Free Response · Estimating derivatives and integrals from tabular data

Rate & Data from Tables AP Calculus AB FRQ: Explained with Practice & Scoring Tips

Given a table of values, you will estimate instantaneous rates of change, approximate definite integrals using Riemann sums, and interpret results in context. Three core theorems act as your navigation system — knowing when to apply each one is the key to full credit.

Intermediate Value Theorem (IVT)If $f$ is continuous on $[a, b]$ and $k$ is between $f(a)$ and $f(b)$ , then there exists $c \in (a, b)$ such that $f(c) = k$ .
Mean Value Theorem (MVT)If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then there exists $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$ .
Fundamental Theorem of Calculus (FTC) $\int_a^b f'(t)\,dt = f(b) - f(a)$
Trapezoidal Rule $\int_a^b f(x)\,dx \approx \sum \frac{f(x_i) + f(x_{i+1})}{2} \cdot \Delta x$
Average Value of a Function $f_{\text{avg}} = \frac{1}{b - a} \int_a^b f(x)\,dx$

Step-by-Step SOP

1
Step 1 — Check intervals and units
Before calculating anything, check whether the $t$ -values are equally spaced (they usually are not on AP exams). Then identify the units: if $R(t)$ is in words per minute, the integral gives total words and the derivative gives words per minute squared.
2
Step 2 — Write out the full setup
Even if the arithmetic is simple, always write the expression first — for example $\frac{R(2) - R(0)}{2 - 0}$ . On the AP exam, a correct answer with no supporting work earns zero points.
3
Step 3 — Substitute and compute
Plug the table values into the expression you set up. For a Left Riemann Sum, use the left endpoint of each subinterval. For the Trapezoidal Rule, average adjacent pairs and multiply by each subinterval width.
4
Step 4 — Interpret with units
Whenever the prompt says 'explain' or 'interpret,' your answer must include: the time interval, the numerical value, the correct units, and a description of what the quantity represents.
5
Step 5 — Choose the right theorem for justification
If the question asks whether a specific function value $f(c)$ must equal some $k$ , cite IVT. If it asks whether a specific derivative value $f'(c)$ must exist, cite MVT. Always verify the theorem's conditions before stating the conclusion.

Practice Questions

1 2 3

Question 1

2025 AP Calculus AB Exam — FRQ 3

Hard

A student starts reading a book at time

t = 0

minutes and continues reading for the next 10 minutes. The rate at which the student reads is modeled by the differentiable function

R

, where

R(t)

is measured in words per minute. Selected values of

R(t)

are given in the table shown.

Values of R(t)

t (minutes)	0	2	8	10
R(t) (words per minute)	90	100	150	162

APart A

Easy

Approximate

R'(1)

using the average rate of change of

R

over the interval

0 \leq t \leq 2

. Show the work that leads to your answer. Indicate units of measure.

Need a Hint?

The average rate of change over

[0, 2]

approximates

R'(1)

because

t = 1

is the midpoint. Use

\frac{R(2) - R(0)}{2 - 0}

Show Solution

Set up the difference quotient

R'(1) \approx \frac{R(2) - R(0)}{2 - 0} = \frac{100 - 90}{2} = \frac{10}{2} = 5

State the units

The units of

R'

are

\frac{\text{words/min}}{\text{min}} = \text{words/min}^2

.

Therefore,

R'(1) \approx 5

words per minute per minute.

AP Scoring Note — P1 + P2

**P1**: Present the answer with the supporting expression

\frac{100-90}{2}

. The expression alone (without the numerical answer) is not sufficient.

**P2**: State correct units — words/min² or 'words per minute per minute.' Writing only 'words/min' loses P2.

BPart B

Medium

Must there be a value

c

, for

0 < c < 10

, such that

R(c) = 155

? Justify your answer.

Need a Hint?

This is an IVT question. Check: (1) Is

R

continuous on

[0,10]

? Why? (2) Does 155 lie strictly between two known values of

R

Show Solution

Establish continuity with justification

R

is differentiable on

[0, 10]

, and differentiability implies continuity. Therefore

R

is continuous on

[0, 10]

Verify the bounding values

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

Since 155 lies strictly between

R(0)

and

R(10)

, the hypothesis of the Intermediate Value Theorem is satisfied.

