Course scope. This guide is shared by Calculus AB and Calculus BC. It covers free-response technique for the whole course, not one unit.
How to use this guide
This guide is for students who know the calculus but lose marks in written answers. Read the verb table, work through the three worked examples, then try “Practise it” before opening the answers.
Every example here is original Marlbridge practice, not past exam material. The marks shown are a suggested Marlbridge rubric to help you judge your own work; they are not official scoring.
The free-response section at a glance
| Part | Questions | Time | Calculator | Share of exam score |
|---|---|---|---|---|
| Part A | 2 | 30 minutes | Graphing calculator required | 16.7% |
| Part B | 4 | 60 minutes | Not permitted | 33.3% |
The section lasts 90 minutes and is half of the exam score. You read the questions in the Bluebook app and write your answers by hand in a paper booklet.
- Shared content. The AB and BC exams have three free-response questions in common, built on AB content.
- Where BC differs. The other BC questions can use BC-only content: parametric, polar and vector-valued functions (Unit 9), series (Unit 10) and the BC-only topics in earlier units. These can appear in either part.
- What is assessed. All four course practices. Justification can count as much as calculation, and communication and notation (Practice 4) is assessed only in this section.
- Contexts. At least two questions are set in a real-world situation, so units matter.
Reading a multi-part question
A question has a stem (the set-up, often with a table or graph) and then several parts: (a), (b), (c) and so on. Before you write:
- Read the stem twice. Underline function names, the interval (such as 0 ≤ t ≤ 12), units, and words like “continuous” or “twice-differentiable”. You will need them for justifications.
- Read every part first. Later parts often reuse earlier results.
- Circle the verb in each part. “Find … Justify your answer” is two tasks: a value and a reason.
- Obey method instructions. “Using a trapezoidal sum” or “Use the tangent line” means another method does not earn the point.
- Check the form wanted. “Do not evaluate” means stop at the expression. “Indicate units” makes units part of the answer.
- Notice new information. A later part may add a fact (“For t > 6, …”). It applies from that part on.
Parts are marked separately. If one part goes wrong, keep going; a later part may need only the stem.
The task verbs and what they require
These are the verbs used in free-response questions, described in Marlbridge’s own words.
| Verb | What you must hand in | How much to write |
|---|---|---|
| Approximate | A numerical estimate using rounded values or an approximation method | The expression used, then the value |
| Calculate / Write an expression | An expression or equation that answers the question; for “calculate”, its value too | The expression always; the value unless told not to evaluate |
| Determine / Find | Values, intervals or solutions found with a definition, theorem or test | Visible steps, then the answer |
| Estimate | An approximate value from a model, graph or table | The model or reading used, then the value |
| Evaluate | The value of an expression at a point or over an interval, correctly rounded | The expression with values substituted, then the value |
| Explain / Give a reason | Reasons for a result, based on a definition or theorem | One or two sentences about this function |
| Identify / Indicate | A piece of information, with no explanation | The answer only, clearly labelled |
| Interpret (in context) | What a value or expression means in the situation, usually with units | One full sentence |
| Interpret (a representation) | Information read from a graph, table, formula or description | The value or feature, and where it came from |
| Justify | The definitions, theorems or tests used, why they apply here, and the result | Two to four sentences, conditions checked |
| Represent | A graph, symbols, words or a table showing a relationship | A labelled sketch or expression |
| Verify | A check that a theorem’s or test’s conditions hold, or that a solution is correct | Each condition stated and shown true |
Group 1: Calculate, Write an expression, Evaluate
These want a value or an expression, with the working shown. The expression is part of the answer.
Short example. Write an expression for the average value of √x on [4, 9], then evaluate it. Average = (1/(9 − 4)) ∫₄⁹ √x dx = (1/5)(2/3)(27 − 8) = 38/15. Had the question said “do not evaluate”, the expression alone would be the full answer.
