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## Word problems involving definite integrals

# Analyzing problems involving definite integrals

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

- [Instructor] The population of a town grows at a rate of r of t is equal to 300 times e to the 0.3t people per year, where time, where t is time in years. At time t is equal to two the town's population is 1,200 people. What is the town's population
at t is equal to seven? 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 is equal to two years, to t is equal to seven years. So we're gonna 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 two to time seven. So this is, we 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,
change in population, population. But they don't want us to, we don't, they don't, they're not asking us for the change in population. They want us to know what
is the town's population if t is equal to seven. So what you would want is, you would want what your
population is at t equals two plus the change in
population from two to seven to get you your population at seven. So they tell us the population
at time t equals two, the town's population is 1,200 people. So if you want the population
at t is equal to seven, it's going to be 1,200 plus how whatever the
change in population is. You take the integral
of the rate function, you are, and this is
the rate of population, this integral is gonna give
you the change in population from time t equals two to t equals seven. So we can see clearly that
is choice D right over here. These other choices, we could
look at 'em really quick. Choice b is just the change in population. Assuming that this, and
this is actually increasing, so this would tell us how much
did the population increase from t two, from t equals
two to t equals seven. So that's not what we want. We want what the actual population is. This is how much the population increases from time zero to time seven. 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 zero, but you can't assume that. Maybe the town got settled by
10 people or by 1,000 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 seven minus the
rate of change of the rate at time two. 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 of t is equal to 0.3t centimeters per minute, where t is the time in minutes. At time t equals zero, 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. Alright 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 wanna take the integral of the rate function. 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
of t right over there, I could say y is equal to r of t, and this is, this is t. And let's see, the first
minute goes from zero to one, second minute goes from one to two, third minute goes from two to three, fourth minute goes from three to four. The rate function, this
actually just looks like a straight up linear rate
function, something like this. And so what is the fourth minute? Well the fourth minute is
the first minute is this one, second, third, fourth. The fourth minute is going from
minute three to minute four. So it's what we wanna do is
the expression that gives us this area right over here
under the rate curve. Well this is our lower
bound's going to be three, and our upper bound is going to be four. 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 the t is equal to four, but no that would be the fifth minute. This would tell us, this would tell us our change
over the first four minutes, not just during the fourth minute. And then this, well this
is just gonna be zero. If you're taking, this is, what is the change in the
value from three to three. Well it didn't change at all, because it's essentially at an instant, there's no time that
passes from three to three. So you can rule all of these out.