Calculus

The definitive reference. Compiled 15 September 2026 from the corpus at corpus/sm/, for candidates targeting a study score of 45+.

Companion to 01-study-design.md, which is the authority on scope, examination structure and the formula sheet. Nothing here contradicts it; where a number in that document and a number recomputed here differ slightly, the recomputation is flagged.

Evidence tags (same convention as 01-study-design.md)

Provenance of every number below. All percentages are VCAA's own published "percentage of students obtaining full marks" figures, carried in [QJSON] as pct. They are not estimates. Counts and medians were computed on corpus/sm/questions.json as it stood on 15 September 2026 (1,732 graded parts, of which 387 are tagged topic == "Calculus", carrying 623 published marks, of which 158 parts are separators — pct ≤ 50). Fifty of the 387 are Northern Hemisphere Timetable parts for which VCAA publishes no mark distribution and, in most cases, no mark allocation in the report; they are counted as parts but contribute nothing to the mark totals or the medians. The corpus is under active revision; these counts are a dated snapshot, and the composition of the separator list in §4 is the thing to re-derive if questions.json is re-tagged. Where the extraction of a paper is lossy — and it often is — the gap is marked rather than reconstructed. The 2011 papers extracted with a CID-shifted font and are unreadable in plain text; the 2024 November papers extracted empty. Both are flagged at the point of use.

One scope warning before anything else. [SD] places kinematics (rectilinear motion) inside the Calculus area of study, but [QJSON] tags most kinematics parts as Space and measurement — 2009 Exam 1 Q7, 2012 Exam 1 Q8, 2015 Exam 1 Q6, 2022 Exam 1 Q8, 2023 Exam 1 Q3a and 2025 Exam 1 Q3b are all tagged that way, while 2021 Exam 1 Q7b and 2024 Exam 1 Q9b are tagged Calculus. This document follows the tag, so the kinematics catalogue lives mostly in the Space and measurement reference. The two tagged-Calculus kinematics types are covered here (Types 33 and 34), and the a = v·dv/dx = d/dx(½v²) machinery is treated in §6 because it is calculus in everything but filing.

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Contents

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1. What the study design puts in this area

1.1 The content, verbatim

There is no "key knowledge for Calculus" — [SD] attaches key knowledge and key skills to the three outcomes, not to areas of study (see 01-study-design.md §1.1). What Calculus has is content dot points, and they are reproduced here in full.

[SD] overview:

"In this area of study students cover the advanced calculus techniques for analytical and numerical differentiation and integration of a broad range of functions, and combinations of functions; and their application in a variety of theoretical and practical situations, including curve sketching, evaluation of arc length, area and volume, differential equations and kinematics, and modelling with differential equations drawing from a variety of fields such as biology, economics and science."

Topic 1 — Differential calculus and integral calculus. [SD]:

This topic includes:
- the relationship between the graph of a function and the graphs of its anti-derivative functions
- derivatives of inverse circular functions
- second derivatives, use of notations f''(x) and d²y/dx², and their application to the analysis of graphs of functions, including points of inflection and concavity
- applications of chain rule to related rates of change and implicit differentiation; for example, implicit differentiation of the relations x² + y² = 9, 3xy² = x + y and x·sin(y) + x²·cos(y) = 1
- techniques of anti-differentiation and for the evaluation of definite integrals:
- anti-differentiation of 1/x to obtain logₑ(x)
- anti-differentiation of 1/√(a² − x²) and a/(a² + x²) for a > 0 by recognition that they are derivatives of corresponding inverse circular functions
- use of the substitution u = g(x) to anti-differentiate expressions
- use of the trigonometric identities sin²(ax) = ½(1 − cos(2ax)) and cos²(ax) = ½(1 + cos(2ax)) in anti-differentiation techniques
- anti-differentiation using partial fractions of rational functions
- integration by parts
- numerical and symbolic integration using technology
- application of integration, areas of regions bounded by curves, arc lengths for parametrically determined curves, surface area of solids of revolution, volumes of solids of revolution of a region about either coordinate axis.

Topic 2 — Differential equations. [SD]:

This topic includes:
- formulation of differential equations from contexts in, for example, chemistry, biology and economics, in situations where rates are involved (including some differential equations whose analytic solutions are not required, but can be solved numerically using technology)
- the logistic differential equation
- verification of solutions of differential equations and their representation using direction (slope) fields
- solution of simple differential equations of the form dy/dx = f(x), dy/dx = g(y) and in general differential equations of the form dy/dx = f(x)·g(y) using separation of variables and differential equations of the form d²y/dx² = f(x)
- numerical solution by Euler's method (first order approximation).

Topic 3 — Kinematics: rectilinear motion. [SD]:

This topic includes:
- use of velocity–time graphs to describe and analyse rectilinear motion
- application of differentiation, anti-differentiation and solution of differential equations to rectilinear motion of a single particle, including the different derivative forms for acceleration a = d²x/dt² = dv/dt = v·dv/dx = d/dx(½v²).

1.2 The Outcome 1 key skills that bear directly on this area

[SD], Outcome 1 key skills (extracted; the full list is in 01-study-design.md §2.1):

• "apply implicit differentiation, by hand in simple cases"
• "use analytic techniques to find derivatives and anti-derivatives by pattern recognition, and apply anti-differentiation to evaluate definite integrals"
• "set up and evaluate definite integrals to calculate areas, volumes, curve lengths (where described parametrically) and surface areas"
• "set up and solve differential equations of specified forms"
• "sketch by hand and describe behaviour of the graphs of specified functions and relations, and identify their key features, including the use of the first and second derivative"

Four load-bearing phrases. "by hand in simple cases" is an Examination 1 flag on implicit differentiation, and the papers honour it: an implicit-differentiation question appears in Examination 1 in nineteen of the twenty years 2006–2025 (the exception is 2020). "by pattern recognition" is the justification for the fact that VCAA marks down unnecessary substitutions — [RPT19 E1] lists among areas of weakness "not recognising and writing down the anti-derivative of standard functions leading to unnecessary use of substitutions in integration problems (Questions 1 and 8)". "set up and evaluate" is two verbs, and VCAA examines them separately — see §3.1. "where described parametrically" closes arc length, and the formula sheet closed with it: the Cartesian form ∫√(1 + (dy/dx)²) dx was on the 2016–2022 sheet and is not on the current one [FS].

1.3 The formula sheet — what is given and what is not

[FS] (current sheet, Examinations 1 and 2, identical calculus block on both). Given:

The single most consequential line in that table is the right-hand column's first row. There is no volume-of-revolution formula on the sheet. π∫y² dx and π∫x² dy are recall, and the reports show the state losing marks on exactly that: [RPT15 E2] on 2015 Exam 2 Section B Q1fii, "A number of students who set up the integral correctly did not include the π in their calculations and obtained 1.7"; [RPT22 E2] on 2022 Exam 2 Section B Q1dii, "some students who correctly included π in their integral earlier did not include it in their evaluation of the volume".

1.4 What the examinations actually test — the counts

Calculus is the largest or joint-largest area of study by examination weight in every era of the archive.

Under the current design Calculus is, for the first time, the single heaviest area of study — consistent with 01-study-design.md §1.7, which puts it at "about 28% of the tagged marks in 2023–2025" and "about 26% across the whole 2006–2025 archive".

Where the marks sit within a paper.

(Parts exceed marks on Examination 1 because the NHT parts carry no published mark allocation. Medians and means are over the parts for which VCAA published a percentage — i.e. the November sittings.)

Read that first row again. The median Calculus part on the technology-free paper is a separator. Half of all Examination 1 Calculus parts in the archive were earned in full by 43.5% of the state or fewer. Nothing else in the subject behaves like this: for written parts in the 2023–2025 November papers, the median full-mark rates by area are Statistics 72%, Algebra 59%, Calculus 56%, Space and measurement 55%, Functions 47%, Discrete mathematics 29% (n = 3). Calculus is middling on Examination 2 and brutal on Examination 1, and the reason is stated plainly across twenty years of reports: it is the area where exactness, by-hand antidifferentiation and algebraic simplification all have to be right at once.

How Calculus is distributed within each paper. On Examination 1 it is essentially never fewer than three questions and often five, spread through the paper rather than blocked. On Examination 2 Section A it occupies a contiguous run of four to six items — Q8–Q11 in 2006, Q9–Q13 in 2012, Q6–Q11 in 2023, Q7–Q11 in 2025 — which is worth knowing because it lets you budget. On Examination 2 Section B, Calculus is the backbone of one or two of the four or five questions: the "tank / fountain / pond" modelling question is a fixture (2011 Q5, 2014 Q4, 2016 Q3, 2018 Q3, 2023 Q4, 2024 Q3, 2025 Q3), and the "curve, area, volume, surface" question is the other (2006 Q1, 2015 Q1, 2017 Q3, 2019 Q1, 2022 Q1, 2023 Q3, 2024 NHT Q2).

1.5 What changed in 2023

[RPT23 E1], verbatim: "New topics tested in 2023 included integration by parts (Question 5), area of a surface of revolution (Question 7), proof by induction (Question 8) and planes (Question 9)."

Four Calculus changes, all visible in the archive:

1. Integration by parts is in. First examined 2023 Exam 1 Q5 ("Evaluate ∫₁² x² logₑ(x) dx", 3 marks, 53% full marks) and immediately again as a reduction formula in 2023 Exam 2 Section A Q10 (33%). [RPT23 E1] listed it among areas of strength in its first outing — which tells you it is mechanically learnable and therefore not a place a 45+ candidate can afford to drop marks.
2. Surface area of revolution is in, with all four formulas supplied [FS]. Examined 2023 Exam 1 Q7 (31%), 2023 Exam 2 Section A Q11 (46%), 2023 Exam 2 Section B Q3b–d, 2024 NHT Exam 1 Q7, 2024 NHT Exam 2 Section B Q2c, 2025 Exam 2 Section A Q9, 2026 NHT Exam 2 Section A Q3 and Section B Q3d.
3. Cartesian arc length is out. The dot point says "arc lengths for parametrically determined curves"; the key skill says "curve lengths (where described parametrically)"; and the Cartesian formula was removed from the sheet. This retires 2016 Exam 1 Q7 (43%) as representative practice — it asks for the arc length of y = ⅓(x² + 2)^{3/2} from x = 0 to x = 2 in Cartesian form. Do it for the algebra (the perfect square under the surd is the point), not as an exam rehearsal.
4. The logistic equation is named. dP/dt = kP(1 − P/M). Examined 2023 Exam 2 Section B Q4 and 2024 NHT Exam 1 Q9.

Everything else in the Calculus content list is continuous from 2006, which is why the whole twenty-year archive is live practice for this area in a way that it is not for Space and measurement.

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2. The complete catalogue of question types

Thirty-four types. For each: the literal VCAA wording template quoted from a real paper, what is actually being tested, the standard method, archive instances with ref and pct, typical marks, and the traps the reports name.

Percentages in the instance lists are VCAA's published full-marks rates. Where a paper's text is unreadable (2011, 2024 November) the instance is still listed with its percentage, and the wording is quoted from a different year.

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Type 1 — Implicit differentiation: equation of the tangent or normal

Wording template — 2025 Exam 1 Q1 [PAPERS]:

"Consider the curve with equation xe^{2y} + y²e^x = 8e⁴. Find the equation of the tangent to the curve at the point (4, −2)." (4 marks)

and 2017 Exam 1 Q1:

"Find the equation of the tangent to the curve given by 3xy² + 2y = x at the point (1, −1)." (3 marks)

and the normal variant, 2016 Exam 1 Q3:

"Find the equation of the line perpendicular to the graph of cos(y) + y sin(x) = x² at (0, −π/2)." (4 marks)

What it is really testing. Not implicit differentiation — that part the state does well. It is testing (a) whether you apply the product rule and chain rule simultaneously without dropping a term, (b) whether you substitute the point before rearranging, and (c) whether you answer the question asked. [RPT25 E1]: "While the implicit differentiation was often performed successfully, arithmetic errors prevented some students from obtaining the correct gradient. A small number of students who successfully found the value of the gradient at the given point neglected to give the equation of the tangent at that point and were not awarded full marks."

Standard method. Differentiate both sides with respect to x, treating every y as y(x); every y-term generates a dy/dx factor; every xy-product needs the product rule. Substitute the given point immediately. Solve the resulting linear equation for dy/dx. Then y − y₁ = m(x − x₁), or m_normal = −1/m.

[RPT18 E1] states the substitution-first principle outright: "Many students attempted to find an expression for dy/dx in terms of x and y. This was not necessary, with a more effective approach being to substitute x = π/6 and y = π/6 immediately following the implicit differentiation." [RPT21 E1] repeats it.

