This video provides a complete solution walkthrough for the 2026 AP Calculus AB Free Response Question #1, which involves a bird nesting problem. The problem tests four key calculus concepts: (1) approximating derivatives using average rate of change, (2) midpoint Riemann sums for definite integrals, (3) interpreting definite integrals in context to find total accumulation, and (4) applying the Intermediate Value Theorem to prove the existence of a solution. The solution demonstrates how to set up and solve each part systematically, including proper unit analysis and mathematical justification.
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2026 AP Calculus AB FRQ #1 AnswersAdded:
Welcome back to the channel, guys.
Today, the AP Calc AB FRQs were released, and so we are going to be going through each of them. So, this is going to be the first FRQ here. We're told that male birds of a certain species arrive at a nesting area over a 30-day period. The rate at which the male birds arrive in the nesting area at time t days is modeled by differentiable function m. So, this is the rate at which they are arriving, right? And then we're told selected values of m of t are shown in the table. Then, first we want to approximate m prime of 7 and 1/2 using the average rate of change from 5 to 10. So, basically, to find the average rate of change, we're going to do y2 - y1 over x2 - x1, basically rise over run. So, what I would say here is that m prime of 7.5 equals 16 minus 7.
That's y2 and y1 over 10 minus 5. And so, there you'll end up just getting that you have 9 over 5, which equals 1.8. And it says, "Indicate units of measure." So, that would be birds per day, right? That's m of t.
But then, we were also dividing by t, right? Like, if you look at this, the 10 minus 5, that unit is days. So, then it's also per day. Right? So, it would be birds per day per day.
Okay, going on to part B, we want to use a midpoint Riemann sum with the three subintervals 0 to 10, 0 to 20, and 20 to 30. So, 10 to 20, and then we have uh or actually, I should actually write it like this. So, then it'll be we could like break it down the middle, and then also then 10 to 20, and then 20 to 30, right? And so, we need to show the work.
So, the midpoint Riemann sum means that we're going to be using the point in the middle of each interval. And so, what we do is we do that number, so the interval from 0 to 30 of m of t dt, that is going to be equivalent to approximately we'll have that value of seven times the length of this subinterval, which is 10.
Plus, we would have six times the length of that subinterval of 10, plus two, which is the middle here, times the length of that subinterval of 10. And so, if you end up adding those all up, it's 70 + 60 + 20, which is 150 birds.
And then, part two is interpret the meaning of this context. Well, when we take the integral of something, basically if this was birds per day, you can think about this as multiplying out the days. So, if we had the unit being birds over day, you could think about taking this integral as what it does effectively is this dt, you could consider is the day. So, we actually cancel out the units.
And so, all of that is to say that what this represents is just the change in the amount of birds at the nesting site.
So, I'm not going to write it all out, but it would be something along the lines of the change in the amount of birds in the nesting site over the 30-day period.
Now, part C says, "The rate at which female birds of the same species arrive at the same nesting area in birds per day is modeled by this function." Says, "For how many female birds of the species arrive at the nesting site from T equals 15 to T equals 45." Sorry about that. It says, "Show the setup for your calculations and round your answer to the nearest integer." So, for this question, what we need to do is find the integral from 15 to 45 basically, right?
How many species arrive? Well, this F is going to be the rate, and basically, just like what I said in the last question, if we have that our actual function is birds per day, taking the integral will tell us the change in birds. So, we would write out the integral from 15 to 45 of f of t dt. It just says show up the set set up for your calculations, and so that should be enough. If you wanted to, you could write the integral from 15 to 45 of then write out this whole thing, but that's not very necessary. And then I could just go in here, do the integral from 15 to 45 of 18 + 16 sign and actually let me put this in parentheses. + 16 sign of we would have pi over 20 * t + 15 dt.
And let's make sure that this is all good. We should not have this here.
It should be here. And we're going to get this, right? Make sure we're in radians as well. Perfect. And so, this gives us and let's make sure that all of this is perfect. 18 + sign of This should be correct, right?
Okay.
So, then we have and I'm just looking at the this to make sure that the numbers are all good.
Yeah, this should be correct. Okay, so then we have and we want the nearest integer, so I guess it's uh 600 and uh 40 two, right?
Okay, great. So, then part D says, "On the interval 15 to t to 30, the difference in the rates by which male and female birds of the species arrive is m of t minus f of t.
Is there a time t in the interval 15 to t to 20 where d of t equals 0?" So, at this this is like an algebra question, to be honest. So, the m of t from 15 to 20 is going to be like here, right? So, somewhere in between six and five, right?
And then also, we're going to have that um for D of for F of T, this is going to be like on our function. So, what I'm thinking to do is can we set this equal to either six or five, right? We know that there is going to be some point We we're basically going to use the IVT.
So, what we'll do is we'll say, "Okay, what is M What is D of 15?" We'll say, "Okay, we have M of 15 minus F of 15."
So, at that point, what I'm going to do is actually change some of this to be I'll say uh this is just F of T equals F of T equals and I'll just write this out.
Equals all of this, right?
And so, also just to check the previous answer, F of T DT. Okay. So, anyway, we could do Okay, so M of 15 is six minus F of 15, so that equals four.
And then we could do F of 20 and M of 20 is five, and so D of 20 equals -2. So, we know, based on the IVT, that every point in between these two are going to be touched. So, what we would say is something along the lines of We would say something along the lines of and let's make this smaller.
15. Yes, by the definition of the IVT, since M of T and F of T and therefore D of T are continuous on the interval 15 to T to 20 and -2 is less than or like -1.7 is less than like whatever other number I got. So, it's less than let's say is less than zero is less than let's say just four.
There must be some value of T inside of the interval such that D of T equals zero. And that is the full answer to FRQ one.
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