2024 AP Calc AB FRQ Answers & Solutions

Question 1

Calculator OK

Coffee Temperature — Table Data

Rate & Data from Tables

Hard

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.

Values of C(t)

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

Part AEasy2 points

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.

Answer

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

degrees Celsius per minute

Full Solution & Work

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 are degrees Celsius per minute.

AP Scoring — 2 Points

**P1**: A response must include a difference and a quotient as supporting work —

\frac{-16}{4}

,

\frac{69-85}{4}

, or equivalent is sufficient.
**P2**: Units of "degrees Celsius per minute" attached to a numerical value. Units alone with no numerical approximation do not earn this point.

Part BMedium3 points

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

\dfrac{1}{12}\int_0^{12} C(t)\,dt

in the context of the problem.

Answer

\int_0^{12} C(t)\,dt \approx 3\cdot C(0) + 4\cdot C(3) + 5\cdot C(7) = 300+340+345=985

Full Solution & Work

Set up 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)

Substitute values and compute

= 3\cdot 100 + 4\cdot 85 + 5\cdot 69 = 300+340+345=985

Interpret the average value expression

\dfrac{1}{12}\int_0^{12} C(t)\,dt

is the **average temperature of the coffee** (in degrees Celsius) **over the time interval from

t=0

to

t=12

**.

AP Scoring — 3 Points

**P1**: Correct form of the left Riemann sum — at least five of the six numerical factors must be correct.
**P2**: Correct estimate, 985 — requires P1 (a fully correct right Riemann sum, e.g.

4\cdot 85+4\cdot 69+5\cdot 55

, earns P1 but not P2).
**P3**: Interpretation must include both "average temperature" and the time interval; units are not required, but incorrect units forfeit this point.

Part CMedium3 points

For

12 \leq t \leq 20

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

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

, where

C'(t)

is measured in degrees Celsius per minute. Find the temperature of the coffee at time

t=20

. Show the setup for your calculations.

Answer

C(20) = C(12) + \displaystyle\int_{12}^{20} C'(t)\,dt = 55-14.670812=40.329

Full Solution & Work

Set up C(20) using C(12) plus accumulated change

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

Substitute the known value and evaluate

= 55 + \int_{12}^{20} \frac{-24.55e^{0.01t}}{t}\,dt = 55 - 14.670812 = 40.329188

State the answer

The temperature of the coffee at time

t=20

is **40.329** degrees Celsius.

AP Scoring — 3 Points

**P1**: A definite integral with integrand

C'(t)

. If the limits of integration are incorrect, this response is not eligible for the third point.
**P2**: Adding

C(12)

(or 55) to a definite integral with lower limit 12.
**P3**: Correct answer

55-14.671

(with supporting work) — an answer of just 40.329 with no supporting work earns no points at all.

Common mistake: A "linkage error" like writing $C(20)=\int_{12}^{20}C'(t)\,dt = 55-14.670812$ (treating the whole right side as equal to the integral alone) still earns the first two points but not the third — keep the equation chain unambiguous.

Part DMedium1 point

For the model defined in part (c), it can be shown that

C''(t) = \dfrac{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 at an increasing rate. Give a reason for your answer.

Answer

Increasing rate, because

C''(t) > 0

on

12 < t < 20

.

Full Solution & Work

Determine the sign of C''(t)

For

12<t<20

:

0.01t < 1

so

e^{0.01t}>0

;

100-t>0

(since

t<20<100

);

t^2>0

. So

C''(t)>0

throughout this interval.

Interpret the sign

Because

C''(t)>0

on

12<t<20

, the rate of change

C'(t)

is **increasing** on this interval — that is, the temperature is changing at an **increasing rate**.

AP Scoring — 1 Point

**P1**: Earned only for a correct answer with a correct reason that references the sign of the **second derivative** of

C

. A reason based on evaluating

C''(t)

at only a single point does not earn this point, and neither does an ambiguous pronoun like "it is positive, so increasing."

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Question 2

Calculator OK

Particle Velocity — Logarithmic Model

Particle Motion

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Hard

A particle moves along the

x

-axis so that its velocity at time

t \geq 0

is given by

v(t) = \ln(t^2-4t+5) - 0.2t

.

Part AEasy2 points

There is one time,

t=t_R

, in the interval

0<t<2

when the particle is at rest (not moving). Find

t_R

. For

0<t<t_R

, is the particle moving to the right or to the left? Give a reason for your answer.

