Worked example: problem involving definite integral (algebraic)

- [Instructor] We are told the population of a town grows at a rate of e to the 1.2t power minus two t people per year where t is the number of years. At t equals two years, the town has 1,500 people. First, they ask us approximately by how many people does the population grow between t equals two and t equals five, and then what is the town's population at t equals five years. If we actually figure out this first question, the second question is actually pretty straight-forward. We figure out the amount that it grows, and then add it to what we were at at t equals two, add it to 1,500. 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. And we've then seen in multiple videos now, let me just draw, 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 so this is showing me how my rate of change changes as a function of time. Let's say it's something like this. So 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. And we have seen in previous videos that if you wanna figure out the change in the thing 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. And so roughly that would be the area under the curve over that very, very small change in time. What we really wanna do is find the area under this curve from t equals two to t equals five. And we have seen multiple times in calculus how to express that. So the definite integral from t is equal to two to t is equal to five of this expression of e to the 1.2t minus two t dt, dt. So, 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 to the 1.2t? Well, let me just try to do it over here. So if I am trying to calculate, let me write it as e to the five, or actually 6/5 t. Say, 12/10 is the same thing as 6/5. 6/5 t dt, so this is an indefinite integral, I'm just trying to figure out the antiderivative here. Well, if I had a 6/5 right over here, then use substitution, or sometimes you would say the inverse chain rule, would be very appropriate. Well, we could put a 6/5 there if we write a 5/6 right over here. 5/6 times 6/5, then 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 5/6, this 5/6 right over here. This antiderivative is pretty straight-forward. Since I have the derivative of 6/5 t right over here, I can find the antiderivative with respect to 6/5 t, which is essentially I'm doing u-substitution. If you have to do u-substitution, you would make that right over there u, and then that and that would be your du. But, needless to say, this would be 5/6 times e to the 6/5 t, and if you're thinking about the indefinite integral, you would then have a plus c here, of course. And you could verify that the derivative of this is indeed e to the 1.2t. So this is going to be equal to, so this part right over here, the antiderivative is 5/6 e to the 6/5 t, and then this part right over here, the antiderivative of two t is t squared, so minus t squared. And we are going to evaluate that at five and two and find the difference. So let's evaluate this at when t is equal to five. Well, you are going to, and let me color-code this a little bit. When t is equal to five you get 5/6 times e to the 6/5 times five is e to the 6th minus five squared, so minus 25. And so from that I want to subtract. When we evaluate it at two we get 5/6 e to the 6/5 times two is the same thing as 12/5, or we could say that's 2.4. 2.4 minus four, two squared is four. And so what do we get? Well, there's a couple of ways that we could do this. So then we could write this as, let me write it this way, we could write this, so we have a 5/6 and a 5/6, so we could write this as 5/6 times e to the sixth, e to the sixth, minus e to the 2.4, minus 'cause we distribute that negative sign, e to the 2.4, e to the 2.4 power. And then we have minus 25, let me just get another color to keep track of it. We have minus 25, and then you have minus negative four. So that would be plus four. So that would be minus 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. And so let's see, if we wanna find e to the sixth power, that's 403, okay, now let me figure out, so minus, I would say, 20, 2.4, that looks like a 24, I'll correct it as soon as I get back to that screen. E to that power, e to the 2.4 power. And I get equals, so what's in parentheses is this number right here, so times 5/6. Times five divided by six is equal to that minus 21. Minus 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. So approximately by how many people does the population grow between t equals two and t equals five? 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 equals five. Well at time two at two years we had 1,500 people. And then we grow over this interval by 306, so plus 306. Well, that's gonna get me to, we deserve a little bit of a drum roll, 1,806 people at t equals five years.