2026 AP Calc AB FRQ Answers & Solutions

Question 1

Calculator OK

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

t

days is modeled by a differentiable function

M

, where

M(t)

is measured in number of birds per day. Selected values of

M(t)

are shown in the table.

Values of M(t)

t (days)	0	5	10	15	20	25	30
M(t) (birds per day)	2	7	16	6	5	2	0

Part AEasy2 points

Approximate

M'(7.5)

using the average rate of change of

M

over the interval

5 \leq t \leq 10

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

Answer

M'(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 = 7.5

is the midpoint of

[5, 10]

, approximate

M'(7.5)

by the average rate of change over that interval:

M'(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)

are

\dfrac{\text{birds/day}}{\text{day}} = \text{birds/day}^2

(birds per day per day).

AP Scoring — 2 Points

**P1**: Correct expression

\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]

,

[10,20]

, and

[20,30]

to approximate

\int_0^{30} M(t)\,dt

. Show the work that leads to your answer.

Answer

\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]

,

[10,20]

,

[20,30]

, each with width

\Delta t = 10

. The midpoints are

t = 5

,

t = 15

, and

t = 25

.

Evaluate M at each midpoint

From the table:

M(5) = 7

,

M(15) = 6

,

M(25) = 2

.

Compute the Riemann sum

\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 \text{ birds}

AP Scoring — 2 Points

**P1**: Correct setup using midpoints

t=5,15,25

with

\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

\int_0^{30} M(t)\,dt

in the context of the problem.

Answer

\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)

is a rate in birds/day, integrating over

[0,30]

days gives the total accumulation — the total number of male birds that arrive at the nesting area from

t = 0

to

t = 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) = \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 = 15

to

t = 45

? Show the setup for your calculations, and round your answer to the nearest integer.

Answer

Approximately

\mathbf{540 + \frac{320}{\pi} \approx 642}

female birds

Full Solution & Work

Set up the integral

\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:

\int_{15}^{45} F(t)\,dt \approx 641.859 \approx 642 \text{ female birds}

Analytical verification (optional)

Let

u = \frac{\pi}{20}(t+15)

,

du = \frac{\pi}{20}dt

:

= \left[18t - \frac{320}{\pi}\cos\!\left(\frac{\pi}{20}(t+15)\right)\right]_{15}^{45}

= 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

\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 < 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)

. where F is the function defined in part C. Is there a time t in the interval

t

in

15 < t < 20

when

D(t) = 0

? Justify your answer.

Answer

Yes, by the Intermediate Value Theorem, there exists a

t \in (15, 20)

such that

D(t) = 0

.

Full Solution & Work

Compute D at the endpoints

At

t = 15

:

M(15) = 6

,

F(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 > 0

.

At

t = 20

:

M(20) = 5

,

F(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 \approx -1.686 < 0

.

Apply the Intermediate Value Theorem

D(t) = M(t) - F(t)

is differentiable on

15 < t < 30

(given), so

D

is continuous on

[15, 20]

. Since

D(15) > 0

and

D(20) < 0

, and

0

is between these values, by the **IVT** there exists at least one

t \in (15, 20)

such that

D(t) = 0

.

AP Scoring — 3 Points

**P1**: Correct values for

D(15) > 0

and

D(20) < 0

(or equivalent sign analysis).
**P2**: States

D

is continuous on

[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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Question 2

Calculator OK

Area and Volume Between Curves

Area & Volume

Open this question on its own page

Hard

The function

f

is defined by

f(x) = 1.43^x + 0.57

, and the function

g

is defined by

g(x) = \dfrac{14x+12}{x+12}

. The graphs of

f

and

g

intersect at the points

(1,2)

and

(a,b)

as shown in Figure 1. For

x \geq 0

, the equation

y = g(x)

can be rewritten as

x = h(y) = \dfrac{12y-12}{14-y}

, where

h

is a function of

y

.The graph of

x = h(y)

and the horizontal line

y = 3.5

are shown in Figure 2.