Answer

t_R = 1.426

(or

1.425

); moving right for

0<t<t_R

.

Full Solution & Work

Solve v(t) = 0 on (0,2)

v(t) = 0 \implies t = 1.425610

So

t_R = 1.426

(or 1.425).

Determine direction on (0, t_R)

For

0<t<t_R

,

v(t)>0

. Therefore, the particle is moving to the **right** on that interval.

AP Scoring — 2 Points

**P1**: The value

t_R=1.426

(or 1.425) — a response that finds no value or an incorrect value of

t_R

is not eligible for the second point.
**P2**: Correct direction (right) with the explanation that

v(t)>0

on

(0,t_R)

.

Part BMedium2 points

Find the acceleration of the particle at time

t=1.5

. Show the setup for your calculations. Is the speed of the particle increasing or decreasing at time

t=1.5

? Explain your reasoning.

Answer

a(1.5)=v'(1.5)=-1

; speed is increasing at

t=1.5

.

Full Solution & Work

Relate acceleration to v'(t)

a(1.5) = v'(1.5) = -1 \text{ (or } -0.999\text{)}

Compare signs of v and a

v(1.5) = -0.076856 < 0

. Because

a(1.5)

and

v(1.5)

have the **same sign** (both negative), the speed of the particle is **increasing** at

t=1.5

.

AP Scoring — 2 Points

**P1**: A response must demonstrate the relationship

v'=a

to earn this point — just stating "

a(1.5)=-1

" with no connection to

v'

is not sufficient.
**P2**: Earned only for a response consistent with a negative velocity at

t=1.5

and the presented value of

a(1.5)

.

Part CMedium3 points

The position of the particle at time

t

is

x(t)

, and its position at time

t=1

is

x(1)=-3

. Find the position of the particle at time

t=4

. Show the setup for your calculations.

Answer

x(4) = x(1) + \displaystyle\int_1^4 v(t)\,dt = -3+0.197117=-2.803

Full Solution & Work

Set up x(4) using x(1) plus accumulated change

x(4) = x(1) + \int_1^4 v(t)\,dt

Evaluate

= -3+0.197117 = -2.802883

The position of the particle at time

t=4

is **

-2.803

** (or

-2.802

).

AP Scoring — 3 Points

**P1**: A definite integral with integrand

v(t)

. If the limits of integration are incorrect, this response is not eligible for the third point.
**P2**: Adding

x(1)

(or

-3

) to a definite integral with lower limit 1.
**P3**: Correct answer

-3+0.197

with supporting work — an answer of just

-2.803

with no supporting work earns no points.

Part DMedium2 points

Find the total distance traveled by the particle over the interval

1 \leq t \leq 4

. Show the setup for your calculations.

Answer

\displaystyle\int_1^4 |v(t)|\,dt = 0.958

Full Solution & Work

Set up the distance integral

\text{Distance} = \int_1^4 |v(t)|\,dt

Evaluate with a calculator

= 0.9581

The total distance traveled by the particle over

1 \leq t \leq 4

is **0.958**.

AP Scoring — 2 Points

**P1**: The integral

\int_1^4 |v(t)|\,dt

, or an equivalent split at the sign changes of

v

(at

t\approx1.425

and

t\approx2.883

, the other root of

v(t)=0

).
**P2**: Due to variations in numerical integration across calculator models, answers of 0.958, 0.959, or 0.96 all earn this point.

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Question 3

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Seawater Depth — Differential Equation

Differential Equations

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Hard

The depth of seawater at a location can be modeled by the function

H

that satisfies the differential equation

\dfrac{dH}{dt} = \dfrac{1}{2}(H-1)\cos\!\left(\dfrac{t}{2}\right)

, where

H(t)

is measured in feet and

t

is measured in hours after noon (

t=0

). It is known that

H(0)=4

.

Part AEasy1 point

A portion of the slope field for the differential equation is provided. Sketch the solution curve,

y=H(t)

, through the point

(0,4)

.

Answer

The curve passes through

(0,4)

, rises to a local max around

t=\pi

, then decreases.

Full Solution & Work

Follow the slope field from (0, 4)

Starting at

(0,4)

, follow the slope-field arrows: since

H>1

and

\cos(t/2)>0

for small

t>0

,

\frac{dH}{dt}>0

initially, so the curve rises. Slopes flatten and turn negative once

t

passes

\pi

(where

\cos(t/2)=0

), so the curve peaks near

t=\pi

and then decreases through the rest of the shown domain (matches the official figure, which peaks around

H\approx 9

near

t\approx 3

).

