2024 AP Calculus AB FRQ Question 6: Area and Volume Between f and g

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2024.

Question 6

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Area and Volume Between f and g

Area & Volume

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

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)

[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

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

x=5

(limits reversed to

5

2

also work, provided the sign is adjusted accordingly).

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