8.3 - Using Accumulation Functions and Definite Integrals in Applied Contexts

Fundamental Principle: Integrating Rates

When you integrate a rate of change, you get the net change in the quantity:

\int_{a}^{b} \text{rate}\,dt = \text{net change}

Same idea as velocity and position: integrating velocity (the rate of change of position) gives displacement (the net change in position).

Key insight: this goes way beyond motion problems. It applies to any context involving a rate.

Example 1: Beach Erosion

Problem: During a storm, sand washes away from a beach and the distance between a road and the water’s edge shrinks. The rate of decrease is

r(t) = 8e^{-0.3t} \text{ feet per hour}

with

t

in hours after the storm began. The water’s edge starts 250 feet from the road. If the storm lasts 5 hours, how far is the water from the road afterward?

Step 1: Set it up. Start with the initial distance and subtract the net change, since the distance is decreasing.

\text{Final} = 250 - \int_{0}^{5} 8e^{-0.3t}\,dt

Step 2: Evaluate the integral.

\int_{0}^{5} 8e^{-0.3t}\,dt = \frac{8}{0.3}\left(1 - e^{-1.5}\right) \approx 20.717

Step 3: Final distance.

250 - 20.717 \approx 229.3 \text{ feet}

Example 2: Writing an Integral Equation (Without Solving)

Problem: A concert venue opens its doors. People enter at a rate of

p(t)

people per minute. When the doors open there are already 50 people inside (staff and VIPs). Write, but do not solve, an equation involving an integral to find the time

x

when there are 300 people inside. Nobody leaves.

50 + \int_{0}^{x} p(t)\,dt = 300

Breaking down each piece:

50: the initial number of people

$\int_{0}^{x} p(t)\,dt$ : the net change, meaning everyone who entered between $t = 0$ and $t = x$

300: the target

$x$ : the unknown time you’re solving for

This “write but do not solve” setup is a very common AP question type.

Example 3: Using Tables with Riemann Sums

Problem: Water flows into a reservoir at a rate of

f(t)

liters per hour.

$t$ (hours)	0	2	5	8	10
$f(t)$ (L/hr)	4000	3500	2800	2000	1500

f

is differentiable and decreasing on

[0, 10]

. At

t = 0

the reservoir holds 8000 liters.

Part (a): What does $\int_{0}^{10} f(t)\,dt$ represent?

The amount of water, in liters, that flowed into the reservoir over the 10 hours. Integrating a rate of flow gives you the total quantity that flowed.

Part (b): Estimate it with a left Riemann sum

The subintervals have different widths:

[0,2]

has

\Delta t = 2

[2,5]

has

\Delta t = 3

[5,8]

has

\Delta t = 3

, and

[8,10]

has

\Delta t = 2

\int_{0}^{10} f(t)\,dt \approx 2(4000) + 3(3500) + 3(2800) + 2(2000)

= 8000 + 10500 + 8400 + 4000 = 30{,}900 \text{ liters}

Part (c): How much water is in the reservoir after 10 hours?

8000 + 30{,}900 = 38{,}900 \text{ liters}

Example 4: Fundamental Theorem Application

Problem: The amount of oil in a tank is

V(t)

gallons,

t

in hours.

$t$ (hours)	0	2	5	8	10
$V(t)$ (gal)	8000	12500	17200	20800	23000

Find

\int_{0}^{10} V'(t)\,dt

and explain what it means.

By the Fundamental Theorem of Calculus:

\int_{0}^{10} V'(t)\,dt = V(10) - V(0) = 23000 - 8000 = 15{,}000 \text{ gallons}

That’s the net change in the oil in the tank over 10 hours. 15,000 gallons were added.

Important Distinction

When the table gives you $V(t)$ (an amount, not a rate): use the Fundamental Theorem.

\int_a^b V'(t)\,dt = V(b) - V(a)

. Just subtract table values.

When the table gives you $V'(t)$ (a rate): use a Riemann sum to approximate the integral. Multiply rate by time for each interval.

Pattern Recognition Guide

Scenario 1: Given a rate, find the total change

\text{Final} = \text{Initial} \pm \int_{t_1}^{t_2} \text{rate}(t)\,dt

Use plus if the quantity is increasing, minus if it’s decreasing.

Phrases that signal a rate: “at a rate of,” “the rate of change is,” “modeled by

r(t)

[units per time].”

Scenario 2: Write an equation to find an unknown time

\text{Initial} + \int_{0}^{x} \text{rate}(t)\,dt = \text{Target}

where

x

is the unknown.

Scenario 3: Estimate from a table (rate given)

\int_{a}^{b} r(t)\,dt \approx \sum (\Delta t_i) \cdot r(t_i)

Left, right, or midpoint, depending on what the problem says.

Scenario 4: Table shows actual values, not a rate

\int_{a}^{b} f'(t)\,dt = f(b) - f(a)

Just subtract the endpoint values.

Example 5: Population Growth

Problem: A town’s population is

P(t)

thousand people,

t

years after 2020. The rate of change is

P'(t) = 0.5e^{0.08t} \text{ thousand people per year}

In 2020 the population was 15,000. What’s the population in 2025?

