Worked example: problem involving definite integral (algebraic) | AP Calculus AB | Khan Academy

We are told the population of a town grows at a rate of ( e^{1.2t} - 2t ) people per year, where ( t ) is the number of years. At ( t = 2 ) years, the town has fifteen hundred people.

So first, they ask us approximately by how many people does the population grow between ( t = 2 ) and ( t = 5 ).

Then, what is the town's population at ( t = 5 ) years? If we actually figure out this first question, the second question is actually pretty straightforward. We figure out the amount that it grows and then add it to what we were at ( t = 2 ) and add it to fifteen hundred.

So pause this video and see if you can figure it out.

The key here is to appreciate that this right over here is expressing the rate of how fast the population is growing. We have been seeing in multiple videos now, let me just draw and do a quick review of this notion of a rate curve.

So those are my axes, and this is my ( t )-axis, my time axis, and this is showing me how my rate of change changes as a function of time.

So let's say it's something like this. Once again, if I said at this time right over here, this is my rate. This doesn't tell me, for example, what my population is. This tells me what is my rate of change of a population.

We have seen in previous videos that if you want to figure out the change in the thing, that the rate is met, that the rate is the rate of change of, say the change in population, you would find the area under the rate curve between those two appropriate times.

And why does that make sense? Well, imagine a very small change in time right over here. If you have a very small change in time and if you assume that your rate is approximately constant over that very small change in time, well then your change in, let's say we're measuring the rate of change of population here, your accumulation you could say is going to be your rate times your change in time, which would be the area of this rectangle.

So that would be roughly the area under the curve over that very, very small change in time.

So what we really want to do is find the area under this curve from ( t = 2 ) to ( t = 5 ). We have seen multiple times in calculus how to express that, so the definite integral from ( t ) is equal to 2 to ( t ) is equal to 5 of this expression ( e^{1.2t} - 2t , dt ).

If we just evaluate that, that will be the answer to this first question. So what is this going to be? Well, let's actually work it out.

So what is the antiderivative of ( e^{1.2t} )? Well, let me just try to do it over here.

If I'm trying to calculate, let me write it as ( e^{\frac{6}{5}t} ) (since ( 1.2t = \frac{6}{5}t )). This is an indefinite integral I'm just trying to figure out the anti-derivative here.

Well, if I had a ( \frac{6}{5} ) right over here, then ( u )-substitution or sometimes you would say the inverse chain rule would be very appropriate.

We could put a ( \frac{6}{5} ) there if we write a ( \frac{5}{6} ) right over here ( \frac{5}{6} \times \frac{6}{5} ) and we can take constants in and out of the integral like this – scaling constants I should say.

Well now, so this is going to be equal to ( \frac{5}{6} e^{\frac{6}{5} t} ). And if you're thinking about the indefinite integral, you would then have a plus ( c ) here of course, and you can verify that the derivative of this is indeed ( e^{1.2t} ).

So this is going to be equal to so this part right over here, the anti-derivative is ( \frac{5}{6} e^{\frac{6}{5}t} ) and then this part right over here, the anti-derivative of ( 2t ) is ( t^2 ), so minus ( t^2 ).

We are going to evaluate that at 5 and 2 and find the difference.

So let's evaluate this at when ( t = 5 ). Well, you are going to let me color code this a little bit. When ( t = 5 ), you get ( \frac{5}{6} e^{\frac{6}{5} \times 5} - 25 ).

And so from that, I want to subtract when we evaluate it at 2, we get ( \frac{5}{6} e^{\frac{6}{5} \times 2} - 4 ).

What do we get? Well, there's a couple of ways that we could do this.

We could write this as ( \frac{5}{6} (e^{6} - e^{2.4}) - 21 ).

So that would be ( -21 ) and I would need a calculator to figure this out.

So let me do that. Let me get my calculator on this computer, and there we go.

So let's see if we want to find ( e^{6} ); that’s 430. Okay, so then now let me figure out ( e^{2.4} ).

And I get equals so what's in parentheses is this number right here, so times ( \frac{5}{6} ), so ( \frac{5}{6} \times 430 - 21 ) is equal to this.

So if I round to the nearest hundredth, it's going to be approximately 306.00.

So this is approximately 306.00.

Approximately by how many people does the population grow between ( t = 2 ) and ( t = 5 )? Well, by approximately 306 people.

Let me write that down: so approximately 306 people.

And they say what is the town's population at ( t = 5 )?

Well at time ( t = 2 ), at two years we had fifteen hundred people and then we grow over this interval by three hundred six.

So ( 1500 + 306 ) is going to get me to deserve a little bit of a drum roll: 1806 people at ( t = 5 ) years.