State the conclusion

By the Intermediate Value Theorem, there must exist at least one value

c

with

0 < c < 10

such that

R(c) = 155

. **Yes**, such a value must exist.

AP Scoring Note — P3 + P4

**P3**: Must state '

R

is differentiable, therefore continuous.' Writing '

R

is continuous' alone — without the reason — does NOT earn P3.

**P4**: Requires all three: continuity stated,

90 < 155 < 162

shown (or any pair bracketing 155), and the conclusion that such a

c

exists.

CPart C

Medium

Use a trapezoidal sum with the three subintervals indicated by the data in the table to approximate

\int_0^{10} R(t)\,dt

. Show the work that leads to your answer.

Need a Hint?

The three subintervals are

[0,2]

[2,8]

[8,10]

with widths 2, 6, 2. Apply

\frac{1}{2}(\Delta t)(f_L + f_R)

to each subinterval.

Show Solution

Identify subintervals and widths

[0,2]

: width

= 2

[2,8]

: width

= 6

[8,10]

: width

= 2

Write the trapezoidal sum

\int_0^{10} R(t)\,dt \approx \frac{R(0)+R(2)}{2}\cdot 2 + \frac{R(2)+R(8)}{2}\cdot 6 + \frac{R(8)+R(10)}{2}\cdot 2

= \frac{90+100}{2}\cdot 2 + \frac{100+150}{2}\cdot 6 + \frac{150+162}{2}\cdot 2

Evaluate

= (95)(2) + (125)(6) + (156)(2)

= 190 + 750 + 312 = 1252

AP Scoring Note — P5 + P6

**P5**: The form must show three terms, each with the

\frac{1}{2}

factor and the correct width. At least 5 of the 6 factors must be correct to earn P5.

**P6**: Requires P5 earned and all six factors correct. Arithmetic errors after a correct setup do not forfeit P6.

DPart D

Hard

A teacher also starts reading at time

t = 0

and continues for 10 minutes. The teacher's reading rate is modeled by

W(t) = -\frac{3}{10}t^2 + 8t + 100

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.

Need a Hint?

Total words read =

\int_0^{10} W(t)\,dt

. Find the antiderivative of

-\frac{3}{10}t^2 + 8t + 100

, then evaluate from 0 to 10.

Show Solution

Set up the definite integral

Total words

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

Find the antiderivative

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

Evaluate from 0 to 10

= \left(-\frac{1}{10}(10)^3 + 4(10)^2 + 100(10)\right) - (0)

= (-100 + 400 + 1000) = 1300

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

AP Scoring Note — P7 + P8 + P9

**P7**: Earned by writing

\int_0^{10} W(t)\,dt

(with or without

dt

).

**P8**: Requires the correct antiderivative

-\frac{1}{10}t^3 + 4t^2 + 100t

.

**P9**: Requires P8 earned. Once you write the correct evaluated expression, subsequent arithmetic errors do not forfeit P9.

Question 2

2024 AP Calculus AB Exam — FRQ 1

Medium

The temperature of coffee in a cup at time

t

minutes is modeled by a decreasing differentiable function

C

, where

C(t)

is measured in degrees Celsius. For

0 \leq t \leq 12

, selected values of

C(t)

are given in the table shown.

Temperature Values of Coffee

t (minutes)	0	3	7	12
C(t) (degrees Celsius)	100	85	69	55

APart A

Easy

Approximate

C'(5)

using the average rate of change of

C

over the interval

3 \leq t \leq 7

. Show the work that leads to your answer and include units of measure.

Need a Hint?

The value

t = 5

is in the interval

[3, 7]

. Use the average rate of change formula:

\frac{C(7) - C(3)}{7 - 3}

Show Solution

Set up the difference quotient

C'(5) \approx \frac{C(7) - C(3)}{7 - 3} = \frac{69 - 85}{4} = \frac{-16}{4} = -4

State the units

The units of

C

are degrees Celsius and

t

is in minutes. Therefore, the units for

C'

are degrees Celsius per minute.

C'(5) \approx -4

degrees Celsius per minute.

AP Scoring Note — P1 + P2

**P1**: Requires a difference and a quotient (e.g.,

\frac{69-85}{7-3}

). A numerical approximation without supporting work does not earn this point.

