ to $t = 9$ . Integrating the rate over this time interval gives the accumulated 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 asks for the average rate of customers entering the store, which is the average value of $R (t)$ over $[3, 9]$ :

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

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

$= 20 customers per hour$

c) Average rate of change

(spoiler)

Here you’re not averaging the rate itself; you’re averaging how the rate changes over time. That’s the average rate of change of $R (t)$ on $[3, 9]$ :

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

$= - 1.667$

The negative sign means the entry rate decreased overall from $t = 3$ to $t = 9$ , by about $1.67$ customers per hour per hour on average.

Average value of a function

Formula: $f_{avg} = \frac{1}{b - a} \int_{a}^{b} f (x) d x$
Units match those of $f (x)$
Concept: continuous analog of arithmetic mean

Geometric interpretation

$f_{avg}$ = height of rectangle with base $b - a$ and same area as under $f (x)$ on $[a, b]$
Area under $f (x)$ : $\int_{a}^{b} f (x) d x$
Rectangle area: $(b - a) \cdot f_{avg}$

Example calculations

Use definite integral to accumulate values
Divide by interval length for average
Example: $f (x) = x^{2}$ on $[1, 3]$ yields $f_{avg} \approx 4.333$

Applications & real-world contexts

Average temperature: integrate $T (t)$ over time, divide by interval length
- Example: $T_{avg} \approx 73.33 3^{\circ}$ F between 3 AM and 7 AM
- Set $T (t) = T_{avg}$ to find when this value occurs

Total accumulation and average rate (flow problems)

Total amount: $\int_{a}^{b} r (t) d t$
- Example: $\int_{0}^{30} r (t) d t = 300$ liters
Average rate: $\frac{1}{b - a} \int_{a}^{b} r (t) d t$
- Example: $r_{avg} = 10$ liters/min

Concavity of accumulation functions

$V (t) = \int_{0}^{t} r (x) d x$
- $V^{'} (t) = r (t)$ , $V^{''} (t) = r^{'} (t)$
Concave up if $r^{'} (t) > 0$ ; concave down if $r^{'} (t) < 0$
- Example: $V (t)$ is concave up on $4 < t < 6$

Interpreting rate functions (customers per hour)

Total customers: integrate rate over time interval
- Example: $\int_{3}^{9} R (t) d t = 120$ customers
Average rate: average value of $R (t)$ over interval
- Example: $R_{avg} = 20$ customers/hour
Average rate of change: $\frac{R ( b ) - R ( a )}{b - a}$
- Example: $- 1.67$ customers/hour² (rate decreasing)

Average value of a function

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

The area under $f (x)$ from $a$ to $b$ is:

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

The average value is:

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

So a rectangle with width $b - a$ and height $f_{avg}$ has area:

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

This is why you can think of $f_{avg}$ as a “flattened” height that would produce the same total accumulated area over the interval.

Examples

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

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

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

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

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

$\approx 4.333$

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 the interval $3 \leq t \leq 7$ is

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

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

$\approx 73.333$

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

To find when the temperature equals this average value, set $T (t)$ equal to $T_{avg}$ :

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

Solving with a calculator gives

$t = 5.449$

That is about 5:27 AM.

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 an accumulation function:

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

By the Fundamental theorem of calculus,

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

and

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

Use the second derivative test for concavity:

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

So we need the sign of $r^{'} (t)$ . Differentiate $r (t)$ :

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

To see where the sign could change, find where $r^{'} (t) = 0$ :

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

$sin (\frac{π t}{3}) = 0$

$\frac{π t}{3} = 0, π, 2 π ...$

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

On the interval $4 < t < 6$ , there are no zeros of $r^{'} (t)$ , so the sign stays consistent throughout that interval. Check a convenient point, such as $t = 4$ :

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

$= - \frac{2 π}{3} (- \frac{3}{2})$

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

Since $r^{'} (t) > 0$ on $4 < t < 6$ , we have $V^{''} (t) > 0$ there, so $V (t)$ is concave up on that interval.

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 asks for the average rate of customers entering the store, which is the average value of $R (t)$ over $[3, 9]$ :

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

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

$= 20 customers per hour$

c) Average rate of change

(spoiler)

Here you’re not averaging the rate itself; you’re averaging how the rate changes over time. That’s the average rate of change of $R (t)$ on $[3, 9]$ :

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

$= - 1.667$

The negative sign means the entry rate decreased overall from $t = 3$ to $t = 9$ , by about $1.67$ customers per hour per hour on average.

Achievable AP Calculus AB

8. Applications of integrals

Our AP Calculus AB course is currently in development and is a work-in-progress.

Average value of a function

6 min read

Font

Discuss

Feedback

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

The area under $f (x)$ from $a$ to $b$ is:

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

The average value is:

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

So a rectangle with width $b - a$ and height $f_{avg}$ has area:

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

This is why you can think of $f_{avg}$ as a “flattened” height that would produce the same total accumulated area over the interval.

Examples

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

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

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

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

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

$\approx 4.333$

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 the interval $3 \leq t \leq 7$ is

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

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

$\approx 73.333$

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

To find when the temperature equals this average value, set $T (t)$ equal to $T_{avg}$ :

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

Solving with a calculator gives

$t = 5.449$

That is about 5:27 AM.

