Using Accumulation Functions and Definite Integrals in Applied Contexts

AP Calculus · Applications of Integration

Definite Integrals in Applied Contexts

The integral of a rate is an amount — and the exam wants that said in a full sentence.

key idea$\int_a^b r(t)\,dt=\text{the \emph{net change} in the quantity whose rate is }r$

Setting up the model

If $r$ is the rate at which something accumulates, its integral is the amount accumulated. The exam almost always asks you to say what the integral means in the context of the problem — a sentence with a number, a unit, and a time span, not just a value.

Writing an interpretation that scores

State what quantity changed — gallons of water, people, degrees, dollars.
State how much, with the correct unit. The unit is the rate's unit multiplied by the unit of the variable of integration.
State over which interval, in the problem's own words: ‘between $t=0$ and $t=6$ minutes’.
Say net change, not total, whenever the rate can be negative. A negative integral means the quantity decreased.
If a starting amount is given, add it: $Q(b)=Q(a)+\displaystyle\int_a^b r\,dt$ turns a net change into an actual amount.

the four things being tested

The integral of a rate is an amount — the units multiply out and the ‘per’ cancels.
$\displaystyle\int_a^b r\,dt$ is a net change; add the initial amount to get a level.
Two competing rates subtract: $\dfrac{dQ}{dt}=\text{in}-\text{out}$, and $Q$ peaks where they are equal.
‘How much’ wants the integral; ‘how fast’ wants the rate itself. Read the question's verb.

gallons per minute $\times$ minutesgallons

people per year $\times$ yearspeople

Integral of a ratean amount, never a rate

Worked examples

EX 1

Interpret an integral in context

the sentence that earns the point

Water enters a tank at $W(t)$ gallons per hour. Interpret $\displaystyle\int_{2}^{5}W(t)\,dt=140$.

Identify the quantity

$W$ is a rate of water entering, so its integral counts water.

Work out the units

$\dfrac{\text{gallons}}{\text{hour}}\times\text{hours}=\text{gallons}$.

Identify the interval

From $t=2$ to $t=5$ hours.

Assemble the sentence

“Between $t=2$ and $t=5$ hours, $140$ gallons of water entered the tank.”

Note what is missing

This says nothing about how much is in the tank — only how much entered. That needs the amount at $t=2$.

EX 2

Net change plus an initial amount

getting to a level

A tank holds $300$ gallons at $t=0$. Water enters at $W(t)$ and leaves at $L(t)$ gallons per hour, with $\displaystyle\int_{0}^{8}W=520$ and $\displaystyle\int_{0}^{8}L=610$. How much is in the tank at $t=8$?

Write the relation

$Q(8)=Q(0)+\displaystyle\int_{0}^{8}\left(W-L\right)dt$.

Combine the integrals

$\displaystyle\int_{0}^{8}W-\int_{0}^{8}L=520-610=-90$ gallons.

Read the sign

Negative, so the tank lost $90$ gallons over the eight hours.

Add the initial amount

$Q(8)=300-90=210$ gallons.

Note the structure

Two rates, one subtraction, one initial value. Nearly every tank problem is exactly this.

EX 3

When is the amount greatest?

in equals out

For a tank with $\dfrac{dQ}{dt}=W(t)-L(t)$, water enters faster than it leaves until $t=6$, after which the reverse holds. When is the tank fullest, and how would you compute that amount?

Identify the critical point

$Q'=W-L$, which is zero exactly when $W=L$ — here at $t=6$.

Test the sign

$Q'\gt 0$ before $t=6$ and $Q'\lt 0$ after.

Classify

Positive to negative is a maximum, so the tank is fullest at $t=6$.

Write the amount

$Q(6)=Q(0)+\displaystyle\int_{0}^{6}\left(W(t)-L(t)\right)dt$.

Note the shortcut you must not take

The maximum is not where $W$ is largest. It is where the two rates cross.

EX 4

Rate versus amount

read the verb

A population grows at $P'(t)=120e^{0.03t}$ people per year. Distinguish what $P'(10)$ and $\displaystyle\int_{0}^{10}P'(t)\,dt$ each tell you.

Evaluate the rate

$P'(10)=120e^{0.3}\approx162$ — units of people per year.

Say what it means

At the instant $t=10$, the population is growing at about $162$ people per year.

Evaluate the integral

$\displaystyle\int_{0}^{10}120e^{0.03t}dt=\left[4000e^{0.03t}\right]_{0}^{10}=4000\left(e^{0.3}-1\right)\approx1400$.

Say what that means

The population grew by about $1400$ people over the first ten years — units of people.

Fix the distinction

One is a speedometer reading, the other an odometer reading. ‘How fast’ and ‘how much’ never share an answer.

Units & interpretation

If $r$ has units of…	…then $\displaystyle\int r\,dt$ has units of	Typical wording
gallons per hour	gallons	“water that entered the tank”
cars per minute	cars	“cars that passed the sensor”
$ per year	$	“revenue over the period”
°C per hour	°C	“change in temperature”
metres per second	metres	“displacement”

Row 4 is the one people mistrust: the integral of a rate of temperature change is a change in temperature, not a temperature. Add it to the starting reading to get an actual value.

Watch out

An interpretation needs number, unit and interval

“It is $140$” earns nothing. “$140$ gallons entered the tank between $t=2$ and $t=5$ hours” earns the point.

Net change is not the final amount

$\int_a^b r$ measures the change. Without the starting value you cannot report a level.

The maximum is where the rates cross

In an inflow-outflow problem the quantity peaks when in $=$ out, not when the inflow is largest.

Do not report a rate when an amount was asked

Evaluating $r$ at an endpoint answers ‘how fast’. If the question says how much, how many, or by how much, integrate.

Course map. Topic 8.3 · LO CHA-4.D, CHA-4.E — interpret the meaning of a definite integral in accumulation problems and determine net change in applied contexts.