**P2**: Units of 'degrees Celsius per minute' must be attached to a numerical value to earn this point.

BPart B

Medium

Use a left Riemann sum with the three subintervals indicated by the data in the table to approximate the value of

\int_{0}^{12} C(t) \, dt

. Interpret the meaning of

\frac{1}{12} \int_{0}^{12} C(t) \, dt

in the context of the problem.

Need a Hint?

A left Riemann sum uses the left-hand endpoint of each subinterval

[t_i, t_{i+1}]

for the height. The average value formula is

\frac{1}{b-a} \int_a^b f(t) \, dt

Show Solution

Calculate the left Riemann sum

\int_{0}^{12} C(t) \, dt \approx (3 - 0) \cdot C(0) + (7 - 3) \cdot C(3) + (12 - 7) \cdot C(7)

= 3(100) + 4(85) + 5(69) = 300 + 340 + 345 = 985

Interpret the meaning

The expression

\frac{1}{12} \int_{0}^{12} C(t) \, dt

represents the **average temperature** of the coffee, in degrees Celsius, over the time interval from

t = 0

t = 12

minutes.

AP Scoring Note — P3 + P4 + P5

**P3**: Form of the left Riemann sum. At least five of the six factors must be correct.

**P4**: Correct numerical approximation (985) pulled from the table.

**P5**: Interpretation must include both 'average temperature' and the specific time interval.

CPart C

Hard

For

12 \leq t \leq 20

, the rate of change of the temperature of the coffee is modeled by

C'(t) = \frac{-24.55e^{0.01t}}{t}

. Find the temperature of the coffee at time

t = 20

. Show the setup for your calculations.

Need a Hint?

Use the Fundamental Theorem of Calculus:

C(20) = C(12) + \int_{12}^{20} C'(t) \, dt

. Note that

C(12) = 55

from the table.

Show Solution

Set up the integral expression

C(20) = C(12) + \int_{12}^{20} C'(t) \, dt = 55 + \int_{12}^{20} \frac{-24.55e^{0.01t}}{t} \, dt

Evaluate using a calculator

Using a graphing calculator:

\int_{12}^{20} C'(t) \, dt \approx -14.670812

C(20) \approx 55 - 14.670812 = 40.329

The temperature of the coffee at

t = 20

is **40.329 degrees Celsius**.

AP Scoring Note — P6 + P7 + P8

**P6**: Correct definite integral setup.

**P7**: Use of the initial condition

C(12) = 55

.

**P8**: Final answer correct to three decimal places. No points awarded for 40.329 without supporting work.

DPart D

Medium

For the model defined in part (C),

C''(t) = \frac{0.2455e^{0.01t}(100-t)}{t^2}

. For

12 < t < 20

, determine whether the temperature of the coffee is changing at a decreasing rate or an increasing rate. Give a reason for your answer.

Need a Hint?

The 'rate of change of temperature' is

C'(t)

. To see if

C'(t)

is increasing or decreasing, look at the sign of its derivative,

C''(t)

Show Solution

Analyze the sign of the second derivative

On the interval

12 < t < 20

:
1.

e^{0.01t}

is always positive.
2.

t^2

is always positive.
3.

(100 - t)

is positive because

t < 20

.

Since all components are positive,

C''(t) > 0

for

12 < t < 20

State the conclusion

Because

C''(t) > 0

on the interval

12 < t < 20

, the rate of change of the temperature (

C'(t)

) is **increasing**.

AP Scoring Note — P9

**P9**: Answer with a reason referencing the sign of the second derivative. Note: Using ambiguous pronouns like 'It is positive' or basing the reason on only a single point (e.g.,

C''(16)

) will not earn this point.

Question 3

2023 AP Calculus AB Exam — FRQ 1

Hard

A customer at a gas station is pumping gasoline into a gas tank. The rate of flow of gasoline is modeled by a differentiable function

f

, where

f(t)

is measured in gallons per second and

t

is measured in seconds since pumping began. Selected values of

f(t)

are given in the table shown.