P(5) = P(0) + \int_{0}^{5} 0.5e^{0.08t}\,dt

\int_{0}^{5} 0.5e^{0.08t}\,dt = \frac{0.5}{0.08}\left(e^{0.4} - 1\right) \approx 3.074

P(5) = 15 + 3.074 = 18.074 \text{ thousand people}

Population in 2025: about 18,074 people

Accumulation Functions: Rate In and Rate Out

A function defined by an integral, like

A(t) = A(0) + \int_{0}^{t} r(s)\,ds

is an accumulation function. Its value at time

t

is the starting amount plus everything that has built up since

t = 0

. By the Fundamental Theorem of Calculus (Unit 6),

A'(t) = r(t)

, so the rate tells you when the amount is going up or down.

A lot of AP problems have something coming in and something going out at the same time. Then the amount changes at the net rate, rate in minus rate out:

A(t) = A(0) + \int_{0}^{t} [\text{rate in}(s) - \text{rate out}(s)]\,ds \qquad A'(t) = \text{rate in}(t) - \text{rate out}(t)

Example 6: Cars In and Out of a Garage

Problem: At 7 AM (

t = 0

) a parking garage holds 120 cars. For

0 \leq t \leq 9

hours, cars enter at a rate of

E(t) = 51 + 24t - 3t^2

cars per hour and leave at a rate of

L(t) = 30 + 6t

cars per hour.

8.3 - Using Accumulation Functions and Definite Integrals in Applied Contexts

Part (a): Write an expression for the number of cars at time $t$

A(t) = 120 + \int_{0}^{t} [E(s) - L(s)]\,ds

Part (b): Is the number of cars going up or down at $t = 8$ ?

A'(8) = E(8) - L(8) = 51 - 78 = -27

A'(8) < 0

, so the number of cars is decreasing at 3 PM, at a rate of 27 cars per hour.

Part (c): When are there the most cars, and how many?

Set the net rate equal to zero:

E(t) - L(t) = 21 + 18t - 3t^2 = -3(t - 7)(t + 1) = 0 \quad \Rightarrow \quad t = 7

(

t = -1

is outside the interval.) Now use the candidates test: check the critical point and both endpoints.

$t$	$A(t)$
$0$	$120$
$7$	$120 + \int_{0}^{7} (21 + 18s - 3s^2)\,ds = 120 + 245 = 365$
$9$	$120 + \int_{0}^{9} (21 + 18s - 3s^2)\,ds = 120 + 189 = 309$

Answer: the garage is fullest at

t = 7

(2 PM), with 365 cars. That matches the graph: cars come in faster than they leave until

t = 7

, and after that the garage starts to empty.

Part (d): What does $\int_{2}^{5} L(t)\,dt$ mean?

\int_{2}^{5} (30 + 6t)\,dt = \left[30t + 3t^2\right]_{2}^{5} = 225 - 72 = 153

153 cars left the garage between 9 AM (

t = 2

) and noon (

t = 5

). Integrating a rate out gives the total amount that went out over that time.

Common Question Types

Type 1: “What does this integral represent?” Answer with something like “the net change in [quantity] from [start] to [end]” or “the total amount of [quantity] that accumulated during [time period].”

Type 2: “Write but do not solve…” Identify the initial value, set up Initial

\pm \int

rate

=

Target, and use a variable for whatever you’re finding.

Type 3: “Estimate using a left/right/midpoint Riemann sum.” Find the subinterval widths (they may not be uniform), pick the right height for each rectangle, and sum width times height.

Type 4: “How much is in the tank after…” Find the net change with an integral or Riemann sum, add it to (or subtract it from) the initial amount, and check your units.

Type 5: “When is the amount greatest (or least)?” Set rate in minus rate out equal to zero, then use the candidates test: check those times and both endpoints with

A(t) = A(0) + \int_0^t (\text{in} - \text{out})\,ds

Calculator Tips

Integrals on a TI-84:

Press MATH, select 9:fnInt(

Enter fnInt(function, variable, lower, upper)

Example: fnInt(8e^(-0.3X), X, 0, 5)

Tables in lists: put time values in L₁, rate values in L₂, then compute the Riemann sum with sum(ΔL₁*L₂), adjusting for left or right.

Recalling: press 2nd + ENTER to pull up the last entry and edit it.

Units Matter

If the rate has units…	The integral has units…
feet per hour	feet
liters per minute	liters
people per day	people
dollars per year	dollars

General rule: integrating a rate makes the “per [time]” disappear, leaving just the quantity.

Key Takeaways

Fundamental principle: $\int \text{rate}\,dt = \text{net change}$

Setup: Initial value $\pm \int$ rate $=$ Final value

Add the integral for increasing quantities, subtract for decreasing ones

“Write but do not solve” means set up the equation with a variable for the unknown

Tables with rates: use Riemann sums

Tables with actual values: use the Fundamental Theorem, $\int_a^b f'(t)\,dt = f(b) - f(a)$

Integrating “per time” removes the time unit

Always interpret an integral as a total change or accumulated amount

Distinguish rate functions from quantity functions in tables

Accumulation function: $A(t) = A(0) + \int_0^t (\text{rate in} - \text{rate out})\,ds$ , and $A'(t) = \text{rate in} - \text{rate out}$

Most or least of something: set rate in equal to rate out, then check those times and the endpoints

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