Analyzing problems involving definite integrals | AP Calculus AB | Khan Academy

The population of a town grows at a rate of ( r(t) = 300 e^{0.3t} ) people per year, where ( t ) is time in years. At time ( t = 2 ), the town's population is 1200 people. What is the town's population at ( t = 7 )?

Which expression can we use to solve the problem?

So they don't want us to actually answer the question; they just want us to set up the expression using some symbols from calculus. So why don't you pause this video and try to think about it?

So let's just remind us what they've given us. They’ve given us the rate function right over here, and so if you want to find a change in population from one time to another time, what you could do is you could take the integral of the rate function from the starting time ( t = 2 ) years to ( t = 7 ) years.

So we're going to take the integral of the rate function, and what this is going to tell us, this is going to tell us the change in population from time 2 to time 7.

So this is, you could say, let me just write this: this is the change (I'll use ( \Delta ) for change) or I'll just, let me just write it out—change in population.

But they don’t want us to—they—we don’t—they’re not asking us for the change in population; they want us to know what is the town's population at ( t = 7 ).

So what you would want is, you would want what your population is at ( t = 2 ) plus the change in population from 2 to 7 to get you your population at 7.

So they tell us the population at time ( t = 2 ); the town's population is 1200 people. So if you want the population at ( t = 7 ), it’s going to be 1200 plus whatever the change in population is.

If you take the integral of the rate function, you are—and this is the rate of population—this integral is going to give you the change in population from time ( t = 2 ) to ( t = 7 ).

So we can see clearly that is choice D right over here. These other choices we could look at them really quick.

Choice B is just the change in population; that's if assuming that this is—and this is actually increasing—so this would tell us how much does the population increase from ( t ) to from ( t = 2 ) to ( t = 7 ).

So that's not what we want. We want what the actual population is. This is how much the population increases from time 0 to time 7.

Now you might say, well wouldn’t that be the town's population? Well, that would be the town's population if they had no people at time 0, but you can’t assume that. Maybe the town got settled by 10 people or by a thousand people or who knows whatever else. So right over there.

And this is taking the derivative of the rate function, which is—it’s actually a little hard to think about what is this. This is the rate of change of the rate at time 7 minus the rate of change of the rate at time 2. So I would rule that one out as well.

Let’s do one more of these.

So here we have the depth of water in a tank is changing at a rate of ( r(t) = 0.3t ) centimeters per minute, where ( t ) is the time in minutes. At time ( t = 0 ), the depth of the water is 35 centimeters. What is the change in the water's depth during the fourth minute?

So let’s pause the video again and see if you can figure this out again or figure out what expression can we use to solve the problem—the problem being what is the change in the water's depth during the fourth minute?

All right, so we’ve just talked about if you’re trying to find the change in a value, you could take the integral of the rate function over the appropriate time.

So we’re talking about during the fourth minute, so we definitely want to take the integral of the rate function, and we just have to think about the bounds.

And all the choices here are taking the integral of the rate function, so really the interesting part is during the fourth minute.

So let me just draw a little number line here, and we can think about what the fourth minute looks like. Or actually, let me just draw the whole thing.

So let’s say this is ( r(t) ) right over there. You could say ( y = r(t) ), and this is ( t ).

And let’s see the first minute goes from 0 to 1, second minute goes from 1 to 2, third minute goes from 2 to 3, fourth minute goes from 3 to 4.

The rate function actually looks just like a straight-up linear rate function, looks something like this.

And so what is the fourth minute? Well, the first minute is this one, second, third, fourth. The fourth minute is going from minute 3 to minute 4.

So what we want to do is the expression that gives us this area right over here under the rate curve—well, the lower bound is going to be 3 and our upper bound is going to be 4.

And so there you have it. It is this first choice. You might have been tempted here if you got a little bit confused—hey, maybe the fourth minute is after we’ve crossed ( t = 4 ), but no, that would be the fifth minute.

This would tell us our change over the first four minutes, not just during the fourth minute.

And then this—well, this is just going to be zero if you’re taking—this is what is the change in the value from 3 to 3? Well, it didn’t change at all in that—in—because it’s essentially at an instant, there's no time that passes from 3 to 3, so you could rule all of these out.