Values of f(t)

t (seconds)	0	60	90	120	135	150
f(t) (gallons per second)	0	0.1	0.15	0.1	0.05	0

APart A

Easy

Using correct units, interpret the meaning of

\int_{60}^{135} f(t)\,dt

in the context of the problem. Use a right Riemann sum with the three subintervals

[60, 90]

[90, 120]

, and

[120, 135]

to approximate the value of

\int_{60}^{135} f(t)\,dt

. Show the work that leads to your answer.

Need a Hint?

For the interpretation: what does integrating a rate give you? State the quantity, units, and time interval. For the Riemann sum: a right sum uses the right endpoint of each subinterval. The subintervals have widths 30, 30, and 15 — they are NOT equal.

Show Solution

Interpret the integral

\int_{60}^{135} f(t)\,dt

represents the total number of gallons of gasoline pumped into the gas tank from time

t = 60

seconds to time

t = 135

seconds.

Set up the right Riemann sum

\int_{60}^{135} f(t)\,dt \approx f(90)\cdot(90-60) + f(120)\cdot(120-90) + f(135)\cdot(135-120)

= (0.15)(30) + (0.1)(30) + (0.05)(15)

Evaluate

= 4.5 + 3.0 + 0.75 = 8.25

gallons

AP Scoring Note — P1 + P2 + P3

**P1 (Interpretation)**: Must mention gallons of gasoline pumped AND the time interval

t = 60

t = 135

. Missing either loses this point.

**P2 (Form)**: At least 5 of the 6 factors in the Riemann sum must be correct. Writing only the products without factors (e.g.,

4.5 + 3.0 + 0.75

) earns P3 but NOT P2.

**P3 (Answer)**: Requires P2 to be earned. Any error in the Riemann sum forfeits P3.

BPart B

Medium

Must there exist a value of

c

, for

60 < c < 120

, such that

f'(c) = 0

? Justify your answer.

Need a Hint?

This is an MVT (or Rolle's Theorem) question — NOT IVT. You need

f'(c) = 0

, which is a derivative condition. Check whether

f(60) = f(120)

, then cite the appropriate theorem.

Show Solution

Establish continuity and differentiability

f

is differentiable on

(0, 150)

, which implies

f

is continuous on

[60, 120]

Check the endpoint values

f(60) = 0.1

and

f(120) = 0.1

, so

f(120) - f(60) = 0.1 - 0.1 = 0

.

The average rate of change of

f

[60, 120]

equals

\frac{f(120) - f(60)}{120 - 60} = \frac{0}{60} = 0

Apply the Mean Value Theorem

By the Mean Value Theorem (or Rolle's Theorem), there must exist at least one value

c

with

60 < c < 120

such that

f'(c) = 0

. **Yes**, such a value must exist.

AP Scoring Note — P1 + P2

**P1**: Must present

f(120) - f(60) = 0

, or equivalently state

f(60) = f(120)

, or show

0.1 - 0.1 = 0

as a numerator. This is required before any theorem can apply.

**P2**: Requires P1, PLUS stating

f

is continuous because

f

is differentiable, PLUS answering yes. You may cite MVT or Rolle's Theorem — but citing IVT here earns zero for P2.

CPart C

Medium

The rate of flow of gasoline can also be modeled by

g(t) = \frac{t}{500}\cos\!\left(\left(\frac{t}{120}\right)^2\right)

for

0 \leq t \leq 150

. Using this model, find the average rate of flow of gasoline over the time interval

0 \leq t \leq 150

. Show the setup for your calculations.

Need a Hint?

Use the average value formula:

f_{\text{avg}} = \frac{1}{b-a}\int_a^b g(t)\,dt

. You need a calculator to evaluate the integral — show the setup first before writing the numerical answer.

Show Solution

Write the average value formula

\text{Average rate} = \frac{1}{150 - 0}\int_0^{150} g(t)\,dt = \frac{1}{150}\int_0^{150} \frac{t}{500}\cos\!\left(\left(\frac{t}{120}\right)^2\right)dt

Evaluate with a calculator

= \frac{1}{150} \cdot 14.3995\ldots \approx 0.0960

gallons per second

AP Scoring Note — P1 + P2

**P1**: Earned by writing the average value formula

\frac{1}{150-0}\int_0^{150} g(t)\,dt

with the correct integrand. The formula may be shown in one or two steps.