2026 AP Calculus AB FRQ Question 2 Graph — Figure 1.

Part AMedium3 points

Let

R

be the region bounded by the graph of

g

, the

x

-axis, the

y

-axis, and the vertical line

x = 1

as shown in Figure 1. Find the area of region

R

. Show the setup for your calculations.

Answer

\text{Area} = \int_0^1 g(x)\,dx \approx 1.513

Full Solution & Work

Set up the integral

The area of region

R

is found by integrating the function

g(x)

from

x = 0

to

x = 1

:

\text{Area} = \int_0^1 g(x)\,dx = \int_0^1 \frac{14x+12}{x+12}\,dx

Analytical verification

To evaluate the integral manually:

- **Step 1: Simplify the integrand**

\frac{14x+12}{x+12} = \frac{14(x+12) - 156}{x+12} = 14 - \frac{156}{x+12}

- **Step 2: Integrate**

\int \left( 14 - \frac{156}{x+12} \right) dx = 14x - 156\ln|x+12|

- **Step 3: Evaluate from 0 to 1**

(14(1) - 156\ln(13)) - (14(0) - 156\ln(12))

= 14 - 156(\ln(13) - \ln(12)) = 14 - 156\ln\left(\frac{13}{12}\right)

14 - 156(0.08004) \approx 1.513

AP Scoring — 3 Points

• **1 pt**: Correct integral setup

\int_0^1 g(x)\,dx

.
• **1 pt**: Correct antiderivative or appropriate use of calculator.
• **1 pt**: Final answer accurate to three decimal places (1.513).

Part BMedium2 points

Region

R

,described in part A is the base of a solid.For this solid,each cross section perpendicular to the

x

-axis is a rectangle whose height is

\frac{1}{3}

times the length of its base in region

R

. Write, but do not evaluate, an integral expression that gives the volume of the solid.

Answer

V = \int_0^1 \frac{1}{3}[g(x)]^2\,dx = \int_0^1 \frac{1}{3}\left(\frac{14x+12}{x+12}\right)^2 dx

Full Solution & Work

Set up cross-sectional area

At each

x

, the base of the rectangle is

g(x)

(height of region

R

). The height of the rectangle is

\frac{1}{3}g(x)

. So the cross-sectional area is:

A(x) = g(x) \cdot \frac{1}{3}g(x) = \frac{1}{3}[g(x)]^2

Write the volume integral

V = \int_0^1 \frac{1}{3}[g(x)]^2\,dx = \int_0^1 \frac{1}{3}\left(\frac{14x+12}{x+12}\right)^2 dx

AP Scoring — 2 Points

**P1**: Integrand

\frac{1}{3}[g(x)]^2

with correct bounds

[0,1]

.
**P2**: Correct integral expression (do not evaluate).

Part CHard4 points

The shaded region in Figure 1 is bounded by the graphs of

f

and

g

on the interval from

x = 0

to

x = a

. Find the area of the shaded region. Show the setup for your calculations.

Answer

\text{Area} = \int_0^a |g(x) - f(x)|\,dx \approx 0.632

Full Solution & Work

Find the second intersection point a

The graphs of

f

and

g

intersect when

f(x) = g(x)

. Given

f(x) = 1.43^x + 0.57

and

g(x) = \frac{14x+12}{x+12}

, use a graphing calculator to solve for the second intersection point

x = a

in the first quadrant:

a \approx 3.256

Set up the integral

The shaded region consists of two parts because the graphs cross at

x = 1

. The total area is the integral of the absolute difference between the functions from

0

to

a

:

\text{Area} = \int_0^{3.256} |f(x) - g(x)|\,dx

Evaluate using the calculator

To calculate precisely, split the integral at the intersection point

x = 1

:

\text{Area} = \int_0^1 (f(x) - g(x))\,dx + \int_1^{3.256} (g(x) - f(x))\,dx \approx 0.632

AP Scoring — 4 Points

• **1 pt**: Correct value for

a

(3.256).
• **1 pt**: Correct integral setup with absolute value or split at

x=1

.
• **1 pt**: Correct identification of upper/lower functions in each interval.
• **1 pt**: Final numerical answer (0.632).