Group 2: Approximate, Estimate
Show the structure of the approximation: the difference quotient, Riemann sum or tangent line.
Short example. f(1.8) = 3.1 and f(2.4) = 4.0. Estimate f′(2). f′(2) ≈ (f(2.4) − f(1.8))/(2.4 − 1.8) = (4.0 − 3.1)/0.6 = 1.5. “1.5” alone hides the method.
Group 3: Determine, Find, Identify, Indicate
“Determine” and “Find” need a visible method. “Identify” and “Indicate” need only the answer. If “Justify” is added, a reason becomes part of the task.
Short example. g is twice differentiable, and g″ changes sign at x = 1 and x = 4 only. Identify the x-coordinates of the points of inflection of g. x = 1 and x = 4. No reason is asked for.
Group 4: Explain, Justify, Verify
A strong justification names the theorem or test, checks that it applies, and states the conclusion for this function.
Short example. g′(x) = (x − 3)eˣ. Does g have a relative minimum at x = 3? Justify. eˣ > 0, so g′ has the sign of x − 3: negative for x < 3, positive for x > 3. g′ changes from negative to positive at x = 3, so g has a relative minimum there.
“Verify” usually means showing that conditions hold, such as continuity and differentiability for the Mean Value Theorem. In BC, it includes checking every condition of a series test.
Group 5: Interpret, Represent
“Interpret” in context turns mathematics into a sentence about the situation, with units. “Interpret” a representation means reading facts from a graph, table or formula. “Represent” asks you to draw or write something: a slope field, a tangent line or (in BC) a Taylor polynomial.
Short example. Velocity v(t) is in metres per second, t in seconds, and ∫₀⁴ v(t) dt = 9. Interpret this. At t = 4 seconds the particle is 9 metres from where it was at t = 0, in the positive direction. It is a change in position, not distance travelled.
Planning your time
Each part gives about 15 minutes per question (30 ÷ 2 and 60 ÷ 4). Use it as a budget.
- Start each part by skimming every question and marking the quick parts.
- Take the sure marks first. If a part stalls for a few minutes, move on and come back.
- Start each part on fresh lines, so you can add a reason or unit later.
- Keep the last few minutes to check units, rounding, and that every “Justify” has a reason.
- BC: series questions can run long because each test’s conditions must be checked.
Showing work on calculations
In the calculator part, if you use the calculator to solve an equation, find a derivative at a point or evaluate a definite integral, you must write the set-up as well as the result: the equation, derivative or integral, in standard notation, not calculator keystrokes. Any other calculator feature needs the mathematical steps shown.
| Situation | Weak | Strong |
|---|---|---|
| Definite integral | “= 153.073” | ∫₀¹⁰ R(t) dt = 153.073 |
| Derivative from a table | “2” | (W(7) − W(3))/(7 − 3) = (32 − 24)/4 = 2 |
| Product rule with given values | “f′(2) = 11” | f′(2) = u′(2)v(2) + u(2)v′(2), values substituted |
| Rounding | 193.07 or 193 | 193.073 |
Rounding. Round or truncate decimal answers to three places after the decimal point. Store intermediate values in the calculator; rounding early can push your final answer out of range.
Exact answers. Without a calculator, exact forms such as 38/15 or √17 are fine. Simplifying further only adds chances for error.
What earns and loses credit
| Earns credit | Loses credit |
|---|---|
| The set-up written before the answer | A bare number, even a correct one |
| A named theorem or test with its conditions checked | “By the theorem” with no conditions |
| Statements about the right function (“f′ changes sign”) | Mixing functions (“f changes sign” when you mean f′) |
| Units in context answers | Missing or wrong units |
| Using your earlier answer consistently in a later part | Leaving a part blank because an earlier one went wrong |
| Correct notation: limits on integrals, dx, + C | “=” between unequal things, lost limits |
| One clear final answer | Two different answers left on the page |
If you change your mind, cross out the old work with one clear line so only one answer remains.
