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2024 AP Calculus AB FRQ Question 4: Region R and Its Accumulation Function

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

Question 4

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Region R and Its Accumulation Function

Graph Analysis & Curve Sketching

Hard
The graph of the differentiable function ff, shown for −6≤x≤7-6 \leq x \leq 7, has a horizontal tangent at x=−2x=-2 and is linear for 0≤x≤70 \leq x \leq 7. Let RR be the region in the second quadrant bounded by the graph of ff, the vertical line x=−6x=-6, and the xx- and yy-axes. Region RR 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 gg is defined by g(x)=∫0xf(t) dtg(x)=\int_0^x f(t)\,dt. Find the values of g(−6)g(-6), g(4)g(4), and g(6)g(6).

Answer

g(−6)=−12g(-6)=-12, g(4)=4g(4)=4, g(6)=3g(6)=3
Full Solution & Work

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

g(−6)=∫0−6f(t) dt=−∫−60f(t) dt=−12g(-6) = \int_0^{-6} f(t)\,dt = -\int_{-6}^0 f(t)\,dt = -12

(since ∫−60f(t) dt\int_{-6}^0 f(t)\,dt is exactly the area of RR, given as 12)

Find g(4) as a triangle area

ff is linear on [0,7][0,7], passing through (0,2)(0,2) and (4,2)(4,2)... reading the graph, the line from (0,2)(0,2) crosses the xx-axis at x=4x=4:
g(4)=∫04f(t) dt=12⋅4⋅2=4g(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)=∫06f(t) dt=12⋅4⋅2−12⋅2⋅1=4−1=3g(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)=−12g(-6)=-12.
**P2**: g(4)=4g(4)=4.
**P3**: g(6)=3g(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)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 gg defined in part (a), find all values of xx in the interval 0≤x≤60 \leq x \leq 6 at which the graph of gg has a critical point. Give a reason for your answer.

Answer

x=4x=4
Full Solution & Work

Apply the Fundamental Theorem of Calculus

g′(x)=f(x)g'(x) = f(x)

Solve g'(x) = 0

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

Therefore, the graph of gg has a critical point at x=4x=4.

AP Scoring — 2 Points

**P1**: Explicitly making the connection g′=fg'=f in this part.
**P2**: Correct answer x=4x=4 with reason — a response reporting any additional critical points in 0<x<60<x<6 does not earn this point.
Part CHard4 points
The function hh is defined by h(x)=∫−6xf′(t) dth(x)=\int_{-6}^x f'(t)\,dt. Find the values of h(6)h(6), h′(6)h'(6), and h′′(6)h''(6). Show the work that leads to your answers.

Answer

h(6)=−1.5h(6)=-1.5, h′(6)=−12h'(6)=-\frac{1}{2}, h′′(6)=0h''(6)=0
Full Solution & Work

Find h(6) using the Fundamental Theorem of Calculus

h(6)=∫−66f′(t) dt=f(6)−f(−6)=−1−0.5=−1.5h(6) = \int_{-6}^6 f'(t)\,dt = f(6)-f(-6) = -1-0.5=-1.5

Find h'(6)

h′(x)=f′(x)  ⟹  h′(6)=f′(6)=−12h'(x)=f'(x) \implies h'(6)=f'(6)=-\frac{1}{2}

(the slope of the linear piece of ff on [0,7][0,7], found from the two labeled points (4,2)→(6,−1)(4,2)\to(6,-1)-consistent slope −12-\frac{1}{2})

Find h''(6)

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

(since ff is linear on [0,7][0,7], its second derivative is identically 0 there)

AP Scoring — 4 Points

**P1**: Uses the Fundamental Theorem of Calculus for h(6)h(6).
**P2**: h(6)=−1.5h(6)=-1.5 with supporting work — a bare answer of h(6)=−1.5h(6)=-1.5 does not earn either of the first two points; h(6)=f(6)−f(−6)h(6)=f(6)-f(-6) earns only the first.
**P3**: States either h′(x)=f′(x)h'(x)=f'(x) or h′(6)=f′(6)h'(6)=f'(6), and provides the answer −12-\frac{1}{2}.
**P4**: h′′(6)=0h''(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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