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## AP®︎/College Calculus AB

### Unit 8: Lesson 3

Using accumulation functions and definite integrals in applied contexts- Area under rate function gives the net change
- Interpreting definite integral as net change
- Worked examples: interpreting definite integrals in context
- Interpreting definite integrals in context
- Analyzing problems involving definite integrals
- Analyzing problems involving definite integrals
- Analyzing problems involving definite integrals
- Worked example: problem involving definite integral (algebraic)
- Problems involving definite integrals (algebraic)

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# Worked examples: interpreting definite integrals in context

AP.CALC:

CHA‑4 (EU)

, CHA‑4.D (LO)

, CHA‑4.D.1 (EK)

, CHA‑4.D.2 (EK)

, CHA‑4.E (LO)

, CHA‑4.E.1 (EK)

Interpreting expressions involving definite integrals in a real-world context.

## Want to join the conversation?

- Why would we use integrals to represent Julia's revenue? Isn't that unnecessarily complicated?(9 votes)
- I think integrals allow us to see the accumulation of her revenue at a certain interval of time.(3 votes)

- what would the definite integral of just k(t) from 0-4 be measuring? Its units would be kg*s.(3 votes)
- The owner of the sauce factory might charge a potential competitor who doesn't have a factory—perhaps because they are just starting out in the business—a usage fee of $1 per hour per kilogram of ketchup produced. The definite integral of k(t) from 0 to 4 would then measure the total fee for those four hours.(1 vote)

## Video transcript

- [Instructor] Julia's
revenue is r of t thousand dollars per month, where t
is the month of the year. Julia had made $3,000 in
the first month of the year, what does three plus
the definite role from one to five of r of t dt equals 19 mean? And we have some choices. So like always, pause the video and see if you can work through it. Alright, now let's work
through this together. So they tell us that she made
$3,000 in the first month and we also see this three
here, so that's interesting. Maybe they represent the same thing, we don't know for sure yet. But let's look at this definite integral. The definite integral from one to five of r of t dt, this is the
area under this rate curve, r of t is the rate atwhich
Julia makes revenue on a monthly basis. So if you take the area
under that rate curve, that's going to give you the net change in revenue from month one to month five, how much that increased. And so if you add that to the amount she made in month one, well that tells you the total she makes from
essentially time zero all the way to month five. And they're saying that is equal to 19. So let's see which of these choices are consistent with that. Julia made an additional $19,000 between months one and five. Choice A would be correct if you didn't see this three over here. Because just the definite
integral is the additional between months one and
five, but that's not what this expression says,
it says three plus this is equal to 19. If it said Julia made
an additional $16,000, well that would make
sense because you could subtract three from both sides
and you'd get that result, but that's not what they're saying. Julia made an average
of $19,000 per month. Well once again, that's also not right because we just said from the beginning, from time zero, all the
way until the fifth month, she made a total of $19,000. Not the average per month is $19,000. Julia made $19,000 in the fifth month. Once again, this is not just saying what happened in the fifth month. This is saying, we have the
$3,000 from the first month and then we have the additional from month between month one and month
five, so that's not that. So this better be our choice. By the end of the fifth
month, Julia had made a total of $19,000. Yes that is correct. She made $3,000 in month
one and then as we go between month one to
the end of month five, to the end of the fifth
month, she has made a total of $19,000. Let's do another one of these. So here we're told the function k of t gives the amount of ketchup in kilograms produced in a sauce
factory by time in hours on a given day. So this is really quantity
is a function of time, it isn't rate, what does
the definite integral from zero to four of k
prime of t dt represent? Once again, pause the video and see if you can work through it. Well k of t is the amount of ketchup as function of time. So k prime of t, that's going to be the rate at which our amount of ketchup is changing a function of time. But once again, when
you're taking the area under the rate curve, that
tells you the net change in the original quantity
in the amount of ketchup. And as the net change between
time zero and time four, so let's see which of these
choices match up to that. The average rate of change of the ketchup production over the first four hours. No, that does not tell us
the average rate of change, there's other ways to calculate that the time it takes to
produce four kilograms of ketchup, so does this
represent the time it takes? To represent four kilograms of ketchup. No, this four is a time right over here. This is gonna tell you how much ketchup gets produced from time zero to time four. The instantaneous rate of
production at t equals four. Now this would be k prime of four, that's not what this integral represents. The amount of ketchup produced over the first four hours, yep
that is exactly right.