2025 AP Calculus AB FRQ Question 4: Accumulation Function from a Graph

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2025.

Question 4

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Accumulation Function from a Graph

Graph Analysis & Curve Sketching

Hard

The continuous function

f

is defined on the closed interval

-6 \leq x \leq 12

. The graph of

f

, consisting of two semicircles (radius 3) and one line segment, is shown in the figure: a lower semicircle from

(-6,0)

down to

(-3,-3)

and back up to

(0,0)

, an upper semicircle from

(0,0)

up to

(3,3)

and back down to

(6,0)

, and a line segment from

(6,0)

(12,3)

. Let

g

be the function defined by

g(x) = \int_6^x f(t)\,dt

2025 AP Calculus AB FRQ Question 4 Graph of f — two semicircles and a line segment — Graph of $f$ on $[-6, 12]$.

Part AEasy2 points

Find

g'(8)

. Give a reason for your answer.

Answer

g'(8) = f(8) = 1

Full Solution & Work

Apply the Fundamental Theorem of Calculus

Since

g(x)=\int_6^x f(t)\,dt

, by FTC Part 1:

g'(x) = f(x)

Evaluate f(8) from the line segment

For

6 \leq x \leq 12

f

is the line through

(6,0)

and

(12,3)

, i.e.

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

f(8) = \frac{1}{2}(8-6) = 1 \implies g'(8) = 1

AP Scoring — 2 Points

**P1**: Considers

g'(x)=f(x)

.
**P2**: Correct answer

g'(8)=1

. (An implied use of FTC, like writing

g'(8)=1

directly by reading the graph, can also earn P2 without P1.)

Part BHard2 points

Find all values of

x

in the open interval

-6 < x < 12

at which the graph of

g

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

Answer

x = -3

x = 3

, and

x = 6

— the graph of

g

has points of inflection at all three.

Full Solution & Work

Relate inflection points of g to f

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

, so

g

has an inflection point wherever

f'

changes sign — equivalently, wherever

f

changes from decreasing to increasing or increasing to decreasing (a local extremum of

f

Read local extrema off the graph of f

x=-3

f

decreases (left semicircle going down) then increases (coming back up) — a local minimum of

f

.
At

x=3

f

increases (right semicircle going up) then decreases (coming back down) — a local maximum of

f

.
At

x=6

f

is still decreasing just before

x=6

(end of the upper semicircle) and increasing just after (start of the line segment) — another local minimum of

f

, even though the graph has a corner there.

Conclusion

f

changes from decreasing to increasing at

x=-3

and

x=6

, and from increasing to decreasing at

x=3

. So the graph of

g

has points of inflection at

x=-3

x=3

, and

x=6

AP Scoring — 2 Points

**P3**: Correct answer — earned *only* for exactly

x=-3, 3, 6

. Naming any additional or different value forfeits both points.
**P4**: A reason tied to the given graph of

f

(e.g. "

f

changes from increasing to decreasing / decreasing to increasing there", or "

f

has a local extremum there"). A reason based on an ambiguous "the function" or "the graph," without specifying

f

, does not earn P4.

Common mistake: The corner at $x=6$ is easy to miss — students often only report $x=-3$ and $x=3$ since those look like the "obvious" smooth extrema, but $f$ still switches from decreasing to increasing there, so $g$ still has an inflection point.

Part CMedium2 points

Find

g(12)

and

g(0)

. Label your answers.

Answer

g(12) = 9

and

g(0) = -\dfrac{9\pi}{2}

Full Solution & Work

Compute g(12) as a signed area

g(12) = \int_6^{12} f(t)\,dt = \frac{1}{2}(6)(3) = 9

(area of the triangle under the line segment from

x=6

x=12

)

Compute g(0) as a signed area

g(0) = \int_6^0 f(x)\,dx = -\int_0^6 f(x)\,dx = -\frac{1}{2}\pi(3)^2 = -\frac{9\pi}{2}

(the upper semicircle from

x=0

x=6

has area

\frac{1}{2}\pi r^2

; the direction of integration flips the sign)

AP Scoring — 2 Points

**P5**:

g(12)=9

, with or without supporting work.
**P6**:

g(0)=-\frac{9\pi}{2}

, with or without supporting work. Both values must be clearly labeled — unlabeled numbers do not earn credit.

Part DHard3 points

Find the value of

x

at which

g

attains an absolute minimum on the closed interval

-6 \leq x \leq 12

. Justify your answer.

Answer

x = 0

Full Solution & Work

Find critical points of g

g

attains an absolute minimum either at a critical point (

g'(x)=f(x)=0

) or an endpoint. From the graph,

f(x)=0

x=0

and

x=6

Evaluate g at all candidates

\begin{array}{c|c} x & g(x) \\\hline -6 & 0 \\ 0 & -9\pi/2 \\ 6 & 0 \\ 12 & 9 \end{array}

(using

g(-6)=\int_6^{-6}f(t)\,dt = -\left(-\frac{9\pi}{2}+\frac{9\pi}{2}\right) = 0

, and the values of

g(0)

g(12)

from part C)

Conclusion

The smallest value in the table is

-\frac{9\pi}{2}

x=0

, so

g

attains its absolute minimum on

[-6,12]

x = 0

AP Scoring — 3 Points

**P7**: Considers

g'(x)=0

f(x)=0

(not earned by simply stating

x=0,6

with no equation).
**P8**: A complete justification — a candidates-test table evaluating

g

-6

0

6

12

(values may be imported from part C), or an equivalent global sign argument.
**P9**: Correct answer

x=0

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