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 an accumulation function:

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

By the Fundamental theorem of calculus,

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

and

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

Use the second derivative test for concavity:

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

So we need the sign of $r^{'} (t)$ . Differentiate $r (t)$ :

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

To see where the sign could change, find where $r^{'} (t) = 0$ :

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

$sin (\frac{π t}{3}) = 0$

$\frac{π t}{3} = 0, π, 2 π ...$

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

On the interval $4 < t < 6$ , there are no zeros of $r^{'} (t)$ , so the sign stays consistent throughout that interval. Check a convenient point, such as $t = 4$ :

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

$= - \frac{2 π}{3} (- \frac{3}{2})$

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

Since $r^{'} (t) > 0$ on $4 < t < 6$ , we have $V^{''} (t) > 0$ there, so $V (t)$ is concave up on that interval.

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 asks for the average rate of customers entering the store, which is the average value of $R (t)$ over $[3, 9]$ :

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

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

$= 20 customers per hour$

c) Average rate of change

(spoiler)

Here you’re not averaging the rate itself; you’re averaging how the rate changes over time. That’s the average rate of change of $R (t)$ on $[3, 9]$ :

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

$= - 1.667$

The negative sign means the entry rate decreased overall from $t = 3$ to $t = 9$ , by about $1.67$ customers per hour per hour on average.

Average value of a function

Formula: $f_{avg} = \frac{1}{b - a} \int_{a}^{b} f (x) d x$
Units match those of $f (x)$
Concept: continuous analog of arithmetic mean

Geometric interpretation

$f_{avg}$ = height of rectangle with base $b - a$ and same area as under $f (x)$ on $[a, b]$
Area under $f (x)$ : $\int_{a}^{b} f (x) d x$
Rectangle area: $(b - a) \cdot f_{avg}$

Example calculations

Use definite integral to accumulate values
Divide by interval length for average
Example: $f (x) = x^{2}$ on $[1, 3]$ yields $f_{avg} \approx 4.333$

Applications & real-world contexts

Average temperature: integrate $T (t)$ over time, divide by interval length
- Example: $T_{avg} \approx 73.33 3^{\circ}$ F between 3 AM and 7 AM
- Set $T (t) = T_{avg}$ to find when this value occurs

Total accumulation and average rate (flow problems)

Total amount: $\int_{a}^{b} r (t) d t$
- Example: $\int_{0}^{30} r (t) d t = 300$ liters
Average rate: $\frac{1}{b - a} \int_{a}^{b} r (t) d t$
- Example: $r_{avg} = 10$ liters/min

Concavity of accumulation functions

$V (t) = \int_{0}^{t} r (x) d x$
- $V^{'} (t) = r (t)$ , $V^{''} (t) = r^{'} (t)$
Concave up if $r^{'} (t) > 0$ ; concave down if $r^{'} (t) < 0$
- Example: $V (t)$ is concave up on $4 < t < 6$

Interpreting rate functions (customers per hour)

Total customers: integrate rate over time interval
- Example: $\int_{3}^{9} R (t) d t = 120$ customers
Average rate: average value of $R (t)$ over interval
- Example: $R_{avg} = 20$ customers/hour
Average rate of change: $\frac{R ( b ) - R ( a )}{b - a}$
- Example: $- 1.67$ customers/hour² (rate decreasing)

Average value of a function

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

The area under $f (x)$ from $a$ to $b$ is:

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

The average value is:

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

So a rectangle with width $b - a$ and height $f_{avg}$ has area:

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

This is why you can think of $f_{avg}$ as a “flattened” height that would produce the same total accumulated area over the interval.

Examples

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

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

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

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

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

$\approx 4.333$

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 the interval $3 \leq t \leq 7$ is

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

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

$\approx 73.333$

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

To find when the temperature equals this average value, set $T (t)$ equal to $T_{avg}$ :

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

Solving with a calculator gives

$t = 5.449$

That is about 5:27 AM.

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 an accumulation function:

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

By the Fundamental theorem of calculus,

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

and

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

Use the second derivative test for concavity:

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

So we need the sign of $r^{'} (t)$ . Differentiate $r (t)$ :

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

To see where the sign could change, find where $r^{'} (t) = 0$ :

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

$sin (\frac{π t}{3}) = 0$

$\frac{π t}{3} = 0, π, 2 π ...$

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

On the interval $4 < t < 6$ , there are no zeros of $r^{'} (t)$ , so the sign stays consistent throughout that interval. Check a convenient point, such as $t = 4$ :

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

$= - \frac{2 π}{3} (- \frac{3}{2})$

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

Since $r^{'} (t) > 0$ on $4 < t < 6$ , we have $V^{''} (t) > 0$ there, so $V (t)$ is concave up on that interval.

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 asks for the average rate of customers entering the store, which is the average value of $R (t)$ over $[3, 9]$ :

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

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

$= 20 customers per hour$

c) Average rate of change

(spoiler)

Here you’re not averaging the rate itself; you’re averaging how the rate changes over time. That’s the average rate of change of $R (t)$ on $[3, 9]$ :

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

$= - 1.667$

The negative sign means the entry rate decreased overall from $t = 3$ to $t = 9$ , by about $1.67$ customers per hour per hour on average.