Worked example 1: Justify and Verify with the Mean Value Theorem (Unit 5)
Question (no calculator). The temperature W(t), in °C, of a liquid is a differentiable function of time t, in minutes.
| t (minutes) | 0 | 3 | 7 | 10 |
|---|---|---|---|---|
| W(t) (°C) | 18 | 24 | 32 | 30 |
(a) Justify that there is a time t, 3 < t < 7, at which W′(t) = 2. (b) Explain why W must have an absolute maximum value on 0 ≤ t ≤ 10.
Weak answer. (a) (32 − 24)/(7 − 3) = 2, so W′(t) = 2 by the MVT. (b) Because it goes up and then down.
What it misses. (a) never says why the theorem applies; the conditions come from “W is differentiable”. (b) describes the table, which says nothing about times between the entries.
Strong answer. (a) W is differentiable, so it is continuous on [3, 7] and differentiable on (3, 7). (W(7) − W(3))/(7 − 3) = (32 − 24)/4 = 2. By the Mean Value Theorem, W′(t) = 2 for some t with 3 < t < 7. (b) W is differentiable, so it is continuous on the closed interval [0, 10]. By the Extreme Value Theorem, W has an absolute maximum on [0, 10].
| Point | What earns it (suggested Marlbridge rubric) |
|---|---|
| 1 | (a) Average rate (32 − 24)/(7 − 3) = 2 |
| 1 | (a) Continuity and differentiability stated, Mean Value Theorem named |
| 1 | (b) Continuity on the closed interval, Extreme Value Theorem named |
Worked example 2: Calculate and Interpret with a rate in context (Unit 8)
Question (calculator required). Rainwater flows into a barrel at R(t) = 12 + 8 sin(0.4t) litres per hour, 0 ≤ t ≤ 10, with t in hours. At t = 0 the barrel holds 40 litres. No water leaves.
(a) How many litres are in the barrel at t = 10? (b) Find R′(6). Using correct units, interpret its meaning in context.
Weak answer. (a) 193.07 litres. (b) R′(6) = −2.36. The water in the barrel is decreasing.
What it misses. (a) has no set-up and only two decimal places. (b) has no units and the wrong meaning: R(6) ≈ 17.404 > 0, so water is still flowing in and the amount is rising. It is the rate that is falling.
Strong answer. (a) 40 + ∫₀¹⁰ R(t) dt = 40 + 153.073 = 193.073 litres. (b) R′(6) = −2.360 litres per hour per hour. At t = 6 hours, the rate at which rainwater flows into the barrel is decreasing at a rate of 2.360 litres per hour per hour.
| Point | What earns it (suggested Marlbridge rubric) |
|---|---|
| 1 | (a) Integral ∫₀¹⁰ R(t) dt written |
| 1 | (a) Adds the starting 40 litres; 193.073 |
| 1 | (b) R′(6) = −2.360 |
| 1 | (b) Meaning: the inflow rate is decreasing, at t = 6, with units |
Worked example 3: Determine a particular solution (Unit 7)
Question (no calculator). y = f(t) is the particular solution of dy/dt = (t + 1)/y with f(0) = 3. Find y = f(t).
Weak answer. y dy = (t + 1) dt, so y²/2 = t²/2 + t, giving y = √(t² + 2t). Adding 3 to fit f(0) = 3: y = √(t² + 2t) + 3.
What it misses. The constant of integration belongs at the integration step. Adding 3 at the end produces a function that does not satisfy the differential equation, so credit stops after the antiderivatives.
Strong answer. Separate: y dy = (t + 1) dt. Integrate: y²/2 = t²/2 + t + C. Use f(0) = 3: 9/2 = 0 + 0 + C, so C = 9/2. So y² = t² + 2t + 9, and since y(0) = 3 > 0, y = √(t² + 2t + 9).
| Point | What earns it (suggested Marlbridge rubric) |
|---|---|
| 1 | Separates the variables |
| 1 | Both antiderivatives correct |
| 1 | Constant of integration included when integrating |
| 1 | Initial condition gives C = 9/2 |
| 1 | Solves for y with the positive root |
Common mistakes
- Writing a calculator result with no integral, derivative or equation shown.