Instances. 2006 Exam 1 Q1a 56% / Q1b 52%; 2007 Exam 1 Q3 43%; 2008 Exam 1 Q2 48%; 2011 Exam 1 Q10 64%; 2012 Exam 1 Q6 45%; 2013 Exam 1 Q6 40%; 2014 Exam 1 Q4 48%; 2015 Exam 1 Q9a 70% / Q9b 62%; 2016 Exam 1 Q3 53%; 2017 Exam 1 Q1 57%; 2018 Exam 1 Q3 46%; 2019 Exam 1 Q10 18%; 2021 Exam 1 Q5 53%; 2022 Exam 1 Q7 34%; 2023 Exam 1 Q4 37%; 2024 Exam 1 Q8a 54%; 2025 Exam 1 Q1 48%.

Typical marks. 3–4 on Examination 1. It is very often Question 1.

Traps the reports name. - Forgetting that the derivative of a constant is 0. [RPT13 E1]: "Some students forgot that the derivative of a constant was 0, so a c remained on the right-hand side after differentiation, meaning that no significant progress was then possible." - Solving for y first. [RPT13 E1]: "Those who attempted to make y the subject often omitted the ±." [RPT11 E1]: "Some students isolated y and then attempted to use the quotient rule. This pathway tended to be messy and unsuccessful." - Expanding a squared bracket before differentiating. [RPT10 E1]: "Students who unnecessarily expanded the term (y − 2)² before differentiating implicitly often did so incorrectly and were more likely to make algebraic and arithmetic errors." - Answering the wrong question. [RPT08 E1]: "A large number of students went beyond what was asked, finding the equation of the normal rather than its gradient. This wastes precious time and runs the risk of unnecessary errors." [RPT16 E1] records the mirror error: "some thought that the gradient of the normal was equal to the reciprocal rather than the negative reciprocal of the gradient of the tangent." - Using a trigonometric identity to expand first. [RPT22 E1] on Q7 (x cos(x + y) = π/48): "A large number of students used a trigonometric identity to expand before differentiating. Only a minority of students who used this approach were able to find the correct answer, with many students finding themselves overwhelmed by the large number of terms produced using this method."

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Type 2 — Implicit differentiation: find the parameter

Wording template — 2013 Exam 1 Q6 [PAPERS]:

"Find the value of c, where c ∈ R, such that the curve defined by y² + 3e^{(x−1)}/(x − 2) = c has a gradient of 2 where x = 1." (4 marks)

and 2009 Exam 1 Q5:

"Consider the family of curves defined by the relation 3x³ − y² + kx + 5y − 2xy = 4 where k ∈ R. a. Verify that every curve in the family passes through the point (0, 4), and find the other point of intersection with the y-axis. b. Find an expression for dy/dx in terms of x, y and k. c. Hence evaluate the gradient of the curve at the point (1, 1)."

What it is really testing. Whether you notice that the extra condition pins the parameter, and therefore that the parameter must be found before the gradient can be evaluated. [RPT09 E1] on Q5c (22%): "The majority of students did not realise that the information given fixed the value of k, which therefore must be found and used, so there were many answers of −7 − k."

Standard method. Differentiate implicitly, keeping the parameter symbolic. Substitute the point into the original relation to get one equation in the parameter, and into the derivative to get a second. Solve.

Instances. 2009 Exam 1 Q5c 22% (2 marks); 2013 Exam 1 Q6 40% (4 marks); 2010 Exam 1 Q9b 33% (3 marks — the negative root must be chosen); 2022 Exam 2 Section A Q10 21% (see Type 3).

Traps. Leaving the parameter in the answer; substituting the point into only one of the two available equations; choosing the wrong root when the point is defined by a quadratic.

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Type 3 — Implicit differentiation as a multiple-choice inequality

Wording template — 2022 Exam 2 Section A Q10 [PAPERS]:

"Consider the curve given by 5x²y − 3xy + y² = 10. The equation of the tangent to this curve at the point (1, m), where m is a real constant, will have a negative gradient when …" (1 mark)

What it is really testing. Implicit differentiation followed by a sign analysis of a rational expression — which value of m makes numerator and denominator opposite in sign. It is the hardest Section A Calculus item of the current design at 21%.

Standard method. Differentiate implicitly, substitute x = 1, y = m, obtain dy/dx as a rational function of m, then solve the inequality dy/dx < 0 by sign table (on Examination 2, by CAS solve).

Instances. 2022 Exam 2 Section A Q10 21%. Related sign-analysis items: 2024 Exam 2 Section B Q1d.i–iii 27% / 27% / 25%.

Traps. Treating the denominator as always positive; forgetting the value of m that makes the denominator zero.

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Type 4 — Derivatives of inverse circular functions

Wording template — 2009 Exam 1 Q10 [PAPERS]:

"Let f(x) = 2 arcsin((1 − x)/2) − π/3. a. State the implied domain and the range of f. b. Find f'(x) giving your answer in the form f'(x) = a/(b√(x(x + c))) where a, b and c are integers." (3 marks)

and 2017 Exam 1 Q6:

"Let f(x) = 1/arcsin(x). Find f'(x) and state the largest set of values of x for which f'(x) is defined." (3 marks)

and 2013 Exam 1 Q4c:

"Find the gradient of the tangent to the graph of y = arccos(1 − 2x) at x = ¼." (2 marks)

What it is really testing. Whether d/dx arcsin(u) = u'/√(1 − u²) is applied with the chain rule and then simplified into the demanded form, and whether you can state a domain restriction that the algebra hides.

Standard method. Write f(x) with the argument identified, apply the sheet derivative [FS], chain-rule the argument, then rationalise. The "in the form …" instruction means you must combine the surds: √(1 − ((1−x)/2)²) becomes ½√(4 − (1 − x)²) = ½√(x(4 − x) + …), and getting this wrong loses the final mark even with a correct derivative.

Instances. 2009 Exam 1 Q10b 23% (3 marks); 2010 Exam 1 Q5 "Given that f(x) = arctan(2x), find f''(π/2)" (3 marks); 2012 Exam 1 Q5 44%; 2013 Exam 1 Q4c; 2017 Exam 1 Q6; 2020 Exam 1 Q6a 60%; 2006 Exam 1 Q5b 28%.

Traps. - arctan(x) is not 1/tan(x). [RPT12 E1] on Q5: "A surprising number of students interpreted arctan(2x) as (tan⁻¹(2x))⁻¹ = 1/tan(2x)." [RPT16 E1] on Q4 records the same confusion. - The derivative of arcsin is not arccos. [RPT09 E1]: "A few students decided that the derivative of arcsin( ) was arccos( )." - Losing the coefficient inside the argument. [RPT12 E1]: "it was disappointing that some were unable to differentiate an arctan function when the x in the argument has a coefficient other than 1." - The range of arctan is (−π/2, π/2) and the range of arcsin is [−π/2, π/2] — questions exploit this. [RPT06 E1] on Q5b (28%): "A large number of students did not seem to recall that inverse tan function has range (−π/2, π/2). … Some students thought that to find the minimum value they had to differentiate rather than consider the range of the function involved."

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Type 5 — Second derivative computed from a first derivative or a differential equation

Wording template — 2017 Exam 2 Section A Q6 [PAPERS]:

"Given that dy/dx = e^x arctan(y), the value of d²y/dx² at the point (0, 1) is …" (1 mark)

and 2013 Exam 2 Section B Q3d.i:

"Given that dN/dt = 0.4N(6 − logₑ(N)), find d²N/dt²." (2 marks)

and 2023 Exam 2 Section B Q4e.i:

"Given that dQ/dt = (11/10)Q(1 − Q/1000), express d²Q/dt² in terms of Q." (1 mark)

What it is really testing. That d²y/dx² = d/dx(dy/dx) requires the product rule and a chain-rule substitution of dy/dx back into itself. It is the single most-failed differentiation type in the archive: 2013 Exam 2 Section B Q3d.i was earned in full by 9% of the state.

Standard method. Differentiate the expression for dy/dx with respect to the independent variable; wherever a y appears, the chain rule produces a factor dy/dx, which you then replace by the original right-hand side. The result is an expression in y (or N, or Q) alone — which is exactly what makes the inflection question that follows tractable.

Instances. 2013 Exam 2 Section B Q3d.i 9%; 2017 Exam 2 Section A Q6 46%; 2019 Exam 2 Section B Q3b.iii 38%; 2020 Exam 2 Section B Q3d 91%; 2023 Exam 2 Section B Q4e.i; 2024 NHT Exam 1 Q9a (logistic d²P/dt²).

Traps. [RPT13 E2]: "A number of students found d²N/dt² in terms of t, while others found d²t/dN² and attempted to invert the result, which showed little understanding of the properties of second derivatives. A common error was to find d/dN(dN/dt)." That last one — differentiating with respect to the wrong variable — is the signature error.

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Type 6 — Points of inflection and concavity

Wording template — 2026 NHT Exam 1 Q3 [PAPERS]:

"Determine the x-coordinate(s) of any point(s) of inflection of the graph of f(x) = x⁵ + x⁴ − x." (3 marks)

and 2024 NHT Exam 1 Q3:

"a. Find the coordinates of the points of inflection of the graph of y = x⁴ − 6x² + 4. b. Hence, or otherwise, find all values of x for which the graph of y = x⁴ − 6x² + 4 is concave up."

and the multiple-choice form, 2017 Exam 2 Section A Q10:

"A function f, its derivative f' and its second derivative f'' are defined for x ∈ R with the following properties. f(a) = 1, f(−a) = −1, f(b) = −1, f(−b) = 1 and f''(x) = (x + a)²(x − b)/g(x), where g(x) < 0. The coordinates of any points of inflection of f(x) are …" (1 mark)

What it is really testing. That f''= 0 is necessary but not sufficient. A repeated factor does not change sign, so it does not give an inflection. 2017 Exam 2 Section A Q10 is the purest test of this in the archive and was answered correctly by 6% of the state — the lowest Calculus percentage of any multiple-choice item ever set.

Standard method. Solve f'' = 0. For each root, check that f'' changes sign across it — by factor multiplicity, by a sign table, or by testing values. Report coordinates if asked (substitute back into f, not f''). "Concave up" means f'' > 0, and is an interval, not a point.

Instances. 2013 Exam 2 Section B Q3d.ii 38%; 2017 Exam 2 Section A Q10 6%; 2019 Exam 2 Section B Q3b.iii 38% (show there is no point of inflection); 2020 Exam 1 Q6b (show an inflection at x = 2); 2021 Exam 2 Section A Q9 38%; 2024 NHT Exam 1 Q3; 2025 Exam 2 Question 2 (Section A) — the counter-example item quoted in 01-study-design.md §1.4; 2026 NHT Exam 1 Q3.

Traps. - Equating the first derivative to zero. [RPT13 E2]: "there was a small group who equated the first derivative to zero." - Failing to justify a non-inflection. [RPT19 E2] on Q3b.iii: "Most students supplied a correct second derivative but not all of them went on to reasonably justify why the graph does not have a point of inflection." - Drawing an inflection as a stationary point. [RPT25 E2]: "Several responses, incorrectly, sketched the point of inflection as a stationary one."

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Type 7 — Concavity as a condition on a parameter

Wording template — 2017 Exam 2 Section A Q8 [PAPERS]:

"Let f(x) = x³ − mx² + 4, where m, x ∈ R. The gradient of f will always be strictly increasing for …" (1 mark)

and 2024 Exam 2 Section B Q1d (paraphrased from [RPT24 E2], the November 2024 paper text being unavailable):

"i. Find the values of k for which the graph of g has exactly one stationary point. ii. … exactly three stationary points. iii. … exactly five stationary points."

What it is really testing. Translating a geometric statement ("gradient strictly increasing", "exactly three stationary points") into an algebraic condition on a parameter, then solving an inequality rather than an equation.

Standard method. "Gradient strictly increasing" ⇔ f'' > 0. "Exactly n stationary points" ⇔ f'(x) = 0 has exactly n real solutions ⇔ a condition on a discriminant or on the range of a rational function.

Instances. 2017 Exam 2 Section A Q8 29%; 2024 Exam 2 Section B Q1d.i 27%, Q1d.ii 27%, Q1d.iii 25%; 2025 Exam 2 Section B Q1d.ii 45.85%.

Traps. [RPT24 E2] on Q1d.ii: "Many students did not include the equality sign." The boundary case is the whole mark.

