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# Worked example: problem involving definite integral (algebraic)

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)

## Video transcript

we are told the population of a town grows at a rate of e to the 1.2 T power minus 2 T people per year where T is the number of years at t equals 2 years the town has 1,500 people so first they ask us approximately by how many people does the population grow between T equals 2 and T equals 5 and then what is the town's population at T equals 5 years and 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 at T equals 2 add it to 1,500 so pause this video and see if you can figure it out so 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 been seen in multiple videos now let me just draw do a quick review of this notion of a rate curve so if 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 so let's say 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 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 and so that would be the roughly that would be the area 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 equals to two T equals five and we have seen multiple times and calculus how to express that so the definite integral from T is equal to 2 to T is equal to 5 of this expression of e to the 1 point 2 t minus 2t 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 one point two T 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 5 or actually six fifths t say 12 tenths is the same thing as six fifths 650 DT so this is an indefinite overall I'm just trying to figure out the antiderivative here well if I had a six fifths right over here then use substitution sometimes you would say the inverse chain rule would be very appropriate well we could put a six fist there if we write a five six right over here five six times six fifths 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 five six this five six right over here this antiderivative is pretty straight forward since I have the derivative of six fifths t right over here I can find the antiderivative with respect to 650 which is essentially I'm doing you substitution if you had to do use substitution you would make that right over there you and then that and that would be your D u but needless to say this would be five six times e to the six fifths 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 to the 1 point 2 T so this is going to be equal to so this part right over here the antiderivative is 5/6 e to the 6 fifths T and then this part right over here the antiderivative of 2 T is T squared so minus T squared and we are going to evaluate that at 5 and 2 and find the difference so let's evaluate this at when T is equal to 5 well you are going to it let me color code this a little bit when T is equal to 5 you get 5 6 times e to the 6 fifths times 5 is e to the 6th minus 5 squared so minus 25 and so from that I want to subtract when we evaluate it at 2 we get 5 6 e to the 6 fifths times 2 is the same thing as 12 fifths or we could say that's two point four two point four 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 that we can write this as let me write it this way we could write this so we have a five six and a five six so we could write this as five six times e to the sixth e to the 6 minus e to the 2.4 - because 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 in another color to keep track of it we have minus 25 and then you have minus negative 4 so that would be plus 4 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 want to find e to the sixth power that's 430 ok so now let me figure out so - I would say twenty two point four it 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 5 divided by 6 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 three hundred six point zero zero so this is approximately three hundred six point zero zero so approximately by how many people does the population grow between T equals two and T equals 5 well by approximately three hundred and six people let me write that down so approximately three hundred and six people and they say what is the town's population at T equals five well at time two at two years we had 1500 people and then we grow over this interval by 306 so plus 306 well that's going to get me to they're a little bit of a drumroll 1806 people at T equals five years
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