Part DHard3 points

Let

T

be the region bounded by the graph of

x = h(y)

, the

y

-axis, and the horizontal line

y = 3.5

as shown in Figure 2. Write, but do not evaluate, an integral expression for the volume of the solid when

T

is revolved about the

y

-axis.

Answer

V = \pi \int_1^{3.5} [h(y)]^2\,dy = \pi \int_1^{3.5} \left(\frac{12y-12}{14-y}\right)^2 dy

Full Solution & Work

Identify the bounds in y

Region

T

is bounded below by

y = 1

(since

h(1) = 0

, the curve starts at the

y

-axis when

y=1

) and above by

y = 3.5

.

Apply the disk method about the y-axis

Revolving about the

y

-axis, each disk has radius

x = h(y)

:

V = \pi \int_1^{3.5} [h(y)]^2\,dy = \pi \int_1^{3.5} \left(\frac{12y-12}{14-y}\right)^2 dy

AP Scoring — 3 Points

**P1**: Correct method (disk/washer about

y

-axis).
**P2**: Correct integrand

\pi[h(y)]^2

.
**P3**: Correct bounds

y = 1

to

y = 3.5

.

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Question 3

No Calculator

Cooling Pie — Differential Equations

Differential Equations

Open this question on its own page

Hard

A pie is taken from a hot oven and put on a table. The internal temperature of the pie at time

t

minutes can be modeled by the function

H

that satisfies the differential equation

\dfrac{dH}{dt} = -\dfrac{1}{15}(H - 20)

, where

H(t)

is measured in degrees Celsius and

H(0) = 75

. For

t > 0

, it is known that

20 < H(t) < 75

.

2026 AP Calculus AB FRQ Question 3 Graph

Part AMedium2 points

Explain why the following could not be a slope field for the differential equation

\dfrac{dH}{dt} = -\dfrac{1}{15}(H-20)

.

Answer

The slope field shows positive slopes for

H > 20

, but the differential equation gives

\frac{dH}{dt} < 0

whenever

H > 20

. Therefore the slope field is incorrect.

Full Solution & Work

Analyze the sign of dH/dt

When

H > 20

:

(H - 20) > 0

, so

\dfrac{dH}{dt} = -\dfrac{1}{15}(H-20) < 0

. Slopes must be **negative** for all

H > 20

.

Identify the contradiction

The shown slope field displays positive (upward) slopes for

H > 20

, which contradicts the differential equation. This is why the slope field could not be correct.

AP Scoring — 2 Points

**P1**: States

\frac{dH}{dt} < 0

when

H > 20

.
**P2**: Connects this to the slope field showing positive slopes, hence the contradiction.

Part BEasy2 points

Find the slope of the line tangent to the graph of

H

at time

t = 0

. Show the work that leads to your answer.

Answer

\dfrac{dH}{dt}\Big|_{t=0} = -\dfrac{1}{15}(75 - 20) = -\dfrac{55}{15} = -\dfrac{11}{3}

Full Solution & Work

Substitute H(0) = 75 into the differential equation

\frac{dH}{dt}\bigg|_{t=0} = -\frac{1}{15}(H(0) - 20) = -\frac{1}{15}(75 - 20) = -\frac{55}{15} = -\frac{11}{3}

AP Scoring — 2 Points

**P1**: Substitutes

H = 75

correctly into the ODE.
**P2**: Correct slope

-\frac{11}{3}

(or equivalent decimal

\approx -3.667

).

Part CMedium2 points

It can be shown that

\dfrac{d^2H}{dt^2} = \dfrac{1}{225}(H-20)

. The line tangent to the graph of

H

at time

t = 0

is used to approximate

H(5)

, the internal temperature of the pie at time

t = 5

. Is this approximation an overestimate or an underestimate for the actual value of

H(5)

? Give a reason for your answer.