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Type 8 — Related rates from a geometric model

Wording template — 2016 Exam 1 Q4 [PAPERS]:

"Chemicals are added to a container so that a particular crystal will grow in the shape of a cube. The side length of the crystal, x millimetres, t days after the chemicals were added to the container, is given by x = arctan(t). Find the rate at which the surface area, A square millimetres, of the crystal is growing one day after the chemicals were added. Give your answer in square millimetres per day." (4 marks)

and 2006 Exam 2 Section B Q1e:

"Hence find an expression for the rate of change of the area, A cm², of the surface of the wine in the upright glass with respect to t, in terms of x. Give your answer in the form dA/dt = ax^b(6 − 3x³)/(1 + x³)^c where a, b and c are constants." (3 marks)

What it is really testing. Constructing a correct chain: dA/dt = (dA/dx)·(dx/dt), where the geometric formula supplies dA/dx and the model supplies dx/dt. Plus — always — whether you wrote down the right geometric formula.

Standard method. Name every variable. Write the geometric relation (A = 6x² for a cube; V = ⅓πr²h with r eliminated by similar triangles for a cone). Differentiate it. Assemble the chain, then substitute the instantaneous values. Never substitute before differentiating.

Instances. 2006 Exam 2 Section B Q1d 75% / Q1e 22% / Q1f 28%; 2009 Exam 2 Section B Q4e 18%; 2011 Exam 2 Section B Q3c (well done); 2014 Exam 2 Section B Q4b 58%; 2016 Exam 1 Q4 40%; 2018 Exam 2 Section B Q3c.i 36% / Q3c.ii 74%; 2024 Exam 2 Section B Q3b 37%.

Typical marks. 2–5.

Traps. - The geometric formula itself. [RPT16 E1]: "Quite a few students made errors with the formula for the surface area of a cube, including A = x², 2x², 4x² or more commonly 3x." - Chain rules built from unrelated variables. [RPT09 E2] on Q4e (18%): "This related rates problem proved to be very difficult for most students. Often chain rule statements were used with variables which were unrelated to the problem." - Ignoring "hence". [RPT06 E2] on Q1e (22%): "Some did not use 'hence' and tried to use dA/dx to find dA/dt, which was a far more complicated approach." - Not converting units. [RPT24 E2] on Q3b (37%): "Students who recognised this as a related rates question managed this well. Some students did not convert the depth measurement to metres."

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Type 9 — Related rates with a rate in and a rate out

Wording template — 2018 Exam 2 Section B Q3 [PAPERS]:

"The fountain is initially empty. A vertical jet of water in the centre fills the fountain at a rate of 0.04 cubic metres per second and, at the same time, water flows out from the bottom of the fountain at a rate of 0.05√h cubic metres per second when the depth is h metres. c. i. Show that dh/dt = (4 − 5√h)/(25π(4h² + 1))." (2 marks)

and 2014 Exam 2 Section B Q4b:

"Water flows in at a constant rate of 0.02 m³/min and flows out at a variable rate of 0.01√h m³/min, where h metres is the depth of the water at any instant. Find the rate, in m/min, at which the depth of the water in the tank is increasing when the depth is 0.25 m." (4 marks)

What it is really testing. dV/dt = (rate in) − (rate out), then dh/dt = (dh/dV)(dV/dt) where dh/dV = 1/(dV/dh).

Standard method. Get V as a function of h (usually a prior "show that" part supplies it). Differentiate to get dV/dh. Invert. Multiply by the net rate. On a "show that", write the product of the derivatives and then show the algebra that turns it into the stated form.

Instances. 2014 Exam 2 Section B Q4b 58%; 2018 Exam 2 Section B Q3c.i 36%; 2018 Exam 2 Section B Q3f 24% (limiting depth from dh/dt = 0); 2025 Exam 2 Section B Q3b 61.55%.

Traps. - Using one rate only. [RPT14 E2]: "Most students attempted this question by using the correct form of the chain rule, but many only used the 'rate in' or the 'rate out', instead of the difference between the rates." - Not showing the algebra on a "show that". [RPT18 E2] on Q3c.i: "Many students moved directly from the product of the derivatives to the required expression, without explicitly showing that their product led to the final (given) answer." - Missing brackets. [RPT18 E2]: "A few students did not understand the importance of brackets when multiplying the derivative expressions."

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Type 10 — Antidifferentiation by pattern recognition into standard forms

Wording template — 2012 Exam 1 Q1 [PAPERS]:

"Find an antiderivative of (6 + x)/(x² + 4)." (2 marks)

and 2006 Exam 1 Q8:

"Find an antiderivative of (2 + 6x)/√(4 − x²)." (4 marks)

What it is really testing. Recognising that a single fraction is a sum of two standard forms, one of which is an inverse-circular antiderivative from [FS] and the other of which is a logₑ or a substitution.

Standard method. Split the numerator: (6 + x)/(x² + 4) = 6/(x² + 4) + x/(x² + 4) → 3 arctan(x/2) + ½logₑ(x² + 4) + c. (2 + 6x)/√(4 − x²) = 2/√(4 − x²) + 6x/√(4 − x²) → 2 arcsin(x/2) − 6√(4 − x²) + c. Note [FS] gives ∫a/(a² + x²) dx = tan⁻¹(x/a) + c — the numerator a, not 1, is why the leading constant so often comes out wrong.

Instances. 2006 Exam 1 Q8 31% (4 marks); 2012 Exam 1 Q1 46% (2 marks); 2021 Exam 1 Q2 63% (3 marks); 2019 Exam 1 Q8 52%.

Traps. - Not splitting at all. [RPT06 E1]: "Many failed to see that the integral should first be split into the sum of two integrals, and many of these proceeded with unproductive substitutions and nonsensical manipulations." - Reaching for partial fractions on an irreducible quadratic. [RPT12 E1]: "A surprising number of students erroneously attempted to use partial fractions, mainly using a difference of squares and sometimes a perfect square. A few attempted to use a difference of squares using complex numbers." - Unnecessary substitution. [RPT21 E1]: "Use of a substitution was unnecessary in this situation and in attempting to use a substitution, some students introduced errors into their working." - The false log rule. [RPT07 E1]: "As in the past, too many students used the incorrect 'log rule' ∫1/f(x) dx = logₑ(f(x))."

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Type 11 — The graph of a function and the graph of its antiderivative

Wording template — 2008 Exam 2 Section A Q12 [PAPERS]:

"The graph of a function f together with the graph of one of its antiderivative functions is shown below. The value of ∫₋₃⁰ f(x) dx is closest to …" (1 mark)

What it is really testing. The first dot point of the Calculus content list — "the relationship between the graph of a function and the graphs of its anti-derivative functions" — read backwards: ∫ₐᵇ f(x) dx = F(b) − F(a), so the value of the definite integral is a difference of heights on the antiderivative graph, not an area to be counted.

Standard method. Identify which curve is F (its stationary points sit at the zeros of f; it is one degree smoother). Read F(0) and F(−3) off the graph. Subtract.

Instances. 2008 Exam 2 Section A Q12 45%; 2021 Exam 2 Section A Q9 38% — [RPT21 E2]: "The antiderivative of the expression in option B is a quartic with a turning point but no point of inflection. Alternatively, the sign of the second derivative changes around x = 3 for option B"; 2015 Exam 2 Section B Q1b.i 91% / Q1b.ii 85%.

Traps. Picking the wrong curve as F; estimating the area under f by counting squares when the antiderivative is drawn right there.

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Type 12 — Integration by substitution, substitution not supplied

Wording template — 2010 Exam 1 Q6 [PAPERS]:

"Evaluate ∫_{π/2}^{3π/4} cos²(2x) sin(2x) dx." (3 marks)

and 2020 Exam 1 Q2:

"Evaluate ∫₋₁⁰ √((1 + x)/(1 − x)) dx. Give your answer in the form a√b + c, where a, b, c ∈ R." (4 marks)

and 2012 Exam 1 Q7:

"Consider the curve with equation y = (x − 1)√(2 − x), 1 ≤ x ≤ 2. Calculate the area of the region enclosed by the curve and the x-axis." (3 marks)

What it is really testing. Choosing the substitution — the only genuinely non-mechanical decision in the whole of VCE integration — and then executing the bookkeeping: du, the new terminals, and the sign.

Standard method. Look for an inner function whose derivative is (a constant multiple of) a factor already present. Set u = g(x), compute du/dx, rewrite dx in terms of du, change the terminals, integrate, evaluate. Never change back to x once the terminals have been changed.

Instances. 2010 Exam 1 Q6 34%; 2011 Exam 1 Q6 47% (u = e^x); 2012 Exam 1 Q7 41% (u = 2 − x); 2015 Exam 1 Q8a 47%; 2020 Exam 1 Q2 28%; 2022 Exam 1 Q9 34%; 2023 Exam 1 Q7 31%; 2025 Exam 1 Q6 38%; 2024 Exam 2 Section A Q9 45%.

Typical marks. 3–4.

Traps. These are the most-repeated sentences in twenty years of reports. - Not changing the terminals. [RPT10 E1]: "Too many students changed the variable correctly but left the terminals unchanged; it should be emphasised that this is not logically correct, even if changing back to the original variable later enables them to obtain a correct answer." [RPT11 E1] repeats the sentence verbatim. [RPT12 E1]: "Several of the students who used the appropriate substitution but kept the x terminals achieved the correct answer by substituting back for x before using the terminals. They could not be awarded full marks due to the inconsistency in their working." - A wrong substitution that "works" slowly. [RPT10 E1]: "Other students tried to use u = sin(2x) or u = cos²(2x), or expanded using double angle formulas and then used u = sin(x) or u = cos(x). Some of these are possible but are very time-consuming, and students who ventured down these paths rarely obtained the correct answer." - Assuming a definite integral must be positive. [RPT10 E1]: "quite a few students got the correct answer but then dropped the negative sign, seemingly assuming that a definite integral had to represent an area." - Notation. [RPT12 E1]: "Common errors included not changing the terminals and an absence of du or continuation of dx throughout, or no indicator at all. … Equals signs must not be placed between quantities that are not equal. A statement such as −4/15 = 4/15 is not valid, nor are statements that equate an indefinite integral with a definite integral." - Splitting instead of substituting. [RPT20 E1] on Q2 (28%): "A number of students split the integral into two … This does not simplify the problem and a substitution is still required in this case."

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Type 13 — Integration by substitution, substitution supplied (Examination 1)

Wording template — 2014 Exam 1 Q5 [PAPERS]:

"a. For the function with rule f(x) = 96 cos(3x) sin(3x), find the value of a such that f(x) = a sin(6x). b. Use an appropriate substitution in the form u = g(x) to find an equivalent definite integral for ∫_{π/36}^{π/12} 96 cos(3x) sin(3x) cos²(6x) dx in terms of u only. c. Hence evaluate ∫_{π/36}^{π/12} 96 cos(3x) sin(3x) cos²(6x) dx, giving your answer in the form kπ, k ∈ Z."

What it is really testing. Note what part b asks for: not the value, but "an equivalent definite integral … in terms of u only". That phrase isolates the change of limits, with the evaluation quarantined into part c, so that a terminal error cannot be hidden by a correct final number.

Standard method. Use part a's identity to simplify first (96 cos 3x sin 3x = 48 sin 6x). Then u = cos(6x), du = −6 sin(6x) dx, terminals u(π/36) = cos(π/6) = √3/2 and u(π/12) = cos(π/2) = 0. The integral becomes 8∫₀^{√3/2} u² du, the sign having been absorbed by swapping the terminals.

Instances. 2014 Exam 1 Q5b 48% / Q5c 45%; 2015 Exam 1 Q8a 47%; 2010 Exam 1 Q7 33% (the derivative is supplied as the hint).

Traps. [RPT14 E1] on Q5b: "the result from Question 5a. was often not used. There were many errors in notation, with dx and du often missing. When performing the required substitution, several students used u = sin(6x), u = sin(3x), u = cos(3x) or u = cos²(6x) rather than u = cos(6x). These attempts led to a more complicated solution and were rarely successful. Many who used u = cos(6x) then stated du/dx = sin(6x), 6 sin(6x) and sometimes 6 sin(x) or hybrids of these. Others failed to change the terminals."

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Type 14 — "Using the substitution …, the definite integral can be expressed as" (Examination 2 Section A)

Wording template — 2025 Exam 2 Section A Q7 [PAPERS]:

"Using the substitution u = cos(θ), ½∫₀^{π/2} sin(2θ)/(1 + cos(θ)) dθ can be expressed as …" (1 mark)

and 2024 NHT Exam 2 Section A Q8:

"Using the substitution u² = x + 1, the definite integral ∫₃⁸ (1/x − 1/(x√(x+1))) dx can be expressed as …"

and 2026 NHT Exam 2 Section A Q9:

"Using the substitution x³ − 1 = u², u ≥ 0, the definite integral ∫₁² √(x³ − 1)/x dx can be expressed as …"

and 2009 Exam 2 Section A Q10:

"Let f : [π, 2π] → R, where f(x) = sin³(x). Using the substitution u = cos(x), the area bounded by the graph of f and the x-axis could be found by evaluating …"

What it is really testing. Exclusively the change of limits and the du algebra. VCAA supplies the substitution so that the only thing left to get wrong is the transformation — and the distractors are built from the four standard errors: unchanged limits, sign dropped, du/dx inverted, and an incomplete substitution that leaves an x behind.

