Worksheets · AP Calculus AB · Unit 6

AP Calculus AB Unit 6 FRQ Practice: Integrals and Accumulation

Four exam-style free-response questions on Riemann sums, accumulation functions, the Fundamental Theorem of Calculus and integration by substitution: one with a graphing calculator, three without. Each is scored out of 9, with a scoring guide and a printable PDF with room to write, free to use in class. Good practice before the Unit 6 Progress Check in AP Classroom. Multiple choice for this unit: Unit 6 worksheet.

Sigma Prep · sigmaprep.io/worksheets/ap-calculus-ab

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AP Calculus AB Unit 6 Free Response: Integrals and Accumulation

Date Version A

Directions: Show your work in the space under each part: the setup, the values you use and your reasoning. A justification needs a mathematical reason, with the conditions of any theorem or test you use checked. An answer with no supporting work may earn no credit. Give decimal answers correct to three decimal places unless a part says otherwise. Each question is worth 9 points. Unless a question says otherwise, the domain of a function f is every real number x for which f(x) is a real number.

Part A: A graphing calculator is required.

Question 1

Set your calculator to radian mode. When you use it to solve an equation, find a derivative at a point or evaluate a definite integral, write down the equation, function or integral you used.

  1. 1.
    A desalination plant produces fresh water at the rate F(t) thousand liters per hour, t hours after the start of a work shift. On 0≤t≤10, F is differentiable and increasing; the table below gives F(t) at selected times. All the fresh water the plant produces flows into its reservoir. At the start of the 16-hour shift, the reservoir holds 335 thousand liters, and no water is released from it during the shift.
    t (hours) 014810F(t) (thousand liters per hour) 1720284355
    1. (a)
      Estimate F′(2.5) from the table, and show the computations that lead to your answer. What does this estimate tell you about the situation? Answer using correct units.
    2. (b)
      Using the table's four subintervals, compute a left Riemann sum for ∫010F(t) dt. Show the computations that lead to your answer. Is the sum greater than or less than the exact value of the integral? Give a reason for your answer.
    1. (c)
      For 0≤t≤15, a model of the rate at which the plant produces fresh water is M(t)=19+44sin⁡ ⁣(0.01t2) thousand liters per hour. Based on M, what is the amount of water in the reservoir at time t=15 hours? Write the integral expression you evaluate, and indicate units of measure.
    2. (d)
      The plant pumps seawater in from the ocean at a rate modeled by Q(t)=21+2ln⁡(t+1) thousand liters per hour, for 0≤t≤10. Is there necessarily a time t, with 1<t<4, at which seawater is pumped in and fresh water is produced at the same rate? Justify your answer.

Part B: No calculator is allowed.

Questions 2–4
  1. 2.
    xyGraph of f−2−10123456789−4−3−2−101234
    A function f is continuous for −1≤x≤8, and its graph, drawn above, is made of six straight segments. Use it to define g(x)=∫6xf(t) dt on the same interval.
    1. (a)
      Identify the open intervals on which the graph of g is concave up. Give a reason for your answer.
    2. (b)
      For Q(x)=g(x) f(x), compute Q′(1). Show the work that leads to your answer.
    1. (c)
      Compute lim⁡x→2g(x)x2−4, showing the work that leads to your answer.
    2. (d)
      Find the average rate of change of g over −1≤x≤4. Is there a value c, −1<c<4, for which g′(c) equals this average rate of change? Justify your answer using the Mean Value Theorem.
  2. 3.
    xyGraph of f−5−4−3−2−10123456−3−2−101234
    A function f is continuous for −4≤x≤5, and its graph, drawn above, is made of five straight segments and one semicircle. Use it to define g(x)=∫−1xf(t) dt on the same interval.
    1. (a)
      Find g′(0). Give a reason for your answer.
    2. (b)
      Over −4<x<5, where does the graph of g change concavity? Give a reason for your answer.
    1. (c)
      Use the graph of f to evaluate g(5) and g(−3). Show the work that leads to your answer.
    2. (d)
      Determine where on −4≤x≤5 the function g takes its absolute minimum. Justify your answer.
  3. 4.
    A wheel of cheese dries as it ages and shrinks while staying in the shape of a cylinder. Its radius at time t is r(t) centimeters, with r a twice-differentiable function; here t is the number of weeks since the cheese was made. The table below gives selected values of r′(t), the rate of change of the radius of the wheel, for 0≤t≤13.
    t (weeks) 0481113r′(t) (centimeters per week) −1.2−1.1−0.7−0.3−0.2
    1. (a)
      Approximate r′′(12) from the values of r′ at t=11 and t=13. Show the computations that lead to your answer, including units of measure.
    2. (b)
      Must r′(t) equal −1 at some time t with 0≤t≤11? Justify your answer.
    1. (c)
      Approximate ∫013r′(t) dt with a left Riemann sum built on the four subintervals of the table. Show the work that leads to your answer.
    2. (d)
      The height of the wheel decreases at a constant rate of 0.4 centimeters per week. At time t=11 weeks, the radius of the wheel is 17 centimeters and the height of the wheel is 13 centimeters. Find the rate at which the volume of the wheel is changing at time t=11, in cubic centimeters per week. (A cylinder with radius r and height h has volume V=πr2h.)

