2022 AP Calculus AB FRQ Question 2: Area and Volume Between f and g

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2022.

Question 2

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Area and Volume Between f and g

Area & Volume

Hard

Let

f

and

g

be the functions defined by

f(x)=\ln(x+3)

and

g(x)=x^4+2x^3

. The graphs of

f

and

g

intersect at

x=-2

and

x=B

, where

B>0

Part AMedium2 points

Find the area of the region enclosed by the graphs of

f

and

g

Answer

≈ 3.604

Full Solution & Work

Find B

Solving

f(x)=g(x)

numerically for the positive root:

B\approx0.782

Set up and evaluate the area integral

[-2,B]

f(x)\geq g(x)

(confirmed by checking

f-g>0

at interior points):

\text{Area}=\int_{-2}^{B}\big(f(x)-g(x)\big)\,dx \approx 3.604

AP Scoring — 2 Points

**P1**: Correct integrand

f(x)-g(x)

(or equivalent) with correct limits

-2

B

.
**P2**: Correct answer, 3.604 (accept 3.603–3.605).

Part BMedium2 points

For

-2\leq x\leq B

, let

h(x)

be the vertical distance between the graphs of

f

and

g

. Is

h

increasing or decreasing at

x=-0.5

? Give a reason for your answer.

Answer

Decreasing, since

h'(-0.5)=-0.6<0

Full Solution & Work

Write h(x) and differentiate

Since

f\geq g

on this interval,

h(x)=f(x)-g(x)=\ln(x+3)-x^4-2x^3

h'(x) = \frac{1}{x+3}-4x^3-6x^2

Evaluate at x = -0.5

h'(-0.5) = \frac{1}{2.5}-4(-0.125)-6(0.25) = 0.4+0.5-1.5 = -0.6

Conclude

Since

h'(-0.5)=-0.6<0

h

is **decreasing** at

x=-0.5

AP Scoring — 2 Points

**P1**: Correct expression for

h'(x)

or a correctly-evaluated

h'(-0.5)

.
**P2**: Correct conclusion ("decreasing") tied to the sign of

h'(-0.5)

Part CHard3 points

The region enclosed by the graphs of

f

and

g

is the base of a solid. Cross sections of the solid taken perpendicular to the

x

-axis are squares. Find the volume of the solid.

Answer

≈ 5.340

Full Solution & Work

Set up the volume integral

Each square cross section has side length

h(x)=f(x)-g(x)

, so area

\big(h(x)\big)^2

V = \int_{-2}^{B} \big(f(x)-g(x)\big)^2\,dx

Evaluate with a calculator

V \approx 5.340

AP Scoring — 3 Points

**P1**: Integrand of the form

(f(x)-g(x))^2

.
**P2**: Correct limits

-2

B

.
**P3**: Correct answer, 5.340 (accept 5.339–5.341).

Part DHard2 points

A vertical line in the

xy

-plane travels from left to right along the base of the solid described in part (c). The vertical line is moving at a constant rate of 7 units per second. Find the rate of change of the area of the cross section above the vertical line with respect to time when the vertical line is at position

x=-0.5

Answer

≈ −9.272 (square units per second)

Full Solution & Work

Write the cross-sectional area as a function of x, then apply the chain rule

A_{\text{cross}}(x) = \big(h(x)\big)^2 \implies \frac{dA_{\text{cross}}}{dt} = 2h(x)\cdot h'(x)\cdot\frac{dx}{dt}

Substitute values at x = -0.5

From parts (b):

h(-0.5)\approx1.104

h'(-0.5)=-0.6

. Given

\frac{dx}{dt}=7

\frac{dA_{\text{cross}}}{dt} = 2(1.104)(-0.6)(7) \approx -9.272

AP Scoring — 2 Points

**P1**: Correct related-rates setup

\frac{dA_{\text{cross}}}{dt}=2h(x)h'(x)\frac{dx}{dt}

.
**P2**: Correct answer,

-9.272

(accept

-9.270

-9.274

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