Answer

Underestimate. Since

\dfrac{d^2H}{dt^2} > 0

for

H > 20

, the graph of

H

is concave up, so the tangent line lies below the curve.

Full Solution & Work

Determine the sign of the second derivative

For

20 < H < 75

:

(H - 20) > 0

, so

\dfrac{d^2H}{dt^2} = \dfrac{(H-20)}{225} > 0

. The graph of

H

is **concave up**.

Conclude overestimate or underestimate

When a function is concave up, the tangent line lies **below** the curve. Therefore the tangent line approximation of

H(5)

is an **underestimate**.

AP Scoring — 2 Points

**P1**: Identifies

\frac{d^2H}{dt^2} > 0

(concave up).
**P2**: States 'underestimate' with concavity reasoning.

Part DHard6 points

Use separation of variables to find an expression for

H(t)

, the particular solution to the given differential equation with initial condition

H(0) = 75

.

Answer

H(t) = 20 + 55e^{-t/15}

Full Solution & Work

Separate variables

\frac{dH}{H - 20} = -\frac{1}{15}\,dt

Integrate both sides

\int \frac{dH}{H-20} = \int -\frac{1}{15}\,dt

\ln|H - 20| = -\frac{t}{15} + C

Exponentiate and solve for H

|H - 20| = e^C \cdot e^{-t/15}

H - 20 = Ae^{-t/15} \quad \text{(where } A = \pm e^C\text{)}

H(t) = 20 + Ae^{-t/15}

Apply the initial condition H(0) = 75

75 = 20 + Ae^{0} \implies A = 55

Write the particular solution

\boxed{H(t) = 20 + 55e^{-t/15}}

Verification: as

t \to \infty

,

H \to 20

✓ (the pie cools to room temperature).

AP Scoring — 6 Points

**P1**: Correct separation of variables.
**P2**: Correct antiderivatives on both sides.
**P3**:

\ln|H-20|

on left side.
**P4**: Correct general solution form

H = 20 + Ae^{-t/15}

.
**P5**: Uses

H(0) = 75

to find

A = 55

.
**P6**: Correct particular solution

H(t) = 20 + 55e^{-t/15}

.

Common mistake: Forgetting the absolute value in $\ln|H-20|$ or dropping the constant of integration are the two most common errors. Both cost points.

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Question 4

No Calculator

Analyzing a Function from its Derivative Graph

Graph Analysis & Curve Sketching

Open this question on its own page

Hard

Let

f

be a twice-differentiable function on

[-4, 4]

with

f(2) = 3

. The graph of

f'

, the derivative of

f

, is shown, passing through the points

(-4, -0.5)

,

(-3, -1)

,

(1, 3)

,

(2, 1.5)

, and

(4, 5)

.

2026 AP Calculus AB FRQ Question 4 Graph

Part AEasy2 points

For

x > 0

, the function g is defined by

g(x) = f(x) - \ln x

. Find

g'(2)

. Show the work that leads to your answer.

Answer

g'(2) = f'(2) - \dfrac{1}{2} = 1.5 - 0.5 = 1

Full Solution & Work

Differentiate g(x)

g'(x) = f'(x) - \frac{1}{x}

Evaluate at x = 2

From the graph of

f'

:

f'(2) = 1.5

.

g'(2) = f'(2) - \frac{1}{2} = 1.5 - 0.5 = 1

AP Scoring — 2 Points

**P1**:

g'(x) = f'(x) - \frac{1}{x}

(correct differentiation).
**P2**: Correct answer

g'(2) = 1

.

Part BMedium2 points

Find all values of

x

on the open interval

(0, 3)

at which the graph of

f

has a point of inflection. Give a reason for your answer.

Answer

f

has a point of inflection at

x = 1

because

f''

changes sign there (from positive to negative on the graph of

f'

).

