Worked examples: interpreting definite integrals in context | AP Calculus AB | Khan Academy

Julia's revenue is r of t thousand dollars per month, where t is the month of the year. Julia had made three thousand dollars in the first month of the year. What does three plus the definite integral from one to five of r of t dt equals 19 mean? We have some choices, so like always, pause the video and see if you can work through it.

All right, now let's work through this together. They tell us that she made three thousand dollars 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 1 to 5 of r of t dt. This is the area under this rate curve. r of t is the rate at which 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. 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 nineteen thousand dollars 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 3 plus this is equal to 19.

If it said Julia made an additional sixteen thousand, well, that would make sense because you could subtract 3 from both sides, and you'd get that result. But that's not what they're saying. Julia made an average of nineteen thousand dollars 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 nineteen thousand dollars, not the average per month is nineteen thousand.

Julia made nineteen thousand dollars in the fifth month. Once again, this is not just saying what happened in the fifth month. This is saying we have the three thousand dollars from the first month, and then we have the additional from month between months 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 nineteen thousand dollars. Yes, that is correct. She made three thousand 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 nineteen thousand dollars.

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 as a function of time; it isn't rate. What does the definite integral from 0 to 4 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 a function of time, so k prime of t—that's going to be the rate at which our amount of ketchup is changing as 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 it's the net change between time 0 and time 4.

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 are other ways to calculate that.

The time it takes to produce four kilograms of ketchup? No, this represents four kilograms of ketchup. No, this four is a time right over here; this is going to tell you how much ketchup gets produced from time 0 to time 4. The instantaneous rate of production at t equals 4? No, this would be k prime of 4; that's not what this integral represents.

The amount of ketchup produced over the first four hours? Yup, that is exactly right.