      End of Version A

Free AP Calculus AB worksheets at sigmaprep.io/worksheets/ap-calculus-abOriginal questions written by Sigma Prep. AP is a registered trademark of the College Board, which is not affiliated with this sheet.

Scoring guide · Version A

Question 1A Riemann sum from a table of a rate, and a model of the rate · calculator · 9 points

(a) 2 points Table estimate of a derivative · Topic 2.3

  • 1 point: F′(2.5)≈F(4)−F(1)4−1=83
  • 1 point: meaning with the time, the value and units: at t=2.5, the rate at which the plant produces fresh water is increasing at about 83 thousand liters per hour per hour

Solution

  1. t=2.5 is the midpoint of the table values t=1 and t=4, so use the average rate of change of F over that interval.
  2. F′(2.5)≈28−203=83 thousand liters per hour per hour.
  3. At time t=2.5 hours, the rate at which the plant produces fresh water is increasing at a rate of approximately 83 thousand liters per hour per hour.

(b) 3 points Riemann or trapezoidal sum · Topic 6.2

  • 1 point: left Riemann sum:
    1(17)+3(20)+4(28)+2(43)
  • 1 point: approximation 275
  • 1 point: underestimate, with the reason: F is increasing

Solution

  1. A left Riemann sum multiplies the width of each subinterval by the value of F at its left end:
    1(17)+3(20)+4(28)+2(43)=275.
  2. The approximation is an underestimate because F is increasing, so on each subinterval the value at the left end is the least value of F there.

(c) 2 points Initial value plus the integral of a rate · Topic 6.7

  • 1 point: setup: 335+∫015M(t) dt
  • 1 point: answer 962.425 thousand liters (units required)

Solution

  1. The amount at time t=15 is the initial amount plus the integral of the rate: 335+∫015M(t) dt.
  2. On the calculator, ∫015M(t) dt=627.425, so the amount is 335+627.425=962.425 thousand liters.

(d) 2 points IVT justification · Topic 1.16

  • 1 point: F and Q are differentiable, so D=F−Q is continuous on 1≤t≤4
  • 1 point: yes, with D(1) and D(4) of opposite signs (−2.386 and 3.781), by the Intermediate Value Theorem

Solution

  1. Let D(t)=F(t)−Q(t); the two rates are equal exactly when D(t)=0.
  2. On the calculator, Q(1)=22.386 and Q(4)=24.219; the table gives F(1)=20 and F(4)=28.
  3. So D(1)=F(1)−Q(1)=20−22.386=−2.386<0 and D(4)=F(4)−Q(4)=28−24.219=3.781>0: D changes sign on [1,4].
  4. F is differentiable (given), and Q is differentiable (it is built from differentiable functions), so D is differentiable and hence continuous on [1,4].
  5. Yes. By the Intermediate Value Theorem, D(t)=0, that is, F(t)=Q(t), at some time t with 1<t<4: seawater is pumped in and fresh water is produced at the same rate.