- Rounding to fewer than three decimal places, or reusing rounded values.
- “By the MVT” or “by the IVT” with no continuity (or differentiability) stated.
- Giving a reason about the wrong function: “f is positive” when you mean f′.
- Reading a negative rate of a rate as “the amount is decreasing”.
- Units missing, or “litres” given for a rate.
- Adding + C after solving for y, or leaving it out.
- (BC) Naming a series test without checking all its conditions.
Practise it
1. (Unit 1) h is continuous on [0, 6] with h(0) = 1, h(2) = 4, h(5) = −1 and h(6) = 2. Justify that h(x) = 3 has at least two solutions in 0 < x < 5.
Model answer
h is continuous on [0, 2] and on [2, 5]. h(0) = 1 < 3 < 4 = h(2) and h(5) = −1 < 3 < 4 = h(2). By the Intermediate Value Theorem, h(c) = 3 for some c in (0, 2) and for some c in (2, 5). The intervals do not overlap, so there are at least two solutions.
2. (Unit 3) Verify that (1, 2) lies on x² + xy + y² = 7. Find dy/dx there and write an equation for the tangent line.
Model answer
1 + 2 + 4 = 7, so the point is on the curve. Implicitly: 2x + y + x(dy/dx) + 2y(dy/dx) = 0, so dy/dx = −(2x + y)/(x + 2y). At (1, 2): dy/dx = −4/5. Tangent: y = 2 − (4/5)(x − 1).
3. (Unit 4) f(2) = 5, f′(2) = −3 and f″(x) > 0 for all x. Approximate f(2.1) with the tangent line at x = 2. Is it an overestimate or an underestimate? Explain.
Model answer
f(2.1) ≈ 5 − 3(0.1) = 4.7. f″ > 0, so the graph of f is concave up and lies above its tangent line. 4.7 is an underestimate.
4. (Unit 6) Water flows into a pond at R(t) litres per hour.
| t (hours) | 0 | 2 | 5 | 6 |
|---|---|---|---|---|
| R(t) (litres per hour) | 40 | 52 | 46 | 30 |
Approximate ∫₀⁶ R(t) dt with a trapezoidal sum on the three subintervals shown. Interpret the integral, with units.
Model answer
2(40 + 52)/2 + 3(52 + 46)/2 + 1(46 + 30)/2 = 92 + 147 + 38 = 277 litres. The integral is the total water, in litres, that flows into the pond from t = 0 to t = 6 hours.
5. (Unit 10, Calculus BC only) Justify that Σ (−1)ⁿ⁺¹/(n² + 1), n = 1 to ∞, converges. Approximate its sum with the first three terms and bound the error.
Model answer
The series alternates with aₙ = 1/(n² + 1) > 0, aₙ₊₁ < aₙ (the denominators increase) and aₙ → 0. By the Alternating Series Test, it converges. S₃ = 1/2 − 1/5 + 1/10 = 0.4. By the alternating series error bound, |error| ≤ a₄ = 1/17 ≈ 0.059.
Where to practise next
Quick questions
Is this page for Calculus AB or Calculus BC?
Both. The free-response section has the same structure and the same task verbs in both courses. Where the BC exam differs, this page says so.
Do I lose credit for writing more than the question asks?
Extra correct work does not hurt, but extra wrong statements can. If you add a reason that is false, a marker may not be able to award the point. Write what the verb asks for, clearly, and stop.
What if I cannot answer part (a) but need its value in part (b)?
Write down a value you will use, say it is your value from part (a) or an assumed value, and carry on. A marker can only credit a method they can see.