AP Scoring — 1 Point

**P1**: The solution curve must pass through

(0,4)

, extend to at least

t=4.5

, and have no obvious conflicts with the given slope lines.

Part BHard3 points

For

0<t<5

, it can be shown that

H(t)>1

. Find the value of

t

, for

0<t<5

, at which

H

has a critical point. Determine whether the critical point corresponds to a relative minimum, a relative maximum, or neither a relative minimum nor a relative maximum of the depth of seawater at the location. Justify your answer.

Answer

t=\pi

is a relative maximum.

Full Solution & Work

Set dH/dt = 0

Because

H(t)>1

,

\frac{dH}{dt}=0

implies

\cos\!\left(\frac{t}{2}\right)=0

, which gives

t=\pi

as a critical point in

0<t<5

.

Sign-analyze dH/dt around t = π

For

0<t<\pi

:

\frac{dH}{dt}>0

. For

\pi<t<5

:

\frac{dH}{dt}<0

. Therefore,

t=\pi

is the location of a **relative maximum** value of

H

.

AP Scoring — 3 Points

**P1**: Considers

\frac{dH}{dt}=0

,

\frac{dH}{dt}>0

,

\frac{dH}{dt}<0

,

\cos(t/2)=0

, or

\cos(t/2)>0

/

<0

.
**P2**: Identifies

t=\pi

, with or without supporting work.
**P3**: Cannot be earned without P1. Earned only for a correct justification and the correct answer "relative maximum." The justification can be a sign analysis of

\frac{dH}{dt}

on either side of

\pi

, or a Second Derivative Test showing

\frac{d^2H}{dt^2}\big|_{t=\pi}<0

.

Part CHard5 points

Use separation of variables to find

y=H(t)

, the particular solution to the differential equation

\dfrac{dH}{dt}=\dfrac{1}{2}(H-1)\cos\!\left(\dfrac{t}{2}\right)

with initial condition

H(0)=4

.

Answer

H(t) = 1+3e^{\sin(t/2)}

Full Solution & Work

Separate variables

\frac{dH}{H-1} = \frac{1}{2}\cos\!\left(\frac{t}{2}\right)dt

Integrate both sides

\int \frac{dH}{H-1} = \int \frac{1}{2}\cos\!\left(\frac{t}{2}\right)dt

\ln|H-1| = \sin\!\left(\frac{t}{2}\right)+C

Apply the initial condition

\ln|4-1| = \sin(0)+C \implies C=\ln 3

Because

H(0)=4

,

H>1

, so

|H-1|=H-1

.

\ln(H-1) = \sin\!\left(\frac{t}{2}\right)+\ln 3

Solve for H

H-1 = e^{\sin(t/2)+\ln 3} = 3e^{\sin(t/2)}

H(t) = 1+3e^{\sin(t/2)}

AP Scoring — 5 Points

**P1**: Correct separation of variables — a response with no separation of variables earns 0 of 5 points.
**P2**: One correct antiderivative (either side).
**P3**: Both correct antiderivatives — presenting

\ln(H-1)

without absolute value symbols still earns this point.
**P4**: Correctly includes the constant of integration and uses the initial condition

H(0)=4

. Requires P1 and at least one of P2/P3.
**P5**: Earned only for the final answer

H(t)=1+3e^{\sin(t/2)}

(or an equivalent form) — requires all 4 previous points.

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Question 4

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Region R and Its Accumulation Function

Graph Analysis & Curve Sketching

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Hard

The graph of the differentiable function

f

, shown for

-6 \leq x \leq 7

, has a horizontal tangent at

x=-2

and is linear for

0 \leq x \leq 7

. Let

R

be the region in the second quadrant bounded by the graph of

f

, the vertical line

x=-6

, and the

x

- and

y

-axes. Region

R

has area 12.

2024 AP Calculus AB FRQ Question 4 Graph of f, with shaded region R — Graph of $f$, with region $R$ (area 12) shaded.

Part AMedium3 points

The function

g

is defined by

g(x)=\int_0^x f(t)\,dt

. Find the values of

g(-6)

,

g(4)

, and

g(6)

.

