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2026 AP Calculus AB FRQ Question 1: Bird Migration Rates

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2026.

Question 1

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Bird Migration Rates

Rate & Data from Tables

Hard
Male birds of a certain species arrive at a nesting area over a thirty-day period. The rate at which the male birds arrive at time tt days is modeled by a differentiable function MM, where M(t)M(t) is measured in number of birds per day. Selected values of M(t)M(t) are shown in the table.

Values of M(t)

t (days)051015202530
M(t) (birds per day)27166520
Part AEasy2 points
Approximate M′(7.5)M'(7.5) using the average rate of change of MM over the interval 5≤t≤105 \leq t \leq 10. Show the work that leads to your answer, and indicate units of measure.

Answer

M′(7.5)≈16−710−5=95=1.8M'(7.5) \approx \dfrac{16 - 7}{10 - 5} = \dfrac{9}{5} = 1.8 birds per day per day
Full Solution & Work

Set up the difference quotient

Since t=7.5t = 7.5 is the midpoint of [5,10][5, 10], approximate M′(7.5)M'(7.5) by the average rate of change over that interval:
M′(7.5)≈M(10)−M(5)10−5=16−75=95=1.8M'(7.5) \approx \frac{M(10) - M(5)}{10 - 5} = \frac{16 - 7}{5} = \frac{9}{5} = 1.8

State the units

The units of M′(t)M'(t) are birds/dayday=birds/day2\dfrac{\text{birds/day}}{\text{day}} = \text{birds/day}^2 (birds per day per day).

AP Scoring — 2 Points

**P1**: Correct expression M(10)−M(5)10−5\frac{M(10)-M(5)}{10-5} with numerical answer 1.8.
**P2**: Correct units — birds per day per day (or birds/day²). Writing only 'birds/day' loses P2.
Part B-iMedium2 points
Use a midpoint Riemann sum with the three subintervals [0,10][0,10], [10,20][10,20], and [20,30][20,30] to approximate ∫030M(t) dt\int_0^{30} M(t)\,dt. Show the work that leads to your answer.

Answer

∫030M(t) dt≈10(7)+10(6)+10(2)=70+60+20=150\int_0^{30} M(t)\,dt \approx 10(7)+10(6)+10(2)=70+60+20=150 birds
Full Solution & Work

Identify the midpoints

The three subintervals are [0,10][0,10], [10,20][10,20], [20,30][20,30], each with width Δt=10\Delta t = 10. The midpoints are t=5t = 5, t=15t = 15, and t=25t = 25.

Evaluate M at each midpoint

From the table: M(5)=7M(5) = 7, M(15)=6M(15) = 6, M(25)=2M(25) = 2.

Compute the Riemann sum

∫030M(t) dt≈10⋅M(5)+10⋅M(15)+10⋅M(25)\int_0^{30} M(t)\,dt \approx 10 \cdot M(5) + 10 \cdot M(15) + 10 \cdot M(25)

=10(7)+10(6)+10(2)=70+60+20=150 birds= 10(7) + 10(6) + 10(2) = 70 + 60 + 20 = 150 \text{ birds}

AP Scoring — 2 Points

**P1**: Correct setup using midpoints t=5,15,25t=5,15,25 with Δt=10\Delta t = 10.
**P2**: Correct final answer of 150.

Common mistake: A common error is using left or right endpoints instead of midpoints — the problem explicitly says 'midpoint Riemann sum.'

Part B-iiEasy1 point
Interpret the meaning of ∫030M(t) dt\int_0^{30} M(t)\,dt in the context of the problem.

Answer

∫030M(t) dt\int_0^{30} M(t)\,dt is the total number of male birds that arrive at the nesting area over the 30-day period.
Full Solution & Work

Interpret rate × time = total

Since M(t)M(t) is a rate in birds/day, integrating over [0,30][0,30] days gives the total accumulation — the total number of male birds that arrive at the nesting area from t=0t = 0 to t=30t = 30 days.

AP Scoring — 1 Point

**P1**: Must reference 'total number of male birds' arriving over 30 days. Omitting 'male' or the time frame loses credit.
Part CMedium3 points
The rate at which female birds of the same species arrive at the same nesting area, in birds per day, is modeled by the function F defined as follows.
F(t)={00≤t<1518+16sin⁡ ⁣(π20(t+15))15≤t≤45F(t) = \begin{cases} 0 & 0 \leq t < 15 \\ 18 + 16\sin\!\left(\dfrac{\pi}{20}(t+15)\right) & 15 \leq t \leq 45 \end{cases}.
How many female birds of this species arrive at the nesting area from t=15t = 15 to t=45t = 45? Show the setup for your calculations, and round your answer to the nearest integer.