Standard method. Compute du (implicit differentiation is normal here: u² = x + 1 gives 2u du = dx). Substitute everything, including surviving xs, which you eliminate through the inverse of the substitution (x = u² − 1). Change both limits. Compare with the options.

Instances. 2009 Exam 2 Section A Q10 23%; 2012 Exam 2 Section A Q13; 2014 Exam 2 Section A Q13 65%; 2016 Exam 2 Section A Q8 73%; 2022 Exam 2 Section A Q7 68%; 2024 NHT Exam 2 Section A Q8; 2025 Exam 2 Section A Q7 79%; 2026 NHT Exam 2 Section A Q9.

Traps. The u² = … form is the hard variant, because the substitution is implicit and dx picks up a factor of u. This type is generally well answered except where the integrand retains an x that must be re-expressed — which is why 2009 Exam 2 Section A Q10 sat at 23%, with 52% of the state choosing option E.

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Type 15 — Integration using partial fractions

Wording template — 2013 Exam 1 Q2 [PAPERS]:

"Evaluate ∫₀¹ (x − 5)/(x² − 5x + 6) dx." (4 marks)

and 2017 Exam 1 Q2:

"Find ∫₁^{√3} 1/(x(1 + x²)) dx, expressing your answer in the form logₑ(a/b), where a and b are positive integers." (4 marks)

and 2026 NHT Exam 1 Q5:

"Evaluate ∫₀³ (x² + 2x)/(x + 1) dx." (4 marks)

What it is really testing. Whether you factorise the denominator correctly, choose the right form of decomposition, keep the modulus signs, and apply log laws at the end without inventing one.

Standard method. If deg(numerator) ≥ deg(denominator), divide first. Factorise the denominator. Choose the form: distinct linear factors → A/(x−a) + B/(x−b); an irreducible quadratic → A/(x−a) + (Bx + C)/(x² + k); a repeated factor → A/(x−a) + B/(x−a)². Solve by substituting the roots. Integrate to logₑ| | terms. Collect with log laws.

Instances. 2007 Exam 1 Q4 35%; 2009 Exam 1 Q8b 31%; 2010 Exam 1 Q7 33%; 2011 Exam 1 Q1 39%; 2013 Exam 1 Q2 47%; 2014 Exam 1 Q6b 37%; 2017 Exam 1 Q2 35%; 2020 Exam 1 Q8 20%; 2021 Exam 1 Q2 63%; 2023 Exam 2 Section B Q4a 48%.

Typical marks. 3–5. This is the most common Examination 1 integration type of the whole archive.

Traps. - Omitting modulus signs. [RPT13 E1]: "The most common error was the lack of modulus signs leading to the logarithms of negative numbers." [RPT07 E1] shows the consequence: "∫(1/(x+1) − 1/(x−1)) dx = logₑ(x + 1) − logₑ(x − 1). The lack of modulus signs in this case leads to logarithms of negative numbers, so a correct answer cannot be properly obtained." - Losing the sign on a reversed linear factor. [RPT10 E1]: "Quite a few students failed to realise that ∫1/(1 − x) dx is −logₑ(1 − x) and not +logₑ(1 − x)." - The wrong decomposition form. [RPT09 E1]: "An alarming number of students found partial fractions of the form a/(4 − x) + b/(4 + x)." [RPT14 E1]: "incorrectly attempting x²/(x² − 4) = A/(x − 2) + B/(x + 2)" — the numerator degree was not reduced first. [RPT17 E1] records both the wrong and the accidentally right: "Several students used partial fractions of the form A/x + B/(1 + x²) … or A/x + (Bx + C)/(1 + x²), which led to correct partial fractions since the value of C was zero." - Factorising the quadratic wrongly. [RPT13 E1]: "quite a few students were unable to successfully factorise the quadratic in the denominator, often giving x² − 5x + 6 = (x − 6)(x + 1)." - Fake log laws. [RPT13 E1]: "the following 'simplification' occurred often: 2logₑ3 − 5logₑ2 = logₑ(2·3/(5·2))." [RPT11 E1] records the parallel error with modulus terms. - The false log rule. [RPT07 E1]: "As in the past, too many students used the incorrect 'log rule' ∫1/f(x) dx = logₑ(f(x))." Named again in [RPT09 E1] and [RPT10 E1].

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Type 16 — Integrals needing a trigonometric identity first

Wording template — 2013 Exam 1 Q9 [PAPERS]:

"The shaded region below is enclosed by the graph of y = sin(x) and the lines y = 3x/π and x = π/3. This region is rotated about the x-axis. Find the volume of the resulting solid of revolution." (4 marks)

and 2008 Exam 1 Q9b:

"Find the exact volume of the solid of revolution formed if the graph [of y = cos⁻¹(x), x ∈ [−1, 1]] is rotated about the y-axis." (3 marks)

and 2021 Exam 1 Q4a:

"The shaded region … is bounded by the graph of y = sin(x) and the x-axis between the first two non-negative x-intercepts of the curve, that is, the interval [0, π]. The shaded region is rotated about the x-axis to form a solid of revolution. Find the volume, V_s, of the solid formed." (3 marks)

What it is really testing. That sin² and cos² are not directly antidifferentiable, and that [SD] names exactly two identities for the job: sin²(ax) = ½(1 − cos(2ax)) and cos²(ax) = ½(1 + cos(2ax)).

Standard method. Square, apply the identity, integrate the resulting constant − cos(2ax) form, evaluate exactly.

Instances. 2008 Exam 1 Q9b 37%; 2011 Exam 1 Q11 46%; 2013 Exam 1 Q9 29%; 2021 Exam 1 Q4a 58%; 2022 Exam 1 Q10b 25%.

Traps. [RPT08 E1]: "A number of students successfully obtained ∫₀^{π/2} cos²(y) dy but then failed to use a double angle formula to evaluate the integral. Those who did use a double angle formula often had a wrong coefficient or sign on the sin(2y) term." [RPT11 E1]: "integrating sin²(x) to get cos²(x) or ⅓sin³(x)" and "giving the integral of cos(2x) as 2sin(2x)". [RPT13 E1]: "using (3x/π − sin x)² as the integrand was common" — the square of the difference instead of the difference of the squares (Type 20).

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Type 17 — Integration by parts

Wording template — 2023 Exam 1 Q5 [PAPERS]:

"Evaluate ∫₁² x² logₑ(x) dx." (3 marks)

and [SAMPLE] Examination 1 Question 11:

"Find ∫ x² cos(2x) dx." (4 marks)

and 2014 Exam 1 Q7 (the pre-2023 disguised form — by parts was not yet content, so VCAA supplied the derivative):

"Consider f(x) = 3x arctan(2x). a. Write down the range of f. b. Show that f'(x) = 3arctan(2x) + 6x/(1 + 4x²). c. Hence evaluate the area enclosed by the graph of g(x) = arctan(2x), the x-axis and the lines x = 1/2 and x = √3/2." (3 marks)

What it is really testing. Correct choice of u and dv/dx — the whole technique is one decision — plus, for logₑ, the recognition that logₑ(x) must be the part you differentiate.

Standard method. [FS]: ∫u(dv/dx) dx = uv − ∫v(du/dx) dx. Choose u to be the factor that gets simpler when differentiated (logₑ x, arctan x, or a power of x against a trigonometric or exponential factor). For ∫x² cos(2x) dx you apply it twice. For a definite integral, evaluate the uv term at the terminals immediately and keep the remaining integral definite.

Instances. 2023 Exam 1 Q5 53%; 2023 Exam 2 Section A Q10 33% (reduction formula — Type 18); 2025 Exam 1 Q6 38% (substitution or by parts); 2014 Exam 1 Q7c 26%; 2007 Exam 2 Section B Q2d 62% / Q2e 66% (by parts in disguise: "show that ∫arctan(x) dx = x arctan(x) − ∫x/(1 + x²) dx").

Traps. [RPT23 E1]: "Some students did not consistently evaluate the definite integral, and some final responses included the independent variable x. A number of students selected the function to differentiate and the function to antidifferentiate incorrectly. Some idiosyncratic methods were also observed." [RPT14 E1] on the disguised form: "Some ignored the word 'hence'."

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Type 18 — Reduction formula via integration by parts

Wording template — 2023 Exam 2 Section A Q10 [PAPERS]:

"If Iₙ = ∫₀¹ (1 − x)ⁿ eˣ dx, where n ∈ N, then for n ≥ 1, Iₙ equals
A. −1 + nIₙ₋₁ B. nIₙ₋₁ C. −1 − nIₙ₋₁ D. −nIₙ₋₁ E. (1 − x)ⁿeˣ + nIₙ₋₁" (1 mark)

What it is really testing. Integration by parts applied symbolically, with the boundary term evaluated and the remaining integral recognised as Iₙ₋₁. Option E is the trap for candidates who apply parts but never evaluate at the limits.

Standard method. u = (1 − x)ⁿ, dv/dx = eˣ. Then Iₙ = [(1 − x)ⁿeˣ]₀¹ + n∫₀¹(1 − x)ⁿ⁻¹eˣ dx = (0 − 1) + nIₙ₋₁ = −1 + nIₙ₋₁.

Instances. 2023 Exam 2 Section A Q10 33% — the only reduction-formula item in the archive, and new content in its first year.

Traps. Leaving a term containing x in an answer that should be a number; a sign error on d/dx(1 − x)ⁿ.

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Type 19 — Areas of regions bounded by curves

Wording template — 2009 Exam 1 Q8 [PAPERS]:

"a. Show that f(x) = (2 + x²)/(4 − x²) can be written in the form f(x) = −1 + 6/(4 − x²). b. Find the exact area enclosed by the graph of f(x) = (2 + x²)/(4 − x²), the x-axis, and the lines x = −1 and x = 1." (3 marks)

and 2015 Exam 1 Q8d, where f(x) = ½arctan(x):

"Find the area enclosed by the graph of f, the x-axis and the line x = √3." (2 marks)

and 2006 Exam 1 Q3b:

"Find the exact area bounded by y = 36/(2x² − 18), the x-axis and the lines x = −2 and x = 2." (4 marks)

What it is really testing. Setting up the correct integral — which curve, which terminals, which variable — then the sign, then the exactness.

Standard method. Sketch. Decide whether to integrate with respect to x or y; an arctan or arccos region is usually easier as a rectangle minus an integral with respect to the other variable. For a region below the axis, either negate the integral, reverse the terminals, or use | | — but do one of them.

Instances. 2006 Exam 1 Q3b; 2008 Exam 1 Q9a 28%; 2009 Exam 1 Q8b 31%; 2010 Exam 1 Q10 23%; 2012 Exam 1 Q7 41%; 2014 Exam 1 Q7c 26%; 2015 Exam 1 Q8d 23%; 2017 Exam 2 Section B Q3c 61%; 2020 Exam 1 Q7b.

Traps. - Sign of an area. [RPT10 E1] on Q10 (23%): "a large number did not recognise that area is a positive quantity, and did not place a negative sign in front of their integral, use absolute values, or reverse the terminals. As the area was below the x-axis, it was essential that one of these techniques be applied — this is assumed knowledge from Mathematical Methods CAS Units 3 and 4. Some students presented a negative answer as the area. Some simply dropped the negative sign or wrote A = −32√2/35 = 32√2/35, which is not a reasonable statement." - Missing the symmetry shortcut. [RPT08 E1] on Q9a (28%): "This question was well done if symmetry was noticed (i.e. the required area was simply half the area of the encompassing rectangle). If symmetry was not noticed, success was rare. Some laborious attempts at integration were seen." - Ignoring a supplied part. [RPT15 E1] on Q8d (23%): "Some students tried to integrate tan(2y) rather than using the information contained in part a. but were usually unsuccessful. … Using a diagram would have been helpful for many students." - Splitting unnecessarily. [RPT10 E1]: "Quite a few split the area into two separate parts unnecessarily, integrating from −1 to the y-axis and adding the integral from the y-axis to 1."

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Type 20 — Area or volume between two curves

Wording template — 2013 Exam 1 Q9 [PAPERS]:

"The shaded region below is enclosed by the graph of y = sin(x) and the lines y = 3x/π and x = π/3."

and 2018 Exam 1 Q9:

"b. Find the x-coordinates of the points of intersection of the curve x² − 2y² = 1 and the line y = x − 1. c. Find the volume of the solid of revolution formed when the region bounded by the curve and the line is rotated about the x-axis." (2 marks)

and 2026 NHT Exam 1 Q4b:

"The shaded region shown above is bounded by the two curves and the line with equation x = 1. This region is rotated about the x-axis to form a solid of revolution. Calculate the volume of this solid." (2 marks)

What it is really testing. ∫(upper − lower) for area, but π∫(outer² − inner²) for volume — and the state persistently conflates them.