Full Solution & Work

Recall: inflection points occur where f'' changes sign

Since

f'' = (f')'

, inflection points of

f

occur where

f'

changes from increasing to decreasing (or vice versa), i.e., where

f'

has a local extremum.

Read the graph of f'

On

(0, 3)

:

f'

increases then decreases, with a local maximum at

x = 1

. At

x = 1

,

f'

changes from increasing to decreasing, so

f''

changes from

+

to

-

.

Conclusion

There is exactly one point of inflection:

x = 1

. At

x = 1

,

f''

changes sign from positive to negative.

AP Scoring — 2 Points

**P1**: Identifies

x = 1

.
**P2**: Justification —

f'

changes from increasing to decreasing at

x = 1

, so

f''

changes sign.

Part CMedium2 points

On

[-4, 4]

, on what open intervals, if any, is the graph of

f

both increasing and concave down? Give a reason for your answer.

Answer

f

is both increasing and concave down on the interval

(1, 3)

.

Full Solution & Work

Condition 1: Increasing ($f'(x) > 0$)

The graph of

f

increases when its derivative

f'

is positive. Based on the provided graph,

f'(x) > 0

on the intervals

(-2, 3)

and

(3, 4)

.

Condition 2: Concave Down ($f''(x) < 0$)

The graph of

f

is concave down when

f'' < 0

, which corresponds to where

f'

is decreasing. From the graph,

f'

decreases on the intervals

(-4, -3)

and

(1, 3)

.

Intersection of Conditions

To satisfy both conditions, we find the intersection of

x \in (-2, 3) \cup (3, 4)

and

x \in (-4, -3) \cup (1, 3)

. The only interval that satisfies both is

(1, 3)

.

AP Scoring — 2 Points

P1: Correct interval

(1, 3)

.
P2: Reason — States that

f

is increasing because

f' > 0

and concave down because

f'

is decreasing on this interval.

Part DHard4 points

On

[-4, 4]

, find the value of

x

at which

f

has an absolute minimum and the value of

x

at which

f

has an absolute maximum. Give reasons for your answers.

Answer

Absolute minimum at

x = -2

; absolute maximum at

x = 4

.

Full Solution & Work

Identify Candidate Points

Absolute extrema can only occur at endpoints

x = -4, 4

or critical points where

f'(x) = 0

. From the graph,

f'(x) = 0

at

x = -2

and

x = 3

.

Analyze the Absolute Minimum

On the interval

[-4, -2]

,

f'(x) < 0

, so

f

is decreasing. On

[-2, 4]

,

f'(x) \ge 0

(except at

x=3

where it is 0), so

f

is increasing. Since the function decreases from

x = -4

to

x = -2

and increases from

x = -2

to

x = 4

, the absolute minimum must occur at

x = -2

.

Analyze the Absolute Maximum

Since

f

is increasing on the entire interval

[-2, 4]

, we only need to compare the endpoints

f(-4)

and

f(4)

. Because the area under

f'

from

x = -2

to

x = 4

is much larger than the magnitude of the area from

x = -4

to

x = -2

, the absolute maximum occurs at the right endpoint

x = 4

.

AP Scoring — 4 Points

P1: Identifies

x = -2

and

x = 3

as critical points.
P2: Justifies absolute minimum at

x = -2

using the sign change of

f'

.
P3: Justifies absolute maximum at

x = 4

by noting

f

increases on

[-2, 4]

.
P4: Correct values for both.

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Question 5

No Calculator

Remote-Controlled Car Motion

Particle Motion

Open this question on its own page

Hard

A remote-controlled toy car moves back and forth along a straight path so that its velocity at time

t

is given by the function

v

, where

v(t)

is measured in feet per second and

t

is measured in seconds.

v(t) = \begin{cases} t^4 - 8t^3 + 16t^2 & 0 \leq t \leq 4 \\ 0 & 4 < t < 6 \\ 10\cos\!\left(\frac{\pi}{3}t\right) - 10 & 6 \leq t \leq 12 \end{cases}

where

v(t)

is in ft/sec and

t

in seconds.
The graph of

v(t)

is shown.