**P2**: The correct answer

\approx 0.096

(or

0.095

). Writing only

\int_0^{150} g(t)\,dt = 0.096

without the

\frac{1}{150}

factor earns neither point.

DPart D

Medium

Using the model

g

defined in part (C), find the value of

g'(140)

. Interpret the meaning of your answer in the context of the problem.

Need a Hint?

Use your calculator to differentiate

g(t)

t = 140

. For the interpretation:

g'(140)

is the rate of change of a rate — state what is changing, at what rate, and at what time. Do NOT use the words 'acceleration' or 'velocity' (those are for motion problems).

Show Solution

Compute g′(140) with a calculator

g'(140) \approx -0.00491 \approx -0.005

gallons per second per second

Interpret in context

At time

t = 140

seconds, the rate at which gasoline is flowing into the tank is **decreasing** at a rate of approximately

0.005

gallon per second per second.

AP Scoring Note — P1 + P2

**P1**: Earned by the correct numerical value

g'(140) \approx -0.005

(or

-0.004

). The value may appear inside the interpretation sentence.

**P2**: The interpretation must include (1) the rate of flow is changing, (2) at the declared value of

g'(140)

, and (3) at

t = 140

seconds.

⚠ Writing 'decreasing at a rate of

-0.005

' loses P2 — the rate itself is

0.005

; 'decreasing' already conveys the direction. Also, using the words 'acceleration' or 'velocity' here loses P2 since this is not a motion problem.

Timed Practice

Ready to challenge yourself?

Take a 20-minute timed mock exam to test your understanding.

Start Timed Test (20m)

Common Pitfalls

⚠
Unequal subinterval widthsAP tables almost always use unequal spacing (e.g., $t = 0, 2, 8, 10$ ). Never assume a fixed $\Delta x$ — compute each subinterval width individually before multiplying.
⚠
Confusing IVT and MVTUse IVT when the question asks whether $f(c) = k$ for some value $k$ . Use MVT when it asks whether $f'(c)$ equals a specific rate. Citing the wrong theorem loses all justification credit.
⚠
IVT requires continuity — with justificationYou must state that $f$ is continuous AND explain why (e.g., ' $R$ is differentiable, therefore continuous'). Simply writing ' $R$ is continuous' without a reason does not earn P3 on the AP rubric.
⚠
Forgetting the $\frac{1}{b-a}$ factor in average valueThe average value formula is $\frac{1}{b-a}\int_a^b f(x)\,dx$ . Omitting the leading coefficient is one of the most frequently penalized errors.
⚠
Unit errors when differentiating or integratingIntegration removes one level of the denominator unit (words/min × min = words). Differentiation adds one level (words/min ÷ min = words/min²). Always track units explicitly.

On this page

Step 1 — Check intervals and units

Step 2 — Write out the full setup

Step 3 — Substitute and compute

Step 4 — Interpret with units

Step 5 — Choose the right theorem for justification

Practice Questions

Set up the difference quotient

State the units

AP Scoring Note — P1 + P2

Establish continuity with justification

Verify the bounding values

State the conclusion

AP Scoring Note — P3 + P4

Identify subintervals and widths

Write the trapezoidal sum

Evaluate

AP Scoring Note — P5 + P6

Set up the definite integral

Find the antiderivative

Evaluate from 0 to 10

AP Scoring Note — P7 + P8 + P9

Set up the difference quotient

State the units

AP Scoring Note — P1 + P2

Calculate the left Riemann sum

Interpret the meaning

AP Scoring Note — P3 + P4 + P5

Set up the integral expression

Evaluate using a calculator

AP Scoring Note — P6 + P7 + P8

Analyze the sign of the second derivative

State the conclusion

AP Scoring Note — P9

Interpret the integral

Set up the right Riemann sum

Evaluate

AP Scoring Note — P1 + P2 + P3

Establish continuity and differentiability

Check the endpoint values

Apply the Mean Value Theorem

AP Scoring Note — P1 + P2

Write the average value formula

Evaluate with a calculator

AP Scoring Note — P1 + P2

Compute g′(140) with a calculator

Interpret in context

AP Scoring Note — P1 + P2

Ready to challenge yourself?