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# 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

the population of a town grows at a rate of R of T is equal to 300 times e to the zero point three T people per year where time T where T is time in years at time T is equal to two the town's population is 1200 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 valve from one time to another time what you could do is you could take the into 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 to two times seven so this is we could say let me just write this this is the change and 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 know that 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 of 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 twelve hundred people so if you want the population at T is equal to seven it's going to be twelve hundred plus how whatever the change in population if you take the integral of the rate function you are and this is the rate of population this integral is going to give you the change in population from time T equals to two T equals seven so we can see clearly that is choice D right over here these other choices we could look at them really quick choice B is just a change in population that if assuming that this is and this is actually increasing so this would tell us how much does the population increase from T to from T equals to 2 T equals 7 so that's not what we want we want what the actual population is this is how much the population increases from time 0 to time 7 now you see 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 0 but you can't assume that maybe the town got settled by 10 people or by a thousand 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 times 7 minus the rate of change of the rate at time 2 so I would rule that one out as well let's do one more of these so here we have the depth of water and a tank is changing at a rate of R of T is equal to zero point three t 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 waters depth during the fourth minute all right 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 want to take the integral of the rate function and we just have to think about the bounds and all the choices here 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 could think about what the fourth minute looks like or actually let me just draw the whole thing so let's say this is our of T right over there you could say Y is equal to R of T and this is this is T and let's see the first minute is 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 and this actually looks just looks like a straight up a linear rate function looks 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 want to do is the expression that gives us this area right over here under the rate curve well this is our lower bound is 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 that 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 going to 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 in that in in because it's essentially at an instant there's no time that passes from three to three so you could rule all of these out
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