2026 AP Calculus AB FRQ Question 5 Graph

Part AEasy2 points

Find the acceleration of the car at time

t = 1

second. Show the work that leads to your answer.

Answer

a(1) = v'(1) = 4(1)^3 - 24(1)^2 + 32(1) = 4 - 24 + 32 = 12

ft/sec²

Full Solution & Work

Differentiate v(t) on [0, 4]

For

0 \leq t \leq 4

:

v(t) = t^4 - 8t^3 + 16t^2

a(t) = v'(t) = 4t^3 - 24t^2 + 32t

Evaluate at t = 1

a(1) = 4(1)^3 - 24(1)^2 + 32(1) = 4 - 24 + 32 = 12 \text{ ft/sec}^2

AP Scoring — 2 Points

**P1**: Correct derivative

4t^3 - 24t^2 + 32t

.
**P2**: Correct answer

a(1) = 12

ft/sec².

Part BEasy1 point

Is the car speeding up or slowing down at time

t = 1

second? Give a reason for your answer.

Answer

The car is speeding up at

t = 1

because

v(1) > 0

and

a(1) > 0

(same sign).

Full Solution & Work

Find v(1)

v(1) = 1 - 8 + 16 = 9 > 0

Compare signs of v and a

Since

v(1) = 9 > 0

and

a(1) = 12 > 0

, velocity and acceleration have the **same sign**. The car is **speeding up**.

AP Scoring — 1 Point

**P1**: 'Speeding up' with justification that

v(1)

and

a(1)

have the same sign.

Part CMedium3 points

Find the distance, in feet, that the car traveled over

0 \leq t \leq 4

seconds.Show the work that leads to your answer.

Answer

\text{Distance} = \int_0^4 |v(t)|\,dt = \int_0^4 v(t)\,dt = \left[\frac{t^5}{5} - 2t^4 + \frac{16t^3}{3}\right]_0^4 = \frac{1024}{5} - 512 + \frac{1024}{3} = \frac{512}{15} \approx 34.133

ft

Full Solution & Work

Check if v changes sign on [0, 4]

Factor:

v(t) = t^2(t^2 - 8t + 16) = t^2(t-4)^2 \geq 0

for all

t

. So

|v(t)| = v(t)

on

[0,4]

— no sign change.

Set up and evaluate the integral

\text{Distance} = \int_0^4 t^2(t-4)^2\,dt = \int_0^4 (t^4 - 8t^3 + 16t^2)\,dt

= \left[\frac{t^5}{5} - 2t^4 + \frac{16t^3}{3}\right]_0^4

= \frac{1024}{5} - 2(256) + \frac{16(64)}{3} = \frac{1024}{5} - 512 + \frac{1024}{3}

Simplify

Common denominator 15:

= \frac{3072}{15} - \frac{7680}{15} + \frac{5120}{15} = \frac{512}{15} \approx 34.133 \text{ feet}

AP Scoring — 3 Points

**P1**: Recognizes distance =

\int_0^4 |v(t)|\,dt

and justifies

v(t) \geq 0

.
**P2**: Correct antiderivative.
**P3**: Correct answer

\frac{512}{15}

ft.

Part DMedium3 points

Find the average velocity of the car over the time interval

6 \leq t \leq 12

seconds. Show the work that leads to your answer.

Answer

\bar{v} = \dfrac{1}{6}\int_6^{12} v(t)\,dt = \dfrac{1}{6}\left[10\cdot\dfrac{3}{\pi}\sin\!\left(\dfrac{\pi}{3}t\right) - 10t\right]_6^{12} = -10

ft/sec

Full Solution & Work

Set up the average velocity formula

\bar{v} = \frac{1}{12-6}\int_6^{12} v(t)\,dt = \frac{1}{6}\int_6^{12}\left[10\cos\!\left(\frac{\pi}{3}t\right) - 10\right]dt

Find the antiderivative

\int \left[10\cos\!\left(\frac{\pi}{3}t\right) - 10\right]dt = \frac{30}{\pi}\sin\!\left(\frac{\pi}{3}t\right) - 10t + C

Evaluate from 6 to 12

\left[\frac{30}{\pi}\sin\!\left(\frac{\pi}{3}t\right) - 10t\right]_6^{12}

At

t=12

:

\frac{30}{\pi}\sin(4\pi) - 120 = 0 - 120 = -120

.
At

t=6

:

\frac{30}{\pi}\sin(2\pi) - 60 = 0 - 60 = -60

.

