2026 AP Calculus AB FRQ Question 2: Area and Volume Between Curves

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

Question 2

Calculator OK

Area and Volume Between Curves

Area & Volume

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

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

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

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

y = 3.5

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