Question 2An accumulation function from a graph: concavity, a product rule, L'Hospital's Rule, the Mean Value Theorem · no calculator · 9 points

(a) 2 points Concavity or inflection points · Topic 6.5

  • 1 point: answer:
    (−1,0)∪(2,4)∪(6,7) (and no other intervals)
  • 1 point: reason: g′=f is increasing on these intervals

Solution

  1. By the Fundamental Theorem of Calculus, g′(x)=f(x), so the graph of g is concave up where g′, that is f, is increasing.
  2. From the graph, f is increasing on (−1,0)∪(2,4)∪(6,7), so the graph of g is concave up there.

(b) 3 points Product rule · Topic 2.8

  • 1 point: product rule:
    Q′(x)=g(x)f′(x)+f(x) g′(x)
  • 1 point: g(1)=52 from the signed areas under the graph of f
  • 1 point: answer 32

Solution

  1. Over [1,6] the graph bounds a trapezoid below the axis on [1,2] (area 52), a triangle below the axis on [2,3] (area 32), a triangle above the axis on [3,4] (area 32), a triangle above the axis on [4,5] (area 32) and a triangle below the axis on [5,6] (area 32). Area below the axis counts as negative, so ∫16f(t) dt=−52−32+32+32−32=−52.    The upper limit 1 is less than the lower limit 6, so g(1)=−∫16f(t) dt=52.
  2. By the Fundamental Theorem of Calculus, g′(1)=f(1)=−2, and f′(1)=−1 is the slope of the segment through x=1.
  3. By the product rule, Q′(x)=g(x)f′(x)+f(x) g′(x), so Q′(1)=52⋅(−1)+(−2)⋅(−2)=32.

(c) 2 points L'Hospital's rule · Topic 4.7

  • 1 point: uses L'Hospital's Rule: both lim⁡x→2(g(x))=0 and lim⁡x→2(x2−4)=0, and the ratio of derivatives g′(x)2x
  • 1 point: answer −34 with justification and limit notation

Solution

  1. g(2)=∫62f(t) dt=0 from the signed areas, and g is continuous, so the numerator tends to 0; the denominator tends to 0 as well.
  2. The derivative of the numerator is g′(x)=f(x) and the derivative of the denominator is 2x.
  3. By L'Hospital's Rule the limit is f(2)4=−34=−34.

(d) 2 points MVT justification · Topic 5.1

  • 1 point: average rate of change g(4)−g(−1)4−(−1)=0−1125=−1110
  • 1 point: yes, with justification: g is differentiable (so continuous) because g′=f

Solution

  1. Over [4,6] the graph bounds a triangle above the axis on [4,5] (area 32) and a triangle below the axis on [5,6] (area 32). Area below the axis counts as negative, so ∫46f(t) dt=32−32=0.    The upper limit 4 is less than the lower limit 6, so g(4)=−∫46f(t) dt=0.
  2. Over [−1,6] the graph bounds a trapezoid below the axis on [−1,0] (area 32), a trapezoid below the axis on [0,2] (area 4), a triangle below the axis on [2,3] (area 32), a triangle above the axis on [3,4] (area 32), a triangle above the axis on [4,5] (area 32) and a triangle below the axis on [5,6] (area 32). Area below the axis counts as negative, so ∫−16f(t) dt=−32−4−32+32+32−32=−112.    The upper limit −1 is less than the lower limit 6, so g(−1)=−∫−16f(t) dt=112.
  3. The average rate of change is g(4)−g(−1)4−(−1)=0−1125=−1110.
  4. Yes. g′(x)=f(x) exists for every x in the interval, so g is differentiable on (−1,4) and continuous on [−1,4]. The Mean Value Theorem therefore guarantees a value c, with −1<c<4, such that g′(c)=−1110.

Question 3An accumulation function from a graph of straight and circular pieces: derivative, inflection points, values, absolute extremum · no calculator · 9 points

(a) 2 points Derivative of an accumulation function · Topic 6.4

  • 1 point: uses the Fundamental Theorem of Calculus: g′(x)=f(x)
  • 1 point: answer g′(0)=12

Solution

  1. f is continuous, so by the Fundamental Theorem of Calculus the derivative of g(x)=∫−1xf(t) dt is g′(x)=f(x).
  2. The graph passes through (0,12), so g′(0)=f(0)=12.