= -120 - (-60) = -60

Compute average velocity

\bar{v} = \frac{-60}{6} = -10 \text{ ft/sec}

The negative sign indicates net motion in the negative direction.

AP Scoring — 3 Points

**P1**: Correct average velocity setup

\frac{1}{6}\int_6^{12} v(t)\,dt

.
**P2**: Correct antiderivative.
**P3**: Correct answer

-10

ft/sec.

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Question 6

No Calculator

Tabular Data, Chain Rule & FTC

Limits, Chain Rule & FTC

Open this question on its own page

Hard

The function

f

is twice differentiable. The table gives values of

f

and

f'

at selected values of

x

.

Values of f and f'

x	0	2	3	6
f(x)	−1	3	8	5
f′(x)	−5	4	9	−2

Part AEasy2 points

Find

\lim_{x \to 2} \dfrac{f(x)}{x}

, or state that it does not exist.

Answer

\lim_{x \to 2} \dfrac{f(x)}{x} = \dfrac{f(2)}{2} = \dfrac{3}{2}

Full Solution & Work

Check if direct substitution applies

Since

f

is differentiable (hence continuous), we can substitute directly:

\lim_{x \to 2} \frac{f(x)}{x} = \frac{f(2)}{2} = \frac{3}{2}

AP Scoring — 2 Points

**P1**: Recognizes direct substitution is valid (no indeterminate form).
**P2**: Correct answer

\frac{3}{2}

.

Part BMedium3 points

Let

g(x) = f(f(x))

. Find

g'(2)

. Show your work.

Answer

g'(2) = f'(f(2)) \cdot f'(2) = f'(3) \cdot 4 = 9 \cdot 4 = 36

Full Solution & Work

Apply the Chain Rule

g'(x) = f'(f(x)) \cdot f'(x)

Evaluate at x = 2

g'(2) = f'(f(2)) \cdot f'(2) = f'(3) \cdot f'(2)

From the table:

f(2) = 3

,

f'(2) = 4

,

f'(3) = 9

.

g'(2) = 9 \cdot 4 = 36

AP Scoring — 3 Points

**P1**: Correct chain rule setup

f'(f(x)) \cdot f'(x)

.
**P2**: Substitutes

f(2) = 3

to get

f'(3)

.
**P3**: Correct answer 36.

Part CHard3 points

Let

h

be differentiable with

h(0) = 10

and

h'(x) = f'(3x)

. Find

h(2)

. Show your work.

Answer

h(2) = h(0) + \int_0^2 f'(3x)\,dx = 10 + \frac{1}{3}[f(6) - f(0)] = 10 + \frac{1}{3}(6) = 12

Full Solution & Work

Use FTC: h(2) = h(0) + integral

h(2) = h(0) + \int_0^2 h'(x)\,dx = 10 + \int_0^2 f'(3x)\,dx

U-substitution: u = 3x

Let

u = 3x

,

du = 3\,dx

, so

dx = \frac{du}{3}

. When

x=0

,

u=0

; when

x=2

,

u=6

:

\int_0^2 f'(3x)\,dx = \frac{1}{3}\int_0^6 f'(u)\,du = \frac{1}{3}[f(6) - f(0)]

Evaluate using table values

= \frac{1}{3}[f(6) - f(0)] = \frac{1}{3}[5 - (-1)] = \frac{1}{3}(6) = 2

h(2) = 10 + 2 = 12

AP Scoring — 3 Points

**P1**: FTC setup

h(2) = 10 + \int_0^2 f'(3x)\,dx

.
**P2**: Correct

u

-substitution giving

\frac{1}{3}[f(6)-f(0)]

.
**P3**: Correct answer

h(2) = 12

.