Answer

g(-6)=-12

,

g(4)=4

,

g(6)=3

Full Solution & Work

Find g(-6) using region R's given area

g(-6) = \int_0^{-6} f(t)\,dt = -\int_{-6}^0 f(t)\,dt = -12

(since

\int_{-6}^0 f(t)\,dt

is exactly the area of

R

, given as 12)

Find g(4) as a triangle area

f

is linear on

[0,7]

, passing through

(0,2)

and

(4,2)

... reading the graph, the line from

(0,2)

crosses the

x

-axis at

x=4

:

g(4)=\int_0^4 f(t)\,dt = \frac{1}{2}\cdot 4\cdot 2 = 4

Find g(6) as a signed sum of triangle areas

g(6) = \int_0^6 f(t)\,dt = \frac{1}{2}\cdot 4\cdot 2 - \frac{1}{2}\cdot 2\cdot 1 = 4-1=3

AP Scoring — 3 Points

**P1**:

g(-6)=-12

.
**P2**:

g(4)=4

.
**P3**:

g(6)=3

.
Supporting work is not required for any of these, but any shown work must be correct to earn the point. Unlabeled values are read left-to-right / top-to-bottom as

g(-6), g(4), g(6)

— a response presenting only 1 or 2 values must label them to earn credit.

Part BMedium2 points

For the function

g

defined in part (a), find all values of

x

in the interval

0 \leq x \leq 6

at which the graph of

g

has a critical point. Give a reason for your answer.

Answer

x=4

Full Solution & Work

Apply the Fundamental Theorem of Calculus

g'(x) = f(x)

Solve g'(x) = 0

g'(x)=f(x)=0 \implies x=4

Therefore, the graph of

g

has a critical point at

x=4

.

AP Scoring — 2 Points

**P1**: Explicitly making the connection

g'=f

in this part.
**P2**: Correct answer

x=4

with reason — a response reporting any additional critical points in

0<x<6

does not earn this point.

Part CHard4 points

The function

h

is defined by

h(x)=\int_{-6}^x f'(t)\,dt

. Find the values of

h(6)

,

h'(6)

, and

h''(6)

. Show the work that leads to your answers.

Answer

h(6)=-1.5

,

h'(6)=-\frac{1}{2}

,

h''(6)=0

Full Solution & Work

Find h(6) using the Fundamental Theorem of Calculus

h(6) = \int_{-6}^6 f'(t)\,dt = f(6)-f(-6) = -1-0.5=-1.5

Find h'(6)

h'(x)=f'(x) \implies h'(6)=f'(6)=-\frac{1}{2}

(the slope of the linear piece of

f

on

[0,7]

, found from the two labeled points

(4,2)\to(6,-1)

-consistent slope

-\frac{1}{2}

)

Find h''(6)

h''(x)=f''(x) \implies h''(6)=f''(6)=0

(since

f

is linear on

[0,7]

, its second derivative is identically 0 there)

AP Scoring — 4 Points

**P1**: Uses the Fundamental Theorem of Calculus for

h(6)

.
**P2**:

h(6)=-1.5

with supporting work — a bare answer of

h(6)=-1.5

does not earn either of the first two points;

h(6)=f(6)-f(-6)

earns only the first.
**P3**: States either

h'(x)=f'(x)

or

h'(6)=f'(6)

, and provides the answer

-\frac{1}{2}

.
**P4**:

h''(6)=0

, with or without supporting work.

Common mistake: Any response that has one or more linkage errors does not earn the point it would have otherwise earned for that step, but stays eligible for subsequent points if those are handled correctly.

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Question 5

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Implicit Curve — Tangent Lines & Related Rates

Implicit Differentiation

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Hard

Consider the curve defined by the equation

x^2+3y+2y^2=48

. It can be shown that

\dfrac{dy}{dx}=\dfrac{-2x}{3+4y}

.

Part AMedium2 points

There is a point on the curve near

(2,4)

with

x

-coordinate 3. Use the line tangent to the curve at

(2,4)

to approximate the

y

-coordinate of this point.

Answer

y \approx 4-\dfrac{4}{19}(3-2)=\dfrac{72}{19}

Full Solution & Work

Find the slope at (2,4)

\left.\frac{dy}{dx}\right|_{(2,4)} = \frac{-2(2)}{3+4(4)} = \frac{-4}{19}

Apply the tangent line approximation

y \approx 4-\frac{4}{19}(3-2) = \frac{72}{19}

AP Scoring — 2 Points

**P1**: Correctly finding

\left.\frac{dy}{dx}\right|_{(2,4)}=-\frac{4}{19}

, even if not labeled or used as a tangent-line slope.
**P2**: Correct approximation, evaluated at

x=3

using a linear approximation through

(2,4)

with slope

-\frac{4}{19}

.