(b) 2 points Concavity or inflection points · Topic 6.5

  • 1 point: answer: x=−3, x=−2, x=2, and x=4 (and no other values)
  • 1 point: reason tied to the graph of f: g′=f changes from increasing to decreasing or from decreasing to increasing

Solution

  1. Since g′(x)=f(x), the graph of g changes concavity exactly where g′′=f′ changes sign, that is, where f switches between increasing and decreasing.
  2. From the graph, f is increasing on (−4,−3)∪(−2,2)∪(4,5) and decreasing on (−3,−2)∪(2,4).
  3. The graph of g has a point of inflection at x=−3, x=−2, x=2, and x=4 because g′=f changes from increasing to decreasing at x=−3 and x=2 and from decreasing to increasing at x=−2 and x=4.

(c) 2 points Values from areas under a graph · Topic 6.6

  • 1 point: g(5)=7
  • 1 point: g(−3)=π2

Solution

  1. Over [−1,5] the graph bounds a triangle above the axis on [−1,1] (area 1), a trapezoid above the axis on [1,2] (area 2), a triangle above the axis on [2,4] (area 3) and a triangle above the axis on [4,5] (area 1). So ∫−15f(t) dt=1+2+3+1=7.    So g(5)=∫−15f(t) dt=7.
  2. Over [−3,−1] the graph bounds a semicircle below the axis on [−3,−1] (area π2). Area below the axis counts as negative, so ∫−3−1f(t) dt=−π2.    The upper limit −3 is less than the lower limit −1, so g(−3)=−∫−3−1f(t) dt=π2.

(d) 3 points Absolute extremum (candidates test) · Topic 5.5

  • 1 point: considers g′(x)=f(x)=0
  • 1 point: justification: a candidates test with the values of g at x=−3, x=−1, and x=4 and at both endpoints

    Also acceptable: A sign argument: g′=f≤0 on (−4,−1) and g′=f≥0 on (−1,5), zero at most at isolated points, so the absolute minimum of g is g(−1)=0.

  • 1 point: answer x=−1

Solution

  1. By the Fundamental Theorem of Calculus, g′(x)=f(x), which is 0 inside the interval at x=−3, x=−1, and x=4; the candidates are these and the endpoints x=−4 and x=5.
  2. From the signed areas under the graph of f (counting from x=−1):
    g(−4)=1+π2, g(−3)=π2, g(−1)=0, g(4)=6, g(5)=7.
  3. On −4≤x≤5, g attains its absolute minimum at x=−1.

Question 4A table of the rate of change of a dimension of a solid, then related rates · no calculator · 9 points

(a) 2 points Table estimate of a derivative · Topic 2.3

  • 1 point: r′′(12) with supporting work:
    r′(13)−r′(11)13−11=0.05
  • 1 point: units with the value: r′′(12)≈0.05 centimeters per week per week

Solution

  1. r′′ is the derivative of r′, and t=12 is the midpoint of the neighboring table times t=11 and t=13.
  2. r′′(12)≈r′(13)−r′(11)13−11=−0.2−(−0.3)2=0.05 centimeters per week per week.

(b) 2 points IVT justification · Topic 1.16

  • 1 point: −1 is between r′(0) and r′(11):
    r′(0)=−1.2<−1<−0.3=r′(11)

    Also acceptable: The Intermediate Value Theorem on [4,8] works too:
    r′(4)=−1.1<−1<−0.7=r′(8).

  • 1 point: yes, by the Intermediate Value Theorem, with r′ continuous because it is differentiable

Solution

  1. From the table, r′(0)=−1.2<−1<−0.3=r′(11), so −1 lies between the two values.
  2. r is twice differentiable, so r′ is differentiable, and a differentiable function is continuous.
  3. Yes. By the Intermediate Value Theorem, r′(t)=−1 at some time t with 0<t<11.

(c) 2 points Riemann or trapezoidal sum · Topic 6.2

  • 1 point: form of the left Riemann sum:
    4r′(0)+4r′(4)+3r′(8)+2r′(11)
  • 1 point: answer −11.9

Solution

  1. A left Riemann sum multiplies the width of each subinterval by the value of r′ at its left end.
  2. ∫013r′(t) dt≈4r′(0)+4r′(4)+3r′(8)+2r′(11)=4(−1.2)+4(−1.1)+3(−0.7)+2(−0.3)=−11.9.