Part D-iEasy1 point

Let

k(x) = \int_0^x t^2 f(t)\,dt

. Find

k'(x)

.

Answer

k'(x) = x^2 f(x)

Full Solution & Work

Apply FTC Part 1

By the Fundamental Theorem of Calculus Part 1:

k'(x) = x^2 f(x)

AP Scoring — 1 Point

**P1**: Correct answer

k'(x) = x^2 f(x)

.

Part D-iiHard3 points

Find

k''(3)

. Show your work.

Answer

k''(3) = 2(3)f(3) + (3)^2 f'(3) = 6(8) + 9(9) = 48 + 81 = 129

Full Solution & Work

Differentiate k'(x) using the Product Rule

k'(x) = x^2 f(x)

k''(x) = 2x\,f(x) + x^2 f'(x)

Evaluate at x = 3

From the table:

f(3) = 8

,

f'(3) = 9

.

k''(3) = 2(3)(8) + (3)^2(9) = 48 + 81 = 129

AP Scoring — 3 Points

**P1**: Correct product rule differentiation of

k'(x)

.
**P2**: Substitutes

f(3) = 8

and

f'(3) = 9

.
**P3**: Correct answer 129.

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2026 AP Calculus AB Free-Response Questions — Full Solutions

Bird Migration Rates

Set up the difference quotient

State the units

AP Scoring — 2 Points

Identify the midpoints

Evaluate M at each midpoint

Compute the Riemann sum

AP Scoring — 2 Points

Interpret rate × time = total

AP Scoring — 1 Point

Set up the integral

Evaluate the integral (Part A — calculator permitted)

Analytical verification (optional)

AP Scoring — 3 Points

Compute D at the endpoints

Apply the Intermediate Value Theorem

AP Scoring — 3 Points

Area and Volume Between Curves

Set up the integral

Analytical verification

AP Scoring — 3 Points

Set up cross-sectional area

Write the volume integral

AP Scoring — 2 Points

Find the second intersection point a

Set up the integral

Evaluate using the calculator

AP Scoring — 4 Points

Identify the bounds in y

Apply the disk method about the y-axis

AP Scoring — 3 Points

Cooling Pie — Differential Equations

Analyze the sign of dH/dt

Identify the contradiction

AP Scoring — 2 Points

Substitute H(0) = 75 into the differential equation

AP Scoring — 2 Points

Determine the sign of the second derivative

Conclude overestimate or underestimate

AP Scoring — 2 Points

Separate variables

Integrate both sides

Exponentiate and solve for H

Apply the initial condition H(0) = 75

Write the particular solution

AP Scoring — 6 Points

Analyzing a Function from its Derivative Graph

Differentiate g(x)

Evaluate at x = 2

AP Scoring — 2 Points

Recall: inflection points occur where f'' changes sign

Read the graph of f'

Conclusion

AP Scoring — 2 Points

Condition 1: Increasing ($f'(x) > 0$)

Condition 2: Concave Down ($f''(x) < 0$)

Intersection of Conditions

AP Scoring — 2 Points

Identify Candidate Points

Analyze the Absolute Minimum

Analyze the Absolute Maximum

AP Scoring — 4 Points

Remote-Controlled Car Motion

Differentiate v(t) on [0, 4]

Evaluate at t = 1

AP Scoring — 2 Points

Find v(1)

Compare signs of v and a

AP Scoring — 1 Point

Check if v changes sign on [0, 4]

Set up and evaluate the integral

Simplify

AP Scoring — 3 Points

Set up the average velocity formula

Find the antiderivative

Evaluate from 6 to 12

Compute average velocity

AP Scoring — 3 Points

Tabular Data, Chain Rule & FTC