Part BHard2 points

Is the horizontal line

y=1

tangent to the curve? Give a reason for your answer.

Answer

No — the only candidate point of tangency,

(0,1)

, is not on the curve.

Full Solution & Work

Find where dy/dx = 0

\frac{dy}{dx} = \frac{-2x}{3+4y}=0 \implies x=0

So if

y=1

is tangent to the curve, the point of tangency must be

(0,1)

.

Check whether (0,1) lies on the curve

0^2+3(1)+2(1)^2 = 5 \neq 48

The point

(0,1)

is not on the curve. Therefore, the horizontal line

y=1

is **not** tangent to the curve.

AP Scoring — 2 Points

**P1**: Considering

\frac{dy}{dx}=0

,

-2x=0

, or

x=0

— or, alternatively, identifying the point

(0,1)

directly.
**P2**: A reason that

y=1

is not tangent to the curve; merely stating "

(0,1)

is not on the curve" without checking it against the equation is insufficient.

Part CMedium1 point

The curve intersects the positive

x

-axis at the point

(\sqrt{48},0)

. Is the line tangent to the curve at this point vertical? Give a reason for your answer.

Answer

No — the denominator of

\dfrac{dy}{dx}

at that point is

3+4(0)=3\neq0

.

Full Solution & Work

Evaluate the denominator at (√48, 0)

\left.\frac{dy}{dx}\right|_{(\sqrt{48},0)} = \frac{-2\sqrt{48}}{3+4(0)}

The denominator,

3+4(0)

, does not equal 0.

Conclude

Since the slope of the tangent line is defined (finite) at

(\sqrt{48},0)

, the tangent line at this point is **not** vertical.

AP Scoring — 1 Point

**P1**: A response does not need to consider the numerator; showing the denominator is nonzero at

(\sqrt{48},0)

is sufficient. Must clearly demonstrate the slope is defined and conclude "no."

Part DHard4 points

For time

t \geq 0

, a particle is moving along another curve defined by the equation

y^3+2xy=24

. At the instant the particle is at the point

(4,2)

, the

y

-coordinate of the particle's position is decreasing at a rate of 2 units per second. At that instant, what is the rate of change of the

x

-coordinate of the particle's position with respect to time?

Answer

\dfrac{dx}{dt}=10

units per second

Full Solution & Work

Differentiate implicitly with respect to t

3y^2\frac{dy}{dt}+2x\frac{dy}{dt}+2y\frac{dx}{dt}=0

Substitute known values

At

(x,y)=(4,2)

with

\dfrac{dy}{dt}=-2

:

3(2)^2(-2)+2(4)(-2)+2(2)\frac{dx}{dt}=0

-24-16+4\frac{dx}{dt}=0

Solve for dx/dt

4\frac{dx}{dt}=40 \implies \frac{dx}{dt}=10 \text{ units per second}

AP Scoring — 4 Points

**P1**: Attempts implicit differentiation of

y^3+2xy=24

with respect to

t

, with at most one error.
**P2**: A correct equation equivalent to

3y^2\frac{dy}{dt}+2x\frac{dy}{dt}+2y\frac{dx}{dt}=0

.
**P3**: Correctly uses

\frac{dy}{dt}=-2

(a response does not need to explicitly declare this — substituting it directly into the implicit equation is sufficient). Using both

+2

and

-2

anywhere in the same response does not earn this point.
**P4**: Cannot be earned without the first 3 points. Earned only for the value 10, with no mistakes in the supporting work.

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Question 6

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Area and Volume Between f and g

Area & Volume

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Hard

The functions

f

and

g

are defined by

f(x)=x^2+2

and

g(x)=x^2-2x

, as shown in the graph.

2024 AP Calculus AB FRQ Question 6 Graph of f and g, with regions R and S — Graphs of $f$ and $g$, with regions $R$ (on $[0,2]$) and $S$ (on $[2,5]$).

Part AMedium2 points

Let

R

be the region bounded by the graphs of

f

and

g

, from

x=0

to

x=2

, as shown in the graph. Write, but do not evaluate, an integral expression that gives the area of region

R

.

