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Course: Calculus 2 > Unit 1

Lesson 1: Accumulations of change introduction

Exploring accumulation of change

Definite integrals are interpreted as the accumulation of quantities. Learn why this is so and how this can be used to analyze real-world contexts.

The definite integral can be used to express information about accumulation and net change in applied contexts. Let's see how it's done.

Thinking about accumulation in a real world context

Say a tank is being filled with water at a constant rate of

5 L/min

(liters per minute) for

6 min

. We can find the volume of the water (in

L

) by multiplying the time and the rate:

\begin{aligned} Volume & = Time \times Rate \\ = 6 min \cdot 5 \frac{L}{min} \\ = 30 \frac{min \cdot L}{min} \\ = 30 L \end{aligned}

Now consider this case graphically. The rate can be represented by the constant function

r_{1} (t) = 5

Each horizontal unit in this graph is measured in minutes and each vertical unit is measured in liters per minute, so the area of each square unit is measured in liters:

Furthermore, the area of the rectangle bounded by the graph of

r_{1}

and the horizontal axis between

t = 0

and

t = 6

gives us the volume of water after

6

minutes:

Now say another tank is being filled, but this time the rate isn't constant:

How can we tell the volume of water in this tank after

6

minutes? To do that, let's think about the Riemann sum approximation of the area under this curve between

t = 0

and

t = 6

. For the sake of convenience, let's use an approximation where each rectangle is

1

minute wide.

We saw how each rectangle represents a volume in liters. Specifically, each rectangle in this Riemann sum is an approximation of the volume of water that was added to the tank at each minute. When we add all the areas, i.e. when all the volumes are accumulated, we get an approximation for the total volume of water after

6

minutes.

As we use more rectangles with smaller widths, we will get a better approximation. If we take this to a limit of accumulating infinite rectangles, we will get the definite integral

\int_{0}^{6} r_{2} (t) d t

. This means that the exact volume of water after

6

minutes is equal to the area bounded by the graph of

r_{2}

and the horizontal axis between

t = 0

and

t = 6

And so, integral calculus allows us to find the total volume after

6

minutes:

\int_{0}^{6} r_{2} (t) d t \approx 24.5 L

Definite integral of the rate of change of a quantity gives the net change in that quantity.

In the example we saw, we had a function that describes a rate. In our case, it was the rate of volume over time. The definite integral of that function gave us the accumulation of volume—that quantity whose rate was given.

Another important feature here was the time interval of the definite integral. In our case, the time interval was the beginning

(t = 0)

and

6

minutes after that

(t = 6)

. So the definite integral gave us the net change in the amount of water in the tank between

t = 0

and

t = 6

These are the two ways we commonly think about definite integrals: they describe an accumulation of a quantity, so the entire definite integral gives us the net change in that quantity.

Why "net change" in the quantity and not simply the quantity?

Using the above example, notice how we weren't told whether there was any amount of water in the tank prior to

t = 0

. If the tank was empty, then

\int_{0}^{6} r_{2} (t) d t \approx 24.5 L

is really the amount of water in the tank after

6

minutes. But if the tank already contained, say,

7

liters of water, then the actual volume of water in the tank after

6

minutes is:

\underset{volume at t = 0}{\underset{⏟}{7}} + \overset{change in volume from t = 0 to t = 6}{\overset{⏞}{\int_{0}^{6} r_{2} (t) d t}}

This is approximately

7 + 24.5 = 31.5 L

Remember: The definite integral always gives us the net change in a quantity, not the actual value of that quantity. To find the actual quantity, we need to add an initial condition to the definite integral.

Problem 1.A

Problem set 1 will walk you through the process of analyzing a context that involves accumulation:

At time

t

, a population of bacteria grows at the rate of

r (t)

grams per day, where

t

is measured in days.

What are the units of the quantity represented by the definite integral $\int_{0}^{8} r (t) d t$ ‍?

Common mistake: Using inappropriate units

As with all applied word problems, units play an important role here. Remember that if

r

is a rate function measured in

\frac{Quantity A}{Quantity B}

, then its definite integral is measured in

Quantity A

For example, in Problem set 1,

r

was measured in

\frac{grams}{day}

, and so the definite integral of

r

was measured in

grams

Problem 2

Eden walked at a rate of

r (t)

kilometers per hour (where

t

is the time in hours).

What does $\int_{2}^{3} r (t) d t = 6$ ‍ mean?

Common mistake: Misinterpreting the interval of integration

For any rate function

r

, the definite integral

\int_{a}^{b} r (t) d t

describes the accumulation of values between $t = a$ ‍ and $t = b$ ‍.

A common mistake is to disregard one of the boundaries (usually the lower one), which results in a wrong interpretation.

For example, in Problem 2, it would be a mistake to interpret

\int_{2}^{3} r (t) d t

as the distance Eden walked in

3

hours. The lower boundary is

2

, so

\int_{2}^{3} r (t) d t

is the distance Eden walked between the

2^{nd}

hour and the

3^{rd}

hour. Furthermore, in cases like that where the time interval is exactly one unit, we usually say "during the

3^{rd}

hour."

Problem 3

Julia's revenue is

r (t)

thousand dollars per month (where

t

is the month of the year). Julia had made

3

thousand dollars in the first month of the year.

What does $3 + \int_{1}^{5} r (t) d t = 19$ ‍ mean?

(Choice A)
Julia made an additional $19$ ‍ thousand dollars between months $1$ ‍ and $5$ ‍.
(Choice B)
Julia made an average of $19$ ‍ thousand dollars per month.
(Choice C)
Julia made $19$ ‍ thousand dollars in the fifth month.
(Choice D)
By the end of the fifth month, Julia had made a total of $19$ ‍ thousand dollars.

Common mistake: Ignoring initial conditions

For a rate function

f

and an antiderivative

F

, the definite integral

\int_{a}^{b} f (t) d t

gives the net change in

F

between

t = a

and

t = b

. If we add an initial condition, we will get an actual value of

F

For example, in Problem 3,

\int_{1}^{5} r (t) d t

represents the change in the amount of money Julia made between the

1^{st}

and the

5^{th}

months. But since we added

3

, which is the amount Julia had at the

1^{st}

month, the expression now represents the actual amount in the

5^{th}

month.

Connection with applied rates of change

In differential calculus, we learned that the derivative

f^{'}

of a function

f

gives the instantaneous rate of change of

f

for a given input. Now we're going the other way! For any rate function

f

, its antiderivative

F

gives the accumulated value of the quantity whose rate is described by

f

	Quantity	Rate
Differential calculus	$f (x)$ ‍	$f^{'} (x)$ ‍
Integral calculus	$F (x) = \int_{a}^{x} f (t) d t$ ‍	$f (x)$ ‍

Problem 4

The function

k (t)

gives the amount of ketchup (in kilograms) produced in a sauce factory by time

t

(in hours) on a given day.

What does $\int_{0}^{4} k^{'} (t) d t$ ‍ represent?

(Choice A)
The amount of ketchup produced over the first $4$ ‍ hours.
(Choice B)
The instantaneous rate of production at $t = 4$ ‍.
(Choice C)
The time it takes to produce $4$ ‍ kilograms of ketchup.
(Choice D)
The average rate of change of the ketchup production over the first $4$ ‍ hours.

Want more practice? Try this exercise.

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