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Course: Integral Calculus>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 (liters per minute) for . We can find the volume of the water (in $\text{L}$) by multiplying the time and the rate:
$\begin{array}{rl}\text{Volume}& =\text{Time}×\text{Rate}\\ & =6\phantom{\rule{0.167em}{0ex}}\text{min}\cdot 5\phantom{\rule{0.167em}{0ex}}\frac{\text{L}}{\text{min}}\\ & =30\frac{\overline{)\text{min}}\cdot \text{L}}{\overline{)\text{min}}}\\ & =30\phantom{\rule{0.167em}{0ex}}\text{L}\end{array}$
Now consider this case graphically. The rate can be represented by the constant function ${r}_{1}\left(t\right)=5$:
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 ${\int }_{0}^{6}{r}_{2}\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$. 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:
${\int }_{0}^{6}{r}_{2}\left(t\right)\phantom{\rule{0.167em}{0ex}}dt\approx 24.5\phantom{\rule{0.167em}{0ex}}\text{L}$

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 $\left(t=0\right)$ and $6$ minutes after that $\left(t=6\right)$. 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 ${\int }_{0}^{6}{r}_{2}\left(t\right)\phantom{\rule{0.167em}{0ex}}dt\approx 24.5\phantom{\rule{0.167em}{0ex}}\text{L}$ 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:
This is approximately .
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 $r\left(t\right)$ grams per day, where $t$ is measured in days.
What are the units of the quantity represented by the definite integral ${\int }_{0}^{8}r\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$?

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 $\frac{\text{Quantity A}}{\text{Quantity B}}$, then its definite integral is measured in $\text{Quantity A}$.
For example, in Problem set 1, $r$ was measured in $\frac{\text{grams}}{\text{day}}$, and so the definite integral of $r$ was measured in $\text{grams}$.
Problem 2
Eden walked at a rate of $r\left(t\right)$ kilometers per hour (where $t$ is the time in hours).
What does ${\int }_{2}^{3}r\left(t\right)\phantom{\rule{0.167em}{0ex}}dt=6$ mean?

Common mistake: Misinterpreting the interval of integration

For any rate function $r$, the definite integral ${\int }_{a}^{b}r\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$ 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 ${\int }_{2}^{3}r\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$ as the distance Eden walked in $3$ hours. The lower boundary is $2$, so ${\int }_{2}^{3}r\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$ is the distance Eden walked between the ${2}^{\text{nd}}$ hour and the ${3}^{\text{rd}}$ hour. Furthermore, in cases like that where the time interval is exactly one unit, we usually say "during the ${3}^{\text{rd}}$ hour."
Problem 3
Julia's revenue is $r\left(t\right)$ 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 $3+{\int }_{1}^{5}r\left(t\right)\phantom{\rule{0.167em}{0ex}}dt=19$ mean?

Common mistake: Ignoring initial conditions

For a rate function $f$ and an antiderivative $F$, the definite integral ${\int }_{a}^{b}f\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$ 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, ${\int }_{1}^{5}r\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$ represents the change in the amount of money Julia made between the ${1}^{\text{st}}$ and the ${5}^{\text{th}}$ months. But since we added $3$, which is the amount Julia had at the ${1}^{\text{st}}$ month, the expression now represents the actual amount in the ${5}^{\text{th}}$ month.

Connection with applied rates of change

In differential calculus, we learned that the derivative ${f}^{\prime }$ 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 $f$.
QuantityRate
Differential calculus$f\left(x\right)$${f}^{\prime }\left(x\right)$
Integral calculus$F\left(x\right)={\int }_{a}^{x}f\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$$f\left(x\right)$
Problem 4
The function $k\left(t\right)$ gives the amount of ketchup (in kilograms) produced in a sauce factory by time $t$ (in hours) on a given day.
What does ${\int }_{0}^{4}{k}^{\prime }\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$ represent?

Want more practice? Try this exercise.

Want to join the conversation?

• In cases where the time interval is exactly one unit (e.g. [2,3]), why do we usually say "during the third hour" instead of "during the second hour"?

EDIT: Specifically, in the example where Eden walks "between the 2nd hour and the 3rd hour", wouldn't it make more sense to say "during the 2nd hour" because the first statement implies that Eden walked from the start of the 2nd hour to the start of the 3rd hour?
• Think about if you were waiting for a friend. Waiting for your friend for the first hour would be from time 0-1, waiting for your friend for the second hour would be 1-2, and waiting for your friend for the third hour would be between 2-3. You may want to find a new friend though if he makes you wait that long.

Time in centuries is similar. Between the years 0-99 AD is the first century, 100-199 is the second century, etc.
• For the the calculation of the integral of r2(t)=6sin(0.3t) under "Thinking about accumulation in a real world context," why is the 6 divided by 0.3?

• Due to the chain rule, when you differentiate you would multiply by 0.3. When you integrate you reverse this process and so you divide by 0.3. You can make a u-substitution to make this easier to understand if you need to.
• For the the calculation of the integral of r2(t)=6sin(0.3t) under "Thinking about accumulation in a real world context," why is the 6 divided by 0.3?
• In problem 4... Why is the function in the integral called k′ instead of k?

k is already a rate function, and in your other rate problems, like problem 2, you didn't make the rate function a derivative when writing the integral notation.
• Note that the problem itself defines the function 𝑘(𝑥). The function in the integral then refers to the derivative of this function 𝑘. The other problems defined the rate function itself, and so is written differently without the prime notation.
• Why is the final problem 0~4 k′(t)dt, when the answer is, "The amount of ketchup produced over the first 4 hours"? To match that answer, shouldn't the integral be, 0~4 k(t)dt (while 0~4 k′(t)dt would be the accumulated rates of change over the first four hours -- or nothing at all if that notation does not apply to integrals)?
• k(t) is not the amount of ketchup produced per hour (as it was the case in all previous examples); k(t) is the amount of ketchup produced during n hours.
That is why k'(t) is now the rate of ketchup production, so that its integral is the "area under the curve", or the total amount of ketchup produced over n hours.
Basically the integral gives you the area under the curve, starting from the rate of change. If the function allows you to calculate the volume (as in this example), you first take its derivative (which gives you the rate of change) and then the integral of its derivative (to go back to the calculation of the volume). We have done a full loop here..
In reality you could have simply found k(t) at t=4 to have the same final answer...
• what is the difference between definate integral calculus and indefinate integral calculus
• In the field of integral calculus we speak of definite integrals and indefinite integrals.

In short, an indefinite integral is a function (𝐹(𝑥) + 𝐶),
while a definite integral is a value ("area under the curve").
• where did the 6sin(0.3) = 6/(0.3) [cos1.8] come from?
• sin(x) is actually being integrated there, which will be covered in future videos.
• What does DT or dx means in this equation?