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### Course: AP®︎/College Calculus AB > Unit 8

Lesson 3: Using accumulation functions and definite integrals in applied contexts- Area under rate function gives the net change
- Interpreting definite integral as net change
- Worked examples: interpreting definite integrals in context
- Interpreting definite integrals in context
- Analyzing problems involving definite integrals
- Analyzing problems involving definite integrals
- Analyzing problems involving definite integrals
- Worked example: problem involving definite integral (algebraic)
- Problems involving definite integrals (algebraic)

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# Worked example: problem involving definite integral (algebraic)

Using definite integral to solve a word problem about the growth in the population of a town.

## Want to join the conversation?

- but why do we get a fraction for the number of people using calculator(2 votes)
- It may be that in places, the calculator uses decimal approximations of fractions. This is why you should always try to evaluate integrals yourself.(2 votes)

- Since we are given the population t(2)=1500, why wouldn't that be used as the lower limit for the definite integral? It seems that using it to calculate a C would provide a more accurate solution?(2 votes)
- We have the rate of change of the population:

𝑃 '(𝑡) = 𝑒^(1.2𝑡) − 2𝑡

What we could do is find the population 𝑃(𝑡) as the*indefinite*integral

𝑃(𝑡) = ∫𝑃 '(𝑡)𝑑𝑡 = (1∕1.2)𝑒^(1.2𝑡) − 𝑡² + 𝐶

Then, since we know 𝑃(2) = 1500

we can use that as the initial condition and find 𝐶:

𝑃(2) = (1∕1.2)𝑒^2.4 − 4 + 𝐶 = 1500

⇒ 𝐶 = 1504 − (1∕1.2)𝑒^2.4 ≈ 1494.81

Thereby,

𝑃(5) ≈ (1∕1.2)𝑒^6 − 25 + 1494.81 ≈ 1806.00

And the change in population between 𝑡 = 2 and 𝑡 = 5 is

𝑃(5) − 𝑃(2) ≈ 1806.00 − 1500 = 306

– – –

Sal's solution goes back to the fundamental theorem of calculus,

which says:

∫[𝑎, 𝑏] 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎)

Thereby,

∫[2, 5] 𝑃 '(𝑡)𝑑𝑡 = 𝑃(5) − 𝑃(2),

which directly gives us the change in population between 𝑡 = 2 and 𝑡 = 5.(2 votes)

- At6:37, Sal said he would correct it, but he never did.(2 votes)
- Why can't you just plug in 5 back into the r(t) function for the second part of the question?(2 votes)
- From3:30-4:00, Sal uses u-substitution to evaluate the indefinite integral of e^1.2t. At4:15, he plugs this value back into the definite integral problem. Why does he not change the bounds of integration after the u-substitution? i.e. from t = [2, 5] to u = [2.4, 6] ?(1 vote)
- Because he back-substituted to continue working with the t-variable instead.(2 votes)

- Do we round to the greater value when writing the result since we're talking about humans? I think I heard something like that is that true or do we do normal rounding?(1 vote)
- At4:24, Does Sal forget to integrate the e^(1.2t)?, i think it should be (1/2.2)e^(2.2t) or (5/11)e^(11/5).(0 votes)
- You're thinking of integrating polynomials. The integral of e^x is e^x, so the integral of e^(6x/5) is (5/6)e^(6x/5)(3 votes)

- The explanation with substitution seems much more complicated than it needs to be. I solved using (e to the 1.2t power divided by 1.2) minus t squared from t = 2 to t= 5 and came up with the same answer. Did I just get lucky?(1 vote)
- Both you and Sal did the same thing. It's just that in your method, you directly used the results of u-substitution, while Sal explicitly showed how the u-sub would work. Plus, you left 1.2 as is while Sal converted it to 5/6.(1 vote)

## Video transcript

- [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.