Standard method. Find the intersection points; those are the terminals. Determine which curve is upper on the interval. For area, ∫(f − g) dx. For volume about the x-axis, π∫(f² − g²) dx — the difference of the squares, never the square of the difference.

Instances. 2013 Exam 1 Q9 29%; 2018 Exam 1 Q9b 70% / Q9c 19%; 2022 Exam 2 Section B Q1d.i 45% / Q1d.ii 37%; 2026 NHT Exam 1 Q4b.

Traps. [RPT22 E2] on Q1d.i: "A significant number of responses incorrectly contained the integrand (f(x) − g(x))², i.e. students stated the square of the difference rather than the difference of the squares." [RPT13 E1] records the same error nine years earlier. [RPT18 E1] on Q9c (19%) notes the elegant alternative most of the state missed: "A small number of students realised that the volume required could be found by finding the volume obtained by rotating the region bounded by the hyperbola, the x-axis and the lines x = 1 and x = 3 about the x-axis and then subtracting the volume of an appropriate cone."

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Type 21 — Volume of revolution about the x-axis

Wording template — 2019 Exam 1 Q8 [PAPERS]:

"Find the volume of the solid of revolution formed when the graph of y = √((1 + 2x)/(1 + x²)) is rotated about the x-axis over the interval [0, 1]." (4 marks)

and 2020 Exam 1 Q8:

"Find the volume, V, of the solid of revolution formed when the graph of y = 2√((x² + x + 1)/((x + 1)(x² + 1))) is rotated about the x-axis over the interval [0, √3]. Give your answer in the form V = 2π(logₑ(a) + b), where a, b ∈ R." (5 marks)

and 2006 Exam 1 Q6:

"The region in the first quadrant enclosed by the coordinate axes, the graph with equation y = e^{−x} and the straight line x = a where a > 0, is rotated about the x-axis to form a solid of revolution. a. Express the volume of the solid of revolution as a definite integral. b. Calculate the volume of the solid of revolution, in terms of a. c. Find the exact value of a if the volume is 5π/18 cubic units."

What it is really testing. Recall of V = π∫ᵃᵇ y² dx — which is not on [FS] — followed by whichever integration technique the squared function demands. The square root in the function definition is deliberate: it clears on squaring, leaving a partial-fractions or standard-form integral.

Standard method. V = π∫ᵃᵇ [f(x)]² dx. Square first. Classify the integrand (split / partial fractions / substitution / double angle). Evaluate exactly.

Instances. 2006 Exam 1 Q6a 70% / Q6b 60% / Q6c 49%; 2007 Exam 1 Q4 35%; 2011 Exam 1 Q11 46%; 2012 Exam 2 Section A Q12 48%; 2012 Exam 2 Section B Q1c 57%; 2013 Exam 1 Q9 29%; 2014 Exam 1 Q6b 37%; 2015 Exam 2 Section B Q1f.i 79% / Q1f.ii 71%; 2017 Exam 1 Q10c 14%; 2018 Exam 1 Q9c 19%; 2019 Exam 1 Q8 52%; 2020 Exam 1 Q8 20%; 2021 Exam 1 Q4a 58%; 2022 Exam 1 Q10b 25%; 2023 Exam 2 Section B Q3a.i 89% / Q3a.ii 85%; 2024 Exam 1 Q5 59%; 2025 Exam 1 Q6 38%; 2025 Exam 2 Section A Q10 69%.

Typical marks. 3–5 on Examination 1; 1–2 for the set-up on Examination 2.

Traps. - Forgetting to square. [RPT15 E2]: "Common errors included not squaring f(x), incorrect terminals and the occasional omission of π and/or dx." - Forgetting π. Recorded in [RPT07 E1], [RPT08 E2], [RPT11 E1], [RPT15 E2], [RPT22 E2] and [RPT23 E2]. [RPT23 E2] on Q3a.ii: "Some students did not include π in their answer, despite it being present in their integral expression in Question 3a.i." - Applying a surface-area formula instead. [RPT24 E1] on Q5: "Some students tried to apply a formula for the surface area of the solid." [RPT23 E2] on Q3a.i: "Some students incorrectly applied a formula for surface area." - Ignoring a supplied earlier part. [RPT14 E1] on Q6b (37%): "Many students did not use the result from Question 6a. … Students are reminded that it is often necessary or beneficial to use the results from earlier parts of a question in the latter parts." - Unnecessary substitution. [RPT19 E1] on Q8 (52%): "Many students who were able to successfully split the integrand used a substitution method to integrate. This was unnecessary and resulted in a loss of marks if not done correctly."

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Type 22 — Volume of revolution about the y-axis

Wording template — 2015 Exam 1 Q5 [PAPERS]:

"Find the volume generated when the region bounded by the graph of y = 2x² − 3, the line y = 5 and the y-axis is rotated about the y-axis." (3 marks)

and 2008 Exam 1 Q9b:

"Find the exact volume of the solid of revolution formed if the graph [of y = cos⁻¹(x)] is rotated about the y-axis." (3 marks)

and 2018 Exam 2 Section B Q3a:

"The curve shown is rotated about the y-axis to form a volume of revolution that is to model a fountain, where length units are in metres. Show that the volume, V cubic metres, of water in the fountain when it is filled to a depth of h metres is given by V = (π/4)(4h³/3 + h)." (2 marks)

and 2024 NHT Exam 2 Section B Q2b.i:

"Write down a definite integral in terms of y that gives the volume of the solid of revolution." (1 mark)

What it is really testing. Inverting the relation — you need x as a function of y — and then integrating with respect to y with y-terminals. [RPT08 E1]: "Many students were unable to start this question, not realising that x = cos(y). Some attempts involved a definite integral with dx instead of dy."

Standard method. Solve for x² in terms of y. V = π∫_{y₁}^{y₂} x² dy. The terminals are y-values; read them off the diagram or from the boundary lines. For a variable upper limit (the fountain), leave it as h and evaluate symbolically.

Instances. 2008 Exam 1 Q9b 37%; 2011 Exam 2 Section B Q3b.ii 65%; 2015 Exam 1 Q5; 2018 Exam 2 Section B Q3a 32%; 2019 Exam 2 Section B Q1e 3%; 2024 NHT Exam 2 Section B Q2b.i; 2025 Exam 2 Section A Q10 69%.

Traps. [RPT11 E2] on Q3b.ii: "Errors included the use of π/2 and 2π instead of π in the definite integral, the omission of π, and the wrong terminals, indicating that students could not determine which region was rotated about the y-axis." [RPT18 E2] on Q3a (32%): "Approximately half of the students were able to either set up an appropriate definite integral or find an antiderivative and attempt to evaluate the constant of integration. Of these, many did not explicitly show that the first part of their response yielded the required volume."

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Type 23 — Arc length of a parametrically defined curve

Wording template — 2017 Exam 1 Q7 [PAPERS]:

"The position vector of a particle moving along a curve at time t is given by r(t) = cos³(t) i + sin³(t) j, 0 ≤ t ≤ π/4. Find the length of the path that the particle travels along the curve from t = 0 to t = π/4." (4 marks)

and 2018 Exam 1 Q10:

"The distance d metres that the particle travels along the curve in three-quarters of a second is given by d = ∫₀^{3/4} √(at² + bt + c) dt. Find a, b and c, where a, b, c ∈ Z." (5 marks)

and 2026 NHT Exam 2 Section B Q3c.ii:

"Find the length of the curve given by y = eˣ − 1 for 0 ≤ x ≤ logₑ(5). Give your answer correct to one decimal place." (1 mark)

What it is really testing. On Examination 1, almost always the perfect square hidden under the surd. (dx/dt)² + (dy/dt)² is engineered to collapse to (something)² so the root can be removed by hand.

Standard method. [FS]: L = ∫_{t₁}^{t₂} √((dx/dt)² + (dy/dt)²) dt. Differentiate both components. Square and add. Look for the perfect square before doing anything else — expand, collect, and try to write it as (A ± B)². Take the positive root, checking the sign of the bracket on the given domain. Integrate.

Instances. 2016 Exam 1 Q7 43% (Cartesian — retired in 2023); 2017 Exam 1 Q7 30%; 2017 Exam 2 Section B Q3e 16%; 2018 Exam 1 Q10; 2020 Exam 1 Q9b 14%; 2020 Exam 2 Section B Q1d 54%; 2026 NHT Exam 2 Section B Q3c.ii.

Traps. - Not seeing the perfect square. [RPT20 E1] on Q9b (14%): "Most students who successfully answered this question were able to identify the perfect square, which allowed the square root in the integrand to be removed. … Few students who tried to write the term inside the square root as a single algebraic fraction were able to see the problem through to the conclusion." - Taking the square root of individual terms. Recorded in both [RPT16 E1] and [RPT17 E1]. - Using the wrong form of the formula. [RPT17 E1] on Q7 (30%): "there were many who did not recognise the appropriate form of the arc length formula. Some incorrect answers involved: finding |r(π/4) − r(0)|, using the formula with dy/dx (sometimes with correct working, except for using dt rather than dx)." - dt omitted. Named in [RPT16 E1], [RPT17 E1] and [RPT20 E2].

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Type 24 — Surface area of a solid of revolution

Wording template — 2023 Exam 1 Q7 [PAPERS]:

"The curve defined by the parametric equations x = t²/4 + 1, y = 3 − t, where 0 ≤ t ≤ 2, is rotated about the x-axis to form an open hollow surface of revolution. Find the surface area of the surface of revolution. Give your answer in the form π((a√b)/c − d) where a, b, c, d ∈ Z⁺." (4 marks)

and 2024 NHT Exam 1 Q7:

"The curve defined by the parametric equations x = 8t and y = t² − 8logₑ(t), where t ∈ [1, 3], is rotated about the y-axis to form a surface of revolution. Find the area of the surface of revolution formed. Give your answer in the form aπ/b where a, b ∈ Z⁺." (4 marks)

and 2023 Exam 2 Section B Q3b.i:

"Express the curved surface area of the solid in the form S = ∫ₐᵇ π√(Ax + B) dx, where a, b, A, B are all positive integers." (2 marks)

and 2025 Exam 2 Section A Q9:

"A parametric curve is given by x = kt, y = e^{kt}, where k is a positive constant. The curve is rotated about the x-axis from t = a to t = b, where b > a, to form a surface of revolution. The area of this surface is given by …" (1 mark)

What it is really testing. Choosing the right one of the four formulas on [FS] — Cartesian/parametric × about-x/about-y — and noticing that the multiplier is 2πy for rotation about x and 2πx for rotation about y. New content in 2023, examined every year since.

Standard method. Pick the formula. Substitute the two derivatives. Simplify under the root (a perfect square is common). Integrate, usually by substitution.

Instances. 2023 Exam 1 Q7 31%; 2023 Exam 2 Section A Q11 46%; 2023 Exam 2 Section B Q3b.i 62% / Q3b.ii 55% / Q3c 38% / Q3d 24%; 2024 NHT Exam 1 Q7; 2024 NHT Exam 2 Section B Q2c; 2025 Exam 2 Section A Q9; 2026 NHT Exam 2 Section A Q3 and Section B Q3d.i–ii.

Traps. - Curved surface only. [RPT23 E2] on Q3c (38%): "Many students found the curved surface area only and did not include one or both ends. Of those who included two ends, errors with an incorrect radius were frequent." - Relying on recognition instead of substituting. [RPT23 E1] on Q7 (31%): "Some students did not use a substitution and instead tried to rely on inspection or recognition to find an antiderivative. This approach was not always successful. Doing an explicit substitution was the more reliable approach." - Brackets in the calculator. [RPT23 E2] on Q3d (24%): "Some errors in the final value appeared to be due to a lack of brackets when entering expressions into a calculator."

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Type 25 — Solving a differential equation by separation of variables

Wording template — 2019 Exam 1 Q1 [PAPERS]:

"Solve the differential equation dy/dx = 2ye^{2x}/(1 + e^{2x}) given that y(0) = π." (4 marks)

and 2016 Exam 1 Q10:

"Solve the differential equation √(2 − x²)·dy/dx = 1/(2 − y), given that y(1) = 0. Express y as a function of x." (5 marks)

and 2022 Exam 1 Q2:

"Solve the differential equation dy/dx = x√(4 − y²) given that y(2) = 0. Give your answer in the form y = f(x)." (3 marks)

and 2017 Exam 1 Q8b:

"Solve the differential equation dy/dx = −x/(1 + y²) with the condition y(−1) = 1. Express your answer in the form ay³ + by + cx² + d = 0, where a, b, c and d are integers." (2 marks)

What it is really testing. Four things in sequence, each independently losable: separate correctly, integrate both sides (usually one side is a standard inverse-circular or log form), evaluate the constant from the initial condition, and rearrange into the demanded form with the right branch.