Answer

\text{Area}=\displaystyle\int_0^2 (f(x)-g(x))\,dx

Full Solution & Work

Set up the integrand

Since

f(x) \geq g(x)

on

[0,2]

(the two parabolas differ by

f(x)-g(x)=2x+2

, positive for

x\geq -1

):

Write the area integral

\text{Area} = \int_0^2 \big(f(x)-g(x)\big)\,dx

AP Scoring — 2 Points

**P1**: An integrand of

f(x)-g(x)

,

|f(x)-g(x)|

,

g(x)-f(x)

, or

|g(x)-f(x)|

in one or more definite integrals (or, equivalently, a difference of definite integrals with integrands

f(x)

and

g(x)

).
**P2**: Earned only for one or more integrals equivalent to

\int_0^2(f(x)-g(x))\,dx

.

Part BHard4 points

Let

S

be the region bounded by the graph of

g

and the

x

-axis, from

x=2

to

x=5

, as shown in the graph. Region

S

is the base of a solid. For this solid, at each

x

the cross section perpendicular to the

x

-axis is a rectangle with height equal to half its base in region

S

. Find the volume of the solid. Show the work that leads to your answer.

Answer

V=\dfrac{414}{5}

Full Solution & Work

Set up the volume integral

V = \int_2^5 \frac{1}{2}\big(g(x)\big)^2\,dx = \int_2^5 \frac{1}{2}(x^2-2x)^2\,dx

Expand and integrate

= \frac{1}{2}\int_2^5 (x^4-4x^3+4x^2)\,dx = \frac{1}{2}\left[\frac{x^5}{5}-x^4+\frac{4x^3}{3}\right]_2^5

Evaluate the antiderivative at the bounds

= \frac{1}{2}\left[\left(\frac{5^5}{5}-5^4+\frac{500}{3}\right)-\left(\frac{32}{5}-16+\frac{32}{3}\right)\right] = \frac{1}{2}\left(\frac{500}{3}-\frac{16}{15}\right) = \frac{414}{5}

AP Scoring — 4 Points

**P1**: An integrand of the form

k(g(x))^2

in a definite integral, for any nonzero constant

k

.
**P2**: Limits of

x=2

and

x=5

in a definite integral with integrand

a(x)\cdot g(x)

for any nonzero factor

a(x)

.
**P3**: A correct antiderivative of

k(x^2-2x)^n

for some integer

n\geq2

.
**P4**: Earned only for the numeric answer

\frac{414}{5}

(equivalent to

\frac{1}{2}\left(\frac{500}{3}-\frac{16}{15}\right)

).

Common mistake: A response that instead uses $f(x)$ in place of $g(x)$ throughout the setup (Volume $=\int_2^5 \frac12(f(x))^2 dx$) can still earn P1 and P2, but is not eligible for P3 or P4 — the cross sections are explicitly built on region $S$, which is bounded by $g$, not $f$.

Part CHard3 points

Write, but do not evaluate, an integral expression that gives the volume of the solid generated when region

S

, as described in part (b), is rotated about the horizontal line

y=20

.

Answer

V=\pi\int_2^5 \Big[20^2-\big(20-g(x)\big)^2\Big]\,dx

Full Solution & Work

Identify outer and inner radii

Rotating about

y=20

: the outer radius (to the

x

-axis, the far boundary of

S

) is

20-0=20

; the inner radius (to

g(x)

, the near boundary) is

20-g(x)

.

Apply the washer method

V = \pi\int_2^5 \Big[20^2-\big(20-g(x)\big)^2\Big]\,dx = \pi\int_2^5\Big[400-\big(20-(x^2-2x)\big)^2\Big]\,dx

AP Scoring — 3 Points

**P1**: An integrand of

20^2-(20-g(x))^2

(or an equivalent expression, in any order/sign combination) in one or more definite integrals.
**P2**: Earned only for an integrand mathematically equivalent to

20^2-(20-g(x))^2

— a response using

f(x)

in place of

g(x)

can still earn P1 but not P2.
**P3**: Earned only for a definite integral including the constant

\pi

and limits

x=2

to

x=5

(limits reversed to

5

to

2

also work, provided the sign is adjusted accordingly).

Study the concept

Learn the theory behind this question type with worked examples and strategy tips.

Practice More

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AP Scoring — 2 Points

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AP Scoring — 1 Point

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AP Scoring — 4 Points

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AP Scoring — 2 Points

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AP Scoring — 4 Points

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AP Scoring — 3 Points

Gary Chang

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