(d) 3 points Related rates · Topic 4.4

  • 1 point: product rule on V=πr2h
  • 1 point: chain rule: the factors drdt and dhdt
  • 1 point: answer −248.2π

Solution

  1. Both r and h change with t, so differentiate V=πr2h with respect to t by the product rule and the chain rule:
    dVdt=2πrhdrdt+πr2dhdt.
  2. At t=11: r=17, h=13, drdt=r′(11)=−0.3 from the table, and dhdt=−0.4.
  3. dVdt=2π(17)(13)(−0.3)+π(17)2(−0.4)=−248.2π cubic centimeters per week.

What does the AP Calculus AB free response ask on Unit 6?

A lot: Unit 6 is about 14% of the free-response points since 2012.

This set has 4 questions in that mix, 1 with a calculator (about 27% of Unit 6 free-response points are in the calculator section), each scored out of 9 with a scoring guide.

Counted from the official AP Calculus AB free-response questions, 2012–2026. No official question is reproduced here.

AP Calculus AB Unit 6 free-response settings by share of exam points

Free-response settings on this set and their share of official AP Calculus AB free-response points, 2012–2026
SettingQuestions on this setShare of Unit 6 free-response pointsShare of all free-response pointsLast on the exam
A Riemann sum from a table of a rate, and a model of the rateCalculator1 of 424%–2026
An accumulation function from a graph of straight and circular pieces: derivative, inflection points, values, absolute extremumNo calculator1 of 414%9%–
An accumulation function from a graph: concavity, a product rule, L'Hospital's Rule, the Mean Value TheoremNo calculator1 of 414%9%–
A table of the rate of change of a dimension of a solid, then related ratesNo calculator1 of 49%5%–

A part counts toward a setting when its primary skill is in Unit 6, in the same section, and the setting asks that skill (in the part’s own setting where there is one), on the main exams 2012–2026, by its points; a part that fits two settings is shared between them.

Multiple-choice practice for this unit: AP Calculus AB Unit 6 Worksheet: Integrals and Accumulation.

Why these kinds of question: AP Calculus AB Unit 6 Review: The Most-Tested Riemann Sum and FTC Questions.

Questions about this free-response set

Unit 6, Integration and Accumulation of Change, has 11 AB topics in the College Board's course description: accumulation as the area under a rate graph; left, right, midpoint and trapezoidal sums; Riemann sums, summation notation and the definite integral; the Fundamental Theorem of Calculus and accumulation functions; the behavior of a function defined by an integral; the properties of definite integrals; evaluating definite integrals with antiderivatives; basic antiderivatives and indefinite integrals; integration by substitution; long division and completing the square; and choosing a technique for an antiderivative. Topics 6.11 to 6.13 (integration by parts, partial fractions and improper integrals) are BC only. The course description weights the unit at 15–20% of the multiple-choice section.

Each question is worth 9 points, spread over its three or four parts. Points are earned for specific pieces of work: a setup, an antiderivative, a value with its units, a conclusion with its reason. The scoring guide lists every point and what earns it, then a worked solution for each part. On the exam the six free-response questions are half of the score.

Only where the exam allows one. On this set, question 1 is Part A, where a graphing calculator is required, and questions 2–4 are Part B, without one. A calculator question has parts that need one (a value to three decimal places, an equation solved on the calculator); its other parts can be done by hand, as on the exam.

We tagged every free-response part from 2012 to 2026 by the skill it tests and the setting of its question. The Unit 6 settings the exam leans on most get the most questions here, each in its own context with its own parts.

No. Every question here is original, written by Sigma Prep in the style of the exam: the settings, functions, numbers and wording are our own, and no official question or scoring guideline is copied or reworded. Every value and statement in the scoring guide was checked by an independent computer-algebra check before the set was published.

Yes. Each version has its own: open it under the questions, or download the PDF, which prints the questions with room to write and then the scoring guide on its own pages. There is also a PDF without the scoring guide to hand out.

AP is a registered trademark of the College Board, which was not involved in making these questions and does not endorse them. All questions and scoring guides are original to Sigma Prep. Frozen 2026-10-24.