Standard method. Write g(y) dy = f(x) dx. Integrate both sides with a single constant. Substitute the initial condition immediately — before rearranging. Then solve for y, choosing the branch consistent with the initial condition.

Instances. 2006 Exam 1 Q2 45%; 2007 Exam 1 Q7b 33% / Q8b 46%; 2009 Exam 1 Q9a 34%; 2010 Exam 2 Section B Q3c 68%; 2013 Exam 1 Q5b 41%; 2016 Exam 1 Q10 14%; 2016 Exam 2 Section B Q3a 42%; 2017 Exam 1 Q8b 36%; 2018 Exam 1 Q8b 23%; 2019 Exam 1 Q1 42%; 2019 Exam 2 Section B Q3b.ii 69%; 2021 Exam 1 Q7a 66%; 2022 Exam 1 Q2 60%; 2022 Exam 2 Section B Q3a.i 85% / Q3a.ii 73%; 2024 Exam 1 Q7 27%; 2025 Exam 2 Section B Q3d 59.3%.

Typical marks. 3–5 on Examination 1. It is the most common single Calculus type on that paper.

Traps. - The constant of integration. Named in [RPT06 E1], [RPT07 E1], [RPT09 E1], [RPT13 E1], [RPT16 E1], [RPT16 E2], [RPT18 E1], [RPT19 E2] and [RPT25 E2]. [RPT16 E2]: "Errors with constants were common among students who added a constant to both sides of the expression before attempting to find its value." [RPT25 E2] on Q3d: "A frequently seen error was not using the initial condition to find the constant of integration." - Choosing the wrong root. [RPT16 E1] on Q10 (14%): "A large number of students, when confronted with a square equals a constant, gave only the positive root. Many gave both roots but did not realise that only the negative root satisfied the initial conditions." [RPT24 E1] on Q7 (27%): "Some students failed to choose the correct sign." - Reciprocating a sum. [RPT09 E1] on Q9a (34%): "several going from dx/dy = (y + 2)² + 4 to dy/dx = 1/(y+2)² + 1/4, which is a typical fraction reciprocal problem." - Not answering in the demanded form. [RPT17 E1] on Q8b (36%): "Most students had the correct integration after separating variables but made no attempt to express the answer with integers as required." [RPT24 E1]: "Some students did not give their answer in the form of y as a function of x as required by the question." - Misreading the initial condition. [RPT07 E1]: "others found an incorrect value for the constant by using (1.1, 1.1) as the initial condition rather than (1, 1)." [RPT16 E1]: "A number of students interpreted y(1) = 0 as x = 0 when y = 1." - Stopping early. [RPT07 E1]: "The other common errors were stopping after finding c, perhaps not comprehending what the question asked."

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Type 26 — Formulating a differential equation from a rate context

Wording template — 2018 Exam 1 Q8a [PAPERS]:

"A tank initially holds 16 L of water in which 0.5 kg of salt has been dissolved. Pure water then flows into the tank at a rate of 5 L per minute. The mixture is stirred continuously and flows out of the tank at a rate of 3 L per minute. Show that the differential equation for Q, the number of kilograms of salt in the tank after t minutes, is given by dQ/dt = −3Q/(16 + 2t)." (1 mark)

and 2025 Exam 2 Section B Q3b:

"Let Q denote the quantity of salt, in kilograms, in the tank at time t minutes. Show that Q satisfies the differential equation dQ/dt = (300 − Q)/150." (1 mark)

and the multiple-choice form, 2023 Exam 2 Section A Q8:

"Initially a spa pool is filled with 8000 litres of water that contains a quantity of dissolved chemical. … 20 litres of well-mixed spa pool water is pumped out every minute while 15 litres of fresh water is pumped in each minute. Let Q be the number of kilograms of chemical that remains dissolved in the spa pool after t minutes. The differential equation relating Q to t is …" (1 mark)

and 2006 Exam 2 Section A Q10:

"A chemical dissolves in a pool at a rate equal to 5% of the amount of undissolved chemical. Initially the amount of undissolved chemical is 8 kg and after t hours x kilograms has dissolved. The differential equation which models this process is …" (1 mark)

What it is really testing. dQ/dt = (rate in) − (rate out), where rate out = (concentration) × (outflow rate), and the concentration has a time-varying denominator whenever inflow and outflow rates differ: V(t) = V₀ + (in − out)t.

Standard method. Write V(t) first. Write concentration = Q/V(t). Rate out = (Q/V(t)) × (outflow rate). Rate in = (incoming concentration) × (inflow rate), which is zero for pure water — and saying so explicitly is worth a mark.

Instances. 2006 Exam 2 Section A Q10 47%; 2007 Exam 2 Section A Q14 37% (logistic form); 2009 Exam 2 Section A Q13 38%; 2011 Exam 2 Section B Q5b 45%; 2016 Exam 2 Section B Q3b 34% / Q3c 40%; 2018 Exam 1 Q8a 44%; 2023 Exam 2 Section A Q8 37%; 2025 Exam 2 Section B Q3b 61.55%.

Traps. - Not stating that the rate in is zero. [RPT18 E1] on Q8a (44%): "This problem required students to recognise a difference of rates. The most common error was a failure to explicitly note that the rate in was zero." - Confusing the amount with the concentration. [RPT16 E2] on Q3b (34%): "Many students did not demonstrate an understanding of what was required by this question. Students frequently found an expression for dy/dt rather than the concentration at time t." 2009 Exam 2 Section A Q13 (38%) is the multiple-choice version of exactly this distinction. - Working backwards from the answer. [RPT11 E2] on Q5b (45%): "Many students tried to work backwards from the differential equation and did not understand where the terms came from. The expression for the 'rate out' was problematic for many." - Starting from the given result. [RPT16 E2] on Q3c (40%): "Some students incorrectly started with the given expression with no explanation of its origin. Students frequently did not seem to realise that work done for Question 3b. was useful here."

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Type 27 — The logistic differential equation

Wording template — 2023 Exam 2 Section B Q4 [PAPERS]:

"A fish farmer releases 200 fish into a pond that originally contained no fish. The fish population, P, grows according to the logistic model, dP/dt = P(1 − P/1000), where t is the time in years after the release of the 200 fish. a. The above logistic differential equation can be expressed as ∫(A/P + B/(1 − P/1000)) dP = ∫dt, where A, B ∈ R. Find the values of A and B." (1 mark)

and 2024 NHT Exam 1 Q9:

"In a population, the number of kangaroos, P, is modelled by the logistic differential equation dP/dt = 0.05P(10 − P/50), where P ∈ [100, 500) … a. Find d²P/dt² in terms of P. b. In the context of the growth of the kangaroo population, what does the point of inflection at A represent? c. Solve the logistic differential equation to show that t = 2logₑ(4P/(500 − P)). d. Hence, or otherwise, find the coordinates of point A."

What it is really testing. Three linked skills: partial-fraction decomposition of 1/(P(1 − P/M)); the sigmoid shape (carrying capacity M as a horizontal asymptote, maximum growth rate at P = M/2, which is the point of inflection); and the contextual interpretation.

Standard method. Separate: ∫dP/(P(1 − P/M)) = ∫k dt. Decompose; the two partial fractions are 1/P and (1/M)/(1 − P/M). Integrate to logs, combine, exponentiate, solve for P, giving P = M/(1 + De^{−kt}). For the inflection, differentiate the logistic right-hand side and set it to zero: P = M/2 always.

Instances. 2007 Exam 2 Section A Q14 37% (the pre-2023 disguised version); 2023 Exam 2 Section B Q4a 48%, Q4b, Q4c, Q4d, Q4e.i–ii, Q4f; 2024 NHT Exam 1 Q9a–d; [SAMPLE] Examination 1 Question 10.

Traps. Getting B wrong because the second factor is (1 − P/1000), not (1000 − P); forgetting that P = M/2 is the population at maximum growth and that the question may want the time; sketching a logistic curve without labelling the horizontal asymptote.

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Type 28 — Verifying a given solution of a differential equation

Wording template — 2016 Exam 2 Section B Q3d [PAPERS]:

"Verify by differentiation and substitution into the left side that y = (t² + 20t + 900)/(6(10 + t)) satisfies the differential equation in part c. Verify that the given solution for y also satisfies the initial condition." (3 marks)

and 2010 Exam 2 Section B Q3a, verifying that P = 20000(4 − 3e^{0.01t}) satisfies both dP/dt = −600e^{0.01t} and P(0) = 20000 (3 marks); and 2008 Exam 2 Section B Q4d.ii:

"Use calculus to verify that the curve with equation 25logₑ(y) − 5y − x + 10logₑ(x) = c …"

What it is really testing. Reading the instruction. "Verify" means substitute and show equality. It does not mean "solve". The reports are unusually blunt about this, and the percentages are the lowest in the area: 2010 Exam 2 Section B Q3a was earned in full by 5% of the state.

Standard method. Differentiate the given solution. Substitute into the left side only. Simplify line by line until it equals the right side. Then, separately, substitute the initial value and confirm the stated result. Both halves are marked.

Instances. 2010 Exam 2 Section B Q3a 5%; 2011 Exam 1 Q2 49%; 2013 Exam 2 Section B Q3a 30%; 2016 Exam 2 Section B Q3d 17%; 2022 Exam 2 Section B Q3d 77%.

Traps. - Solving instead of verifying. [RPT10 E2]: "Many students attempted to solve the differential equation, despite this being asked for in part c. of this question. Some students who solved instead of verifying the solution by substitution, verified the initial condition." [RPT13 E2]: "Many students employed a variety of approaches different to the required method of substitution." - Skipping the initial condition. [RPT16 E2] on Q3d (17%): "It was not always clear how expressions for the left side simplified to the right side. Verification that the given solution satisfied the initial conditions was often absent."

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Type 29 — Differential equations of the form d²y/dx² = f(x)

Wording template — 2010 Exam 1 Q7 [PAPERS]:

"Consider the differential equation d²y/dx² = 4x/(1 − x²)², −1 < x < 1, for which dy/dx = 3 when x = 0, and y = 4 when x = 0. Given that d/dx(12x²/…) = 4x/(1 − x²)², find the solution of this differential equation." (3 marks)

and 2008 Exam 1 Q6:

"The curve with equation y = f(x) passes through the point P(π/8, 2) and has a gradient of −1 at this point. Find the exact gradient of the curve at x = π/12 given that f''(x) = −sec²(2x)." (3 marks)

and 2022 Exam 1 Q9:

"Given that f'(x) = cos(2x)/sin³(2x) and f(π/8) = 3/4, find f(x)." (4 marks)

and 2026 NHT Exam 1 Q4a:

"Consider the differential equation 2·d²y/dx² − dy/dx − y = 0. Two solutions of the differential equation have the form y = e^{mx}, where m ∈ R. Show that m = −½ or m = 1." (2 marks)

What it is really testing. Integrating twice, with two constants, evaluated from two conditions — and, in 2008 Exam 1 Q6, noticing that only one integration is needed because the question asks for the gradient, not the function.

Standard method. Antidifferentiate f(x) to get dy/dx + c₁. Apply the gradient condition. Antidifferentiate again to get y + c₂. Apply the point condition. Where a hint derivative is supplied, use it rather than re-deriving it.

Instances. 2008 Exam 1 Q6 47%; 2010 Exam 1 Q7 33%; 2011 Exam 1 Q2 49%; 2022 Exam 1 Q9 34%; 2009 Exam 1 Q6 73% ("Find all real values of m such that y = e^{mx} is a solution of d²y/dx² − 3dy/dx − 10y = 0"); 2026 NHT Exam 1 Q4a.

Traps. [RPT08 E1] on Q6 (47%): "many then made errors in their attempt to find the value of c. Some completely omitted c … A significant number of students attempted to integrate again to find f(x)" — more work than asked. [RPT10 E1] on Q7 (33%): "A number of students ignored the fact that the first derivative was given and wasted a great deal of time and effort attempting to independently determine the result. … A large number of students omitted the x term from their final answer."