Answer

Approximately 540+320π≈642\mathbf{540 + \frac{320}{\pi} \approx 642} female birds
Full Solution & Work

Set up the integral

Total birds=∫1545F(t) dt=∫1545[18+16sin⁡ ⁣(π20(t+15))]dt\text{Total birds} = \int_{15}^{45} F(t)\,dt = \int_{15}^{45} \left[18 + 16\sin\!\left(\frac{\pi}{20}(t+15)\right)\right] dt

Evaluate the integral (Part A — calculator permitted)

Using a graphing calculator:
∫1545F(t) dt≈641.859≈642 female birds\int_{15}^{45} F(t)\,dt \approx 641.859 \approx 642 \text{ female birds}

Analytical verification (optional)

Let u=π20(t+15)u = \frac{\pi}{20}(t+15), du=π20dtdu = \frac{\pi}{20}dt:
=[18t−320πcos⁡ ⁣(π20(t+15))]1545= \left[18t - \frac{320}{\pi}\cos\!\left(\frac{\pi}{20}(t+15)\right)\right]_{15}^{45}

=18(30)−320π[cos⁡(3π)−cos⁡(3π2)]=540+320π(1)=540+320π≈641.86= 18(30) - \frac{320}{\pi}[\cos(3\pi) - \cos(\frac{3\pi}{2})] = 540 + \frac{320}{\pi}(1) = 540 + \frac{320}{\pi} \approx 641.86

*(Use calculator for full accuracy.)*

AP Scoring — 3 Points

P1: Correct integral setup ∫1545F(t) dt\int_{15}^{45} F(t)\,dt.
P2: Correct antiderivative or clear calculator evaluation.
P3: Correct answer rounded to nearest integer (642).
Part DHard3 points
On the interval 15<t<3015 < t < 30, the difference in the rates at which male and female birds of this species arrive at the nesting area can be modeled by the differentiable function D(t)=M(t)−F(t)D(t) = M(t) - F(t). where F is the function defined in part C. Is there a time t in the interval tt in 15<t<2015 < t < 20 when D(t)=0D(t) = 0? Justify your answer.

Answer

Yes, by the Intermediate Value Theorem, there exists a t∈(15,20)t \in (15, 20) such that D(t)=0D(t) = 0.
Full Solution & Work

Compute D at the endpoints

At t=15t = 15: M(15)=6M(15) = 6, F(15)=18+16sin⁡(π20⋅30)=18+16sin⁡(3π2)=18−16=2F(15) = 18 + 16\sin(\frac{\pi}{20}\cdot 30) = 18 + 16\sin(\frac{3\pi}{2}) = 18 - 16 = 2, so D(15)=6−2=4>0D(15) = 6 - 2 = 4 > 0.

At t=20t = 20: M(20)=5M(20) = 5, F(20)=18+16sin⁡(π20⋅35)=18+16sin⁡(7π4)=18−82≈18−11.314=6.686F(20) = 18 + 16\sin(\frac{\pi}{20}\cdot 35) = 18 + 16\sin(\frac{7\pi}{4}) = 18 - 8\sqrt{2} \approx 18 - 11.314 = 6.686, so D(20)=5−6.686≈−1.686<0D(20) = 5 - 6.686 \approx -1.686 < 0.

Apply the Intermediate Value Theorem

D(t)=M(t)−F(t)D(t) = M(t) - F(t) is differentiable on 15<t<3015 < t < 30 (given), so DD is continuous on [15,20][15, 20]. Since D(15)>0D(15) > 0 and D(20)<0D(20) < 0, and 00 is between these values, by the **IVT** there exists at least one t∈(15,20)t \in (15, 20) such that D(t)=0D(t) = 0.

AP Scoring — 3 Points

**P1**: Correct values for D(15)>0D(15) > 0 and D(20)<0D(20) < 0 (or equivalent sign analysis).
**P2**: States DD is continuous on [15,20][15,20] with justification (differentiable → continuous).
**P3**: Explicitly cites the IVT and states the conclusion.

Common mistake: Many students forget to verify continuity or cite IVT by name. Both are required for full credit.

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