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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 ${5\text{L/min}}$ (liters per minute) for ${6\text{min}}$ . We can find the volume of the water (in $\text{L}$ ) by multiplying the time and the rate:

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 $6$ 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.

*this*tank afterWe 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 $6$ minutes.

*accumulated*, we get an approximation for the total volume of water afterAs 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 ${\int}_{0}^{6}{r}_{2}(t){\textstyle \phantom{\rule{0.167em}{0ex}}}dt$ . 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:

### 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 $(t=0)$ and $6$ minutes after that $(t=6)$ . So the definite integral gave us the $t=0$ and $t=6$ .

*time interval*of the definite integral. In our case, the time interval was the beginning*net change*in the amount of water in the tank betweenThese 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 $t=0$ . If the tank was empty, then $\int}_{0}^{6}{r}_{2}(t){\textstyle \phantom{\rule{0.167em}{0ex}}}dt\approx 24.5{\textstyle \phantom{\rule{0.167em}{0ex}}}\text{L$ 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:

*prior*toThis is approximately $7+24.5=31.5\text{L}$ .

**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.*

### 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 ${\text{grams}}$ .

### Common mistake: Misinterpreting the interval of integration

For any rate function $r$ , the definite integral ${\int}_{a}^{b}r(t){\textstyle \phantom{\rule{0.167em}{0ex}}}dt$ 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 ${\int}_{2}^{3}r(t){\textstyle \phantom{\rule{0.167em}{0ex}}}dt$ as the distance Eden walked in $3$ hours. The lower boundary is $2$ , so ${\int}_{2}^{3}r(t){\textstyle \phantom{\rule{0.167em}{0ex}}}dt$ 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."

### Common mistake: Ignoring initial conditions

For a rate function $f$ and an antiderivative $F$ , the definite integral ${\int}_{a}^{b}f(t){\textstyle \phantom{\rule{0.167em}{0ex}}}dt$ 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, ${\int}_{1}^{5}r(t){\textstyle \phantom{\rule{0.167em}{0ex}}}dt$ 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 $f$ .

Quantity | Rate | |
---|---|---|

Differential calculus | ||

Integral calculus |

*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?(31 votes)- 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.(185 votes)

- 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?

Sorry, the answer given under Ryan's comment is not loading. Can someone please explain it? Thank you(24 votes)- 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.(26 votes)

- 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?(16 votes)
- 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.(7 votes)- 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.(7 votes)

- 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)?(5 votes)
- 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...(10 votes)

- what is the difference between definate integral calculus and indefinate integral calculus(4 votes)
- 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").(9 votes)

- where did the 6sin(0.3) = 6/(0.3) [cos1.8] come from?(6 votes)
- sin(x) is actually being integrated there, which will be covered in future videos.(5 votes)

- What does DT or dx means in this equation?(3 votes)
- dt, dx and any d-something in calculus means "a small change in this thing" In fact it means so small it's basically 0(6 votes)

- in problem 3, "between the 1st and the 5th months" means 4 months, right? From month 0 to month 1 is the 1st month.

in "Common mistake: Ignoring initial conditions" section, it also uses the term antiderivative, which hasn't been taught in this course.(3 votes)- Ahh not quite. When we say months 1 to 5, we mean the ending of month 1 to the end of month 5 (which does still account for 4 months though). The "3" in the question is the income from the end of month 0 the end of month 1 though. It's kinda confusing notation, so I hope I explained it well enough😅

The antiderivative is pretty much the inverse of a derivative. If you apply the derivative to f(x), you get f'(x). And if you apply the antiderivative to f'(x), you get f(x) again (Pretty much how inverses work). The future lessons should cover it in depth (especially when you reach the "Fundamental Theorem of Calculus"). Just remember that the integral and the antidervative are different things. Many use them synonymously(5 votes)

- I'm confused by the last part. There
**hasn't**been any lessons on antiderivatives previously. It would be better to describe antiderivatives before this lesson, at least a basic idea of it not using integral calculus, just the concept with plain English. I hope there will be more clarification in the next lessons, and I hope you put more explanations or better yet, videos for the future generations. 🖐(3 votes)- It is true that there weren't any lessons before this article, which justifies your confusion. Basically an antiderivative is the reverse of a derivative. So if g(x) = f'(x), then the antiderivative of f(x) = g(x). Also, maybe it would have helped you if it was explained that antiderivatives are the same thing as indefinite integrals.

This video explains what antiderivatives are: https://www.khanacademy.org/math/ap-calculus-bc/bc-integration-new/bc-6-7/v/antiderivatives-and-indefinite-integrals

I hope this helps! :D(2 votes)