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Type 30 — Slope (direction) fields: identify the differential equation

Wording template — 2025 Exam 2 Section A Q8 [PAPERS]:

"Consider the direction field below. The direction field best represents the differential equation
A. dy/dx = x² − y B. dy/dx = x − y² C. dy/dx = y − x D. dy/dx = x − y" (1 mark)

and 2013 Exam 2 Section A Q12:

"The differential equation that best represents the above direction field is …"

and 2008 Exam 2 Section A Q9:

"The direction (slope) field for a certain first order differential equation is shown above. The differential equation could be …"

and 2010 Exam 2 Section A Q11:

"A direction field for the volume of water, V megalitres, in a reservoir t years after 2010 is shown below. According to this model, for k > 0, dV/dt is equal to …"

What it is really testing. Reading structure off a picture: symmetry in x (an even power), symmetry in y, where the slope is zero, where it is undefined, and the sign in each quadrant.

Standard method. [RPT25 E2] states it as well as any textbook: "In this direction field, it can be seen that the gradients for positive x values are the same as those for negative x values, indicating the x value is squared. When x = 0 and y > 0, the gradient is negative, indicating A is the best response." Generalised: (i) test x → −x and y → −y symmetry; (ii) test the axes (x = 0, y = 0); (iii) test one off-axis point.

Instances. 2006 Exam 2 Section A Q11 40%; 2008 Exam 2 Section A Q9 40%; 2010 Exam 2 Section A Q11 40%; 2013 Exam 2 Section A Q12 67%; 2023 Exam 2 Section A Q7 68%; 2024 NHT Exam 2 Section A; 2025 Exam 2 Section A Q8 80%.

Traps. Confusing the x and y roles — the persistent one. [RPT07 E1]: "drew the constant gradients vertically rather than horizontally (mixing up the x and y variables)."

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Type 31 — Slope fields: sketch the field, or sketch a solution curve on it

Wording template — 2007 Exam 1 Q8 [PAPERS]:

"a. Sketch the slope field of the differential equation dy/dx = (1 + y²)/2 for y = −2, −1, 0, 1, 2 at each of the values x = −2, −1, 0, 1, 2 on the axes below. b. If y = −1 when x = 0, solve the differential equation given in part a. to find y in terms of x. c. Sketch the graph of the solution curve found in part b. on the slope field in part a."

and 2017 Exam 1 Q8a:

"A slope field representing the differential equation dy/dx = −x/(1 + y²) is shown below. Sketch the solution curve of the differential equation corresponding to the condition y(−1) = 1 on the slope field above and, hence, estimate the positive value of x when y = 0. Give your answer correct to one decimal place." (2 marks)

and 2015 Exam 2 Section A Q13:

"The direction field for a certain differential equation is shown above. The solution curve to the differential equation that passes through the point (−2.5, 1.5) could also pass through …" (1 mark)

and 2023 Exam 2 Section A Q7:

"The direction field for a differential equation is shown above. On a certain solution curve of this differential equation, y = 2 when x = −1. The value of y on the same solution curve when x = 1.5 is closest to …"

What it is really testing. That a solution curve is tangent to the field everywhere — it never crosses a tick. And that "hence" means read the answer off your own sketch.

Standard method. To sketch a field: evaluate dy/dx at each grid point and draw a short segment of that gradient centred on the grid point. To sketch a solution curve: start at the given point, follow the ticks, keep the curve smooth, respect symmetry, and extend it across the whole domain shown.

Instances. 2007 Exam 1 Q8a 29% / Q8c 13%; 2015 Exam 2 Section A Q13 47%; 2017 Exam 1 Q8a 17%; 2023 Exam 2 Section A Q7 68%.

Traps. [RPT17 E1] on Q8a (17%) gives the complete list: "Several curves crossed the slope ticks rather than following them. Errors included: the final curve not being symmetrical; the curve not passing through (−1, 1); giving the value for x as around 1.2 (the value of the y-intercept); finding an approximate value from the solution in part b. even though this was inconsistent with the student's graph (part a. used the word 'hence'). Many graphs were almost flat between x = −0.5 and x = 0.5, resulting in missing the desired y-intercept. Some drew the graph just to the x-intercepts rather than for the whole domain." [RPT07 E1] on Q8a (29%): "Many students drew line segments that did not pass through the points of intersection of the gridlines."

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Type 32 — Euler's method

Wording template — 2007 Exam 1 Q7a [PAPERS]:

"Use Euler's method to find y₂ if dy/dx = 1/x, given that y₀ = y(1) = 1 and h = 0.1. Express your answer as a fraction." (2 marks)

and 2018 Exam 2 Section B Q3e:

"After 25 seconds the depth has risen to 0.4 m. Using Euler's method with a step size of five seconds, find an estimate of the depth 30 seconds after the fountain began to fill. Give your answer in metres, correct to two decimal places." (2 marks)

and 2025 Exam 2 Section B Q3c:

"Using Euler's method with a step size of 15 minutes, find Q(30), the approximate quantity of salt in the tank after 30 minutes. Give your answer in kilograms, correct to two decimal places." (2 marks)

and 2017 Exam 2 Section A Q9:

"Consider dy/dx = 2x² + x + 1, where y(1) = y₀ = 2. Using Euler's method with a step size of 0.1, an approximation to y(0.8) = y₂ is given by …" (1 mark)

What it is really testing. The recursion yₙ₊₁ = yₙ + h·f(xₙ, yₙ) — supplied on [FS] — applied with the correct h (including a negative h when the target lies to the left of the start), and, on Examination 2, shown.

Standard method. Tabulate n, xₙ, yₙ, f(xₙ, yₙ) — one row per step. On Examination 1, keep exact fractions throughout.

Instances. 2007 Exam 1 Q7a 36%; 2009 Exam 1 Q9b 44%; 2017 Exam 2 Section A Q9 45%; 2018 Exam 2 Section B Q3e 25%; 2023 Exam 2 Section A Q6 69% (Euler embedded in pseudocode); 2025 Exam 2 Section B Q3c 38.93%.

Traps. - Using f' instead of f. [RPT09 E1] on Q9b (44%): "A large proportion of students used f'(2) instead of f(2)." - Not showing the method. [RPT18 E2] on Q3e (25%): "Many students did not explicitly demonstrate their use of Euler's method." [RPT25 E2] on Q3c (38.93%): "Euler's method needed to be shown in some form to be awarded both marks. Several responses simply included the answer and did not show the development. A tabulated approach was acceptable as long as both Q(15) and the final answer were shown." - Fraction arithmetic. [RPT07 E1] on Q7a (36%): "Quite a few of the students who started well were unable to complete the question successfully, often due to fraction errors which were again prevalent. Very often the lowest common denominator chosen for 10 and 11 was 121." - A wrong recursion. [RPT07 E1]: "Typical errors in the application of Euler's method included y₁ = y₀ + h/x₁; y₁ = y₀ + h/x₀²; and y₁ = y₀ + h·x₀·logₑ(x₀)." - A negative step. 2017 Exam 2 Section A Q9 runs from x = 1 to x = 0.8; the step is −0.1.

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Type 33 — Calculus of parametrically defined curves

Wording template — 2016 Exam 2 Section A Q7 [PAPERS]:

"Given that x = sin(t) − cos(t) and y = ½sin(2t), then dy/dx in terms of t is …" (1 mark)

and 2019 Exam 2 Section B Q1c:

"i. Express dy/dx in terms of sin(t). ii. State the limiting value of dy/dx as t approaches π/2."

and 2026 NHT Exam 1 Q7:

"A curve is described by the parametric equations x = t² + 1/t² and y = t² − 1/t², where t ∈ R\{0}. a. Find the coordinates of any points on the curve that have a vertical tangent. b. Find the Cartesian equation of the curve."

and 2012 Exam 1 Q9c:

"Show that at time t = 1, dy/dx = (1 + √3)/(1 − √3)." (2 marks)

What it is really testing. dy/dx = (dy/dt)/(dx/dt) — and the discipline not to eliminate the parameter.

Standard method. Differentiate each component with respect to t. Divide. Simplify with identities. Answer in terms of t unless told otherwise. A vertical tangent is dx/dt = 0 with dy/dt ≠ 0.

Instances. 2011 Exam 1 Q5 51%; 2012 Exam 1 Q9c 35%; 2012 Exam 2 Section B Q1d 47%; 2013 Exam 2 Section B Q1b 41%; 2015 Exam 2 Section B Q3a 64%; 2016 Exam 2 Section A Q7 37%; 2019 Exam 2 Section B Q1c.i 61% / Q1c.ii 61%; 2020 Exam 2 Section B Q1b.i 56%; 2023 Exam 2 Section A Q9; 2026 NHT Exam 1 Q7a.

Traps. - Eliminating the parameter. [RPT11 E1] on Q5: "Most students attempted to eliminate the parameter, with varying degrees of success. … Students who attempted to express y in terms of x explicitly were confronted with a more difficult differentiation process. … Some put t in terms of x and then substituted this into the expression for y. This led to a very difficult differentiation, which rarely led to a successful outcome." [RPT15 E2] and [RPT19 E2] repeat the warning. - Answering in x. [RPT15 E2] on Q3a: "Many students attempted a chain rule relation, but a number of these had x instead of t in what otherwise would have been a correct answer." - Giving only one value of t. [RPT13 E2] on Q1b (41%): "a number of students gave only one answer for t, and others gave extra solutions outside the specified domain." - Not using earlier derivatives. [RPT12 E1] on Q9c (35%): "Many students did not use their previously calculated values of dx/dt and dy/dt for t = 1, which was previously calculated in part a., but set about repeating the differentiation."

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Type 34 — Converting a definite integral into the parameter

Wording template — 2019 Exam 2 Section B Q1e [PAPERS]:

"The portion of the curve given by y = x² − 2x for x ∈ [2, 4] is rotated about the y-axis to form a solid of revolution. Write down, but do not evaluate, a definite integral in terms of t that gives the volume of the solid formed." (2 marks)

and 2006 Exam 2 Section B Q1b:

"Use the substitution u = 1 + x³ to write down a definite integral which represents the volume of the glass in terms of u." (2 marks)

What it is really testing. That a change of variable in a definite integral changes four things: the integrand, the differential, and both terminals. 2019 Exam 2 Section B Q1e was earned in full by 3% of the state — the lowest full-marks rate of any multi-mark Calculus part in the archive.

Standard method. V = π∫x² dy. With x = sec(t) + 1 and y = tan(t): dy = sec²(t) dt, x² = (sec t + 1)², terminals y = 0 ⇒ t = 0 and y = 8 ⇒ t = arctan(8). Hence V = π∫₀^{arctan(8)} (sec(t) + 1)² sec²(t) dt. [RPT19 E2] records that arccos(1/√65) was also accepted for the upper terminal.

Instances. 2019 Exam 2 Section B Q1e 3%; 2006 Exam 2 Section B Q1b; 2014 Exam 1 Q5b 48%; the Section A family at Type 14.

Traps. [RPT19 E2]: "Very few students answered this question correctly. The most common incorrect answer was an integral in terms of x. Of those that attempted to give an integral in terms of t, most simply replaced dx with dt." That last clause is the whole failure mode: substituting the differential symbol without substituting the differential.

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3. The standard wordings

VCAA writes examinations from a small, stable vocabulary. Each recurring sentence carries a precise instruction about what work is required, and — critically — about what work is not. Misreading them is the single largest cause of avoidable loss in this area of study. [RPT16 E1] and [RPT17 E1] list "not reading the question carefully enough — this included not answering the question, proceeding further than required or not giving the answer in the specified form" as the first area of weakness on the whole paper, both years.

3.1 "Write down a definite integral that gives …"

Where it appears. 2008, 2009, 2010, 2013, 2014, 2015, 2016, 2017 NHT, 2018, 2018 NHT, 2021, 2021 NHT, 2022 NHT, 2023, 2023 NHT, 2024 NHT, 2025, 2026 NHT — eighteen distinct Examination 2 papers in the corpus contain the exact string "Write down a definite integral" [PAPERS].

Canonical forms.

What it demands. A complete, correct integral expression: the constant (π or 2π), the integrand, both terminals, and the differential. Nothing else. There is no credit for evaluating it — and no penalty either, except time.

Why it is set as its own part. It separates set-up from evaluation so that the two can be marked independently. The design is deliberate and it works: on 2022 Exam 2 Section B Q1d the set-up part scored 45% and the evaluation part 37% — the evaluation being harder despite being a single button press, because students who had lost π at the set-up stage compounded the error.

The variable specification is the whole question. "In terms of x", "in terms of y", "in terms of t" and "in terms of u" each demand a different, complete change of variable. 2019 Exam 2 Section B Q1e asked for t and 79% of the state scored zero [RPT19 E2].

