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2025 AP Calculus AB FRQ Question 3: Reading Rates — Table Data

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2025.

Question 3

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Reading Rates — Table Data

Rate & Data from Tables

Hard
A student starts reading a book at time t=0t = 0 minutes and continues reading for the next 10 minutes. The rate at which the student reads is modeled by the differentiable function RR, where R(t)R(t) is measured in words per minute. Selected values of R(t)R(t) are given in the table shown.

Values of R(t)

t (minutes)02810
R(t) (words per minute)90100150162
Part AEasy2 points
Approximate R′(1)R'(1) using the average rate of change of RR over the interval 0≤t≤20 \leq t \leq 2. Show the work that leads to your answer. Indicate units of measure.

Answer

R′(1)≈100−902−0=5R'(1) \approx \dfrac{100-90}{2-0} = 5 words per minute per minute
Full Solution & Work

Set up the difference quotient

R′(1)≈R(2)−R(0)2−0=100−902=102=5R'(1) \approx \frac{R(2)-R(0)}{2-0} = \frac{100-90}{2} = \frac{10}{2} = 5

State the units

The units are words/minuteminute=words per minute per minute\dfrac{\text{words/minute}}{\text{minute}} = \text{words per minute per minute} (or words/min²).

AP Scoring — 2 Points

**P1**: The answer, supported by the difference quotient 100−902−0\frac{100-90}{2-0} using values from the table. Presenting the quotient with no numerical follow-through does not earn this point on its own.
**P2**: Correct units — "words per minute per minute" (or words/minute²), whether or not attached to a numerical value.
Part BMedium2 points
Must there be a value cc, for 0<c<100 < c < 10, such that R(c)=155R(c) = 155? Justify your answer.

Answer

Yes. Since RR is differentiable (hence continuous) and R(0)=90<155<162=R(10)R(0)=90 < 155 < 162=R(10), by the Intermediate Value Theorem there is a c∈(0,10)c \in (0,10) with R(c)=155R(c)=155.
Full Solution & Work

Establish continuity

RR is given to be differentiable on [0,10][0,10], so RR is continuous on [0,10][0,10].

Compare 155 to the endpoint values

R(0)=90<155<162=R(10)R(0) = 90 < 155 < 162 = R(10)

Apply the Intermediate Value Theorem

Since RR is continuous on [0,10][0,10] and 155 lies between R(0)R(0) and R(10)R(10), the **Intermediate Value Theorem** guarantees a value cc, with 0<c<100<c<10, such that R(c)=155R(c)=155.

AP Scoring — 2 Points

**P3**: States that RR is continuous *because* RR is differentiable — simply asserting "RR is continuous" with no justification does not earn this point.
**P4**: Indicates R(0)<155R(0)<155 (or equivalently uses R(2)<155R(2)<155 or R(8)<155R(8)<155) and R(10)>155R(10)>155, and answers "yes." (P4 does not require P3 to be earned.)

Common mistake: You don't have to name the Intermediate Value Theorem to earn credit, but if you do name a theorem, it must be the correct one — citing the Mean Value Theorem here is a common and costly mix-up.

Part CMedium2 points
Use a trapezoidal sum with the three subintervals indicated by the data in the table to approximate the value of ∫010R(t) dt\int_0^{10} R(t)\,dt. Show the work that leads to your answer.

Answer

∫010R(t) dt≈90+1002(2)+100+1502(6)+150+1622(2)=190+750+312=1252\int_0^{10} R(t)\,dt \approx \dfrac{90+100}{2}(2) + \dfrac{100+150}{2}(6) + \dfrac{150+162}{2}(2) = 190+750+312 = 1252
Full Solution & Work

Set up the trapezoidal sum

∫010R(t) dt≈R(0)+R(2)2(2−0)+R(2)+R(8)2(8−2)+R(8)+R(10)2(10−8)\int_0^{10} R(t)\,dt \approx \frac{R(0)+R(2)}{2}(2-0) + \frac{R(2)+R(8)}{2}(8-2) + \frac{R(8)+R(10)}{2}(10-8)

Substitute values and compute

=90+1002(2)+100+1502(6)+150+1622(2)= \frac{90+100}{2}(2) + \frac{100+150}{2}(6) + \frac{150+162}{2}(2)

=190+750+312=1252= 190 + 750 + 312 = 1252

AP Scoring — 2 Points

**P5**: The form of a trapezoidal sum — three terms, each a product of two factors where one factor is a 12\frac{1}{2} (sum of endpoints) ×\times width. At least five of the six numeric factors must be correct.
**P6**: Correct answer 1252, with supporting work. To be eligible for P6, P5 must be earned. A fully correct left or right Riemann sum (1080 or 1424) earns P5 but **not** P6 — it isn't a trapezoidal sum.
Part DMedium3 points
A teacher also starts reading at time t=0t = 0 minutes and continues reading for the next 10 minutes. The rate at which the teacher reads is modeled by the function WW defined by W(t)=−310t2+8t+100W(t) = -\dfrac{3}{10}t^2 + 8t + 100, where W(t)W(t) is measured in words per minute. Based on the model, how many words has the teacher read by the end of the 10 minutes? Show the work that leads to your answer.

Answer

∫010W(t) dt=1300\int_0^{10} W(t)\,dt = 1300 words
Full Solution & Work

Set up the integral

∫010W(t) dt=∫010(−310t2+8t+100)dt\int_0^{10} W(t)\,dt = \int_0^{10}\left(-\frac{3}{10}t^2+8t+100\right)dt

Find the antiderivative

=[−110t3+4t2+100t]010= \left[-\frac{1}{10}t^3 + 4t^2 + 100t\right]_0^{10}

Evaluate

=(−110(1000)+4(100)+100(10))−0=(−100+400+1000)=1300= \left(-\frac{1}{10}(1000) + 4(100) + 100(10)\right) - 0 = (-100+400+1000) = 1300

The teacher has read **1300 words** by the end of the 10 minutes.

AP Scoring — 3 Points

**P7**: Correct integral (definite or indefinite) with integrand W(t)W(t).
**P8**: Correct antiderivative −110t3+4t2+100t-\frac{1}{10}t^3+4t^2+100t.
**P9**: Correct final answer 1300, with P8 required to be eligible.

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