Related phrasings that mean the same thing. - 2006 Exam 1 Q6a: "Express the volume of the solid of revolution as a definite integral." (70%) - 2017 Exam 2 Section B Q3 (two consecutive parts): "Write down an equation involving definite integrals that can be used to determine a." (2 marks) then "Hence, find the value of a, correct to two decimal places." (1 mark) - 2017 Exam 2 Section B Q3d.i: "Write down an expression involving a definite integral that gives the time taken for the skydiver to reach a speed of 30 m s⁻¹." - 2018 Exam 2 Section B Q3d: "Express the time taken for the depth to reach 0.25 m as a definite integral and evaluate this integral correct to the nearest tenth of a second." (both verbs in one part, 2 marks, 56%) - 2014 Exam 2 Section B Q4c: "By using an appropriate definite integral, find the time it takes for the tank to fill." (2 marks, 32%) - 2017 Exam 2 Section B Q1c.i: "[write] π∫₀ᵃ(f(x))² dx = π∫ₐ³(f(x))² dx" — an equation of two definite integrals defining an unknown terminal (71%).

The time-as-an-integral idiom is worth isolating. When a question gives dh/dt = F(h) and asks for a time, the answer is t = ∫_{h₁}^{h₂} 1/F(h) dh — reciprocate and integrate with respect to the dependent variable. [RPT14 E2] on Q4c (32%): "Many students did not attempt this question. A number attempted to find t in terms of h, instead of using a definite integral. Some attempted to integrate dh/dt with respect to h."

3.2 "Using a suitable substitution" / "Using the substitution u = …"

Where it appears. "Suitable substitution" occurs in 2006, 2007, 2008, 2010, 2012, 2015, 2016, 2017, 2018, 2018 NHT, 2019, 2019 NHT, 2020, 2021 NHT and 2023 NHT [PAPERS]. "Using the substitution" with an explicit substitution occurs in 2009, 2014, 2022, 2022 NHT, 2024 NHT, 2025 and 2026 NHT.

Two distinct instructions, and the difference matters.

What VCAA does when it supplies the substitution. It removes the only creative step and then makes the bookkeeping the entire question. Every distractor in the Section A family is a specific bookkeeping failure. In the u² = x + 1 variants (2024 NHT Exam 2 Section A Q8) and x³ − 1 = u² (2026 NHT Exam 2 Section A Q9), the substitution is given implicitly, so differentiating it is itself a step: 2u du = dx and 3x² dx = 2u du.

The corollary a 45+ candidate must internalise. If VCAA has not supplied a substitution, and the integrand is a standard form, then a substitution is probably wrong. [RPT19 E1] lists among areas of weakness: "not recognising and writing down the anti-derivative of standard functions leading to unnecessary use of substitutions in integration problems (Questions 1 and 8)." [RPT21 E1]: "Use of a substitution was unnecessary in this situation and in attempting to use a substitution, some students introduced errors into their working."

3.3 "Hence" and "Hence, or otherwise"

"Hence" alone is mandatory. It means: use the immediately preceding result. Examples: 2006 Exam 1 Q1b "Hence find the exact value of dy/dx when y = 1"; 2006 Exam 2 Section B Q1e "Hence find an expression for the rate of change of the area"; 2009 Exam 1 Q5c "Hence evaluate the gradient of the curve at the point (1, 1)"; 2014 Exam 1 Q5c and Q7c "Hence evaluate …"; 2017 Exam 1 Q8a "and, hence, estimate the positive value of x when y = 0"; 2017 Exam 2 Section B Q3 "Hence, find the value of a, correct to two decimal places."

The reports mark the instruction. [RPT06 E2] on Q1e (22%): "Some did not use 'hence' and tried to use dA/dx to find dA/dt, which was a far more complicated approach." [RPT14 E1] on Q7c (26%): "Some ignored the word 'hence'." [RPT17 E1] on Q8a (17%): students who computed the answer from part b rather than reading it off the graph were penalised because "part a. used the word 'hence'".

"Hence, or otherwise" is permissive but signposted. It appears in 2023 Exam 2 twice — Q3b.ii "Hence or otherwise, find the curved surface area of the solid correct to three decimal places" and Q4e.ii "Hence or otherwise, find the size of the fish population in pond 2 and the value of t when the rate of growth of the population is a maximum" — and in 2024 NHT Exam 1 Q3b and Q9d, and 2022 Exam 1 Q1b. It licenses an alternative route, but the "hence" route is always the short one, and on a 1-mark part there is no time for the alternative.

A further wrinkle worth knowing. [RPT17 E1], on show-that questions generally: "Students should be reminded that they can use a given value in the remaining part(s) of a question whether they were able to derive it or not." A failed "show that" does not forfeit the rest of the question.

3.4 "Show that" — and what counts as showing

Frequency. "Show that" occurs 105 times across the corpus text files [PAPERS]. [RPT23 E2], [RPT16 E2] and [RPT10 E2] count three, four and seven instances respectively on a single Examination 2.

What it demands. [RPT17 E1]: "showing a given result. This was required in Question 10a. In such questions, the onus is on students to include sufficient relevant working to demonstrate that they know how to derive the result." [RPT18 E1]: "students are reminded that they must present sufficient evidence that the result has been shown."

The reports are candid that this is routinely gamed. [RPT11 E1]: "as often happens in a 'show that' question, some students were unable to do the relevant algebra yet somehow still managed to give the result stated." [RPT12 E1] on Q9c (35%): "As often happens in a 'show that' type of question, some students were unable to do any convincing algebra, yet still managed to obtain the result stated." [RPT13 E1] on Q5a: "Some students who did have a constant of integration were unable to evaluate it. Nevertheless, this did not stop them from 'showing' the given result."

The mechanical rule. On an n-mark "show that", write at least n distinct lines of algebra between the starting point and the stated result. [RPT08 E2]: "As it was a 'show that' question, the most common error was the omission of a suitable intermediate expression." [RPT25 E2] on 2025 Exam 2 Section B Q3b: "Many responses did not show sufficient development of the formula to be awarded the method marks."

Variants. - "Use integration to show that …" — 2025 Exam 1 Q3a, 2025 Exam 1 Q4a ("Use integration to show that E(T) = ½"). - "Use calculus to show that …" — 2006 Exam 2 Section B, 2026 NHT Exam 2 Section B Q3b.i ("Given that the volume of the solid of revolution is π logₑ(k), use calculus to show that k satisfies the equation k² − 4k + 3 = 0"). - "Use calculus to solve …" — 2025 Exam 2 Section B Q3d. [RPT25 E2]: "'Use calculus' means students are required to show the steps needed to find the solution to gain all 3 marks." On a technology-active paper this phrase forbids deSolve. - "Use implicit differentiation to find …" — 2023 Exam 1 Q4. The method is prescribed. - "Verify by differentiation and substitution into the left side …" — 2016 Exam 2 Section B Q3d. The direction of work is prescribed.

3.5 How VCAA asks for an expression rather than an evaluation

This is the family of instructions that tells you to stop.

"Find an expression for" versus "find the value of" is the commonest misread in Section B. 2006 Exam 2 Section B Q1e asks for dA/dt "in terms of x"; the next part asks for the exact value of x at the maximum. Two parts, two different objects.

"In the form" is not decoration. [RPT16 E1] and [RPT17 E1] both list "not giving the answer in the specified form" among the paper's leading weaknesses. [RPT17 E1] on Q8b (36%): "Most students had the correct integration after separating variables but made no attempt to express the answer with integers as required." [RPT09 E1] on Q10b (23%): "A high proportion of students either ignored the instruction regarding integers or did not know how to proceed." Recent examples of the demanded forms: V = 2π(logₑ(a) + b) (2020), ay³ + by + cx² + d = 0 (2017), Q = a/(16 + 2t)^{b/c} (2018), π((a√b)/c − d) (2023), aπ/b (2024 NHT), logₑ(a/b) (2017), a/(b + cπ) (2018), π√a/b (2023).

3.6 The exactness instruction, and its negation

Every Examination 1 front page since 2010 carries [PAPERS]:

"Unless otherwise specified, an exact answer is required for each question."

and the 2006–2009 papers the equivalent "A decimal approximation will not be accepted if an exact answer is required to a question."

The negation is always explicit, and always precise about the precision: "correct to two decimal places" (291 occurrences of "correct to" across the corpus [PAPERS]), "correct to the nearest integer", "correct to one decimal place", "correct to the nearest tenth of a second", "correct to three decimal places". [RPT24 E2]: "Answers must be left in exact form unless a specific number of decimal places is required." [RPT23 E2] on Q3b.ii (55%): "Incorrect rounding … was a frequent final response. Students are reminded to set their calculators to display sufficient decimal places." [RPT25 E2] on Q3f: "Several responses left the answer as a decimal, rather than in exact form as required by the question." [RPT15 E2] on Q1e: "A significant number of students did not give their answer correct to three decimal places."

3.7 The technology-selection wordings (Examination 2 only)

[RPT25 E2] on Section A Q11: "Use the DEsolve functionality on CAS to solve the given differential equation and then find the domain of the solution. Alternatively, use separation of variables to solve the differential equation manually." [RPT25 E2] on Section B Q1b.i: "Students must make sure variables are defined if they are being used in formulas." [RPT24 E2] on Q1b.i: "Students were expected to write an expression for … within the definite integral instead of stating the generic formula. Many students made transcription errors when transferring their answer from their CAS to the script."

The rule that follows: on Examination 2, a written response must show the instantiated integral — π∫₀¹(2 − sin²x)dx, not V = π∫y²dx.

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4. The separators in this area

Definition. A separator is a question part with pct ≤ 50 — fewer than half the state earned full marks. Of the 387 Calculus parts in [QJSON], 158 are separators, carrying 372 of the area's 623 published marks. That is 41% of the parts and 60% of the marks: separators are concentrated in the multi-mark written questions, not the one-mark multiple choice.

All 158 are listed below, grouped by the question type from §2, then ordered by year. Format: `ref` — pct% — description.

4.1 Implicit differentiation (15)

• 2007 Exam 1 Q3 — 43% — tangent to x³ − 2x²y + 2y² = 2 at P(2, 3); product rule on the middle term
• 2008 Exam 1 Q2 — 48% — gradient of the normal to 3x² + 2xy + y² = 11 at the first-quadrant point where x = 1
• 2008 Exam 2 Section B Q4d.ii — 39% — implicit differentiation of a log relation to show a given dy/dx
• 2009 Exam 1 Q5c — 22% — "hence evaluate the gradient at (1, 1)"; the parameter k must be found first
• 2010 Exam 1 Q9b — 33% — gradient on a conic at x = 2; the negative y-root is required
• 2012 Exam 1 Q6 — 45% — gradient of the tangent to xy² + y + (logₑ(x − 2))² = 14 at (3, 2)
• 2013 Exam 1 Q6 — 40% — find c such that y² + 3e^{x−1}/(x − 2) = c has gradient 2 at x = 1
• 2014 Exam 1 Q4 — 48% — gradient of the normal to y = −3e^{3x}e^y at (1, −3)
• 2014 Exam 2 Section B Q2c — 9% — tangent of gradient 1 to a circle; implicit route generated too many variables
• 2018 Exam 1 Q3 — 46% — gradient of 2x²sin(y) + xy = π²/18 at (π/6, π/6), in the form a/(b + cπ)
• 2019 Exam 1 Q10 — 18% — implicit differentiation with the answer required in a specified form
• 2022 Exam 1 Q7 — 34% — gradient of x cos(x + y) = π/48 at (π/24, 7π/24)
• 2022 Exam 2 Section A Q10 — 21% — values of m for which the tangent to 5x²y − 3xy + y² = 10 at (1, m) has negative gradient
• 2023 Exam 1 Q4 — 37% — x arcsin(y²) = π; dy/dx at (√6, 1/√2) in the form π√a/b
• 2025 Exam 1 Q1 — 48% — equation of the tangent to xe^{2y} + y²e^x = 8e⁴ at (4, −2)

4.2 Inverse circular functions: derivatives and range (3)

• 2006 Exam 1 Q5b — 28% — minimum a for which y = tan⁻¹(x − 1) + a tan(π/8) > 0 for all x; requires the range of arctan
• 2009 Exam 1 Q10b — 23% — f'(x) for an arcsin composite, required in the form a/(b√(x(x + c)))
• 2012 Exam 1 Q5 — 44% — y = arctan(2x); find a such that d²y/dx² = ax(dy/dx)²

4.3 Second derivatives, concavity and points of inflection (12)

• 2013 Exam 2 Section B Q3d.i — 9% — d²N/dt² from dN/dt = 0.4N(6 − logₑ N) by chain and product rules
• 2013 Exam 2 Section B Q3d.ii — 38% — coordinates of the point of inflection from d²N/dt² = 0
• 2017 Exam 2 Section A Q6 — 46% — d²y/dx² at (0, 1) given dy/dx = e^x arctan(y)
• 2017 Exam 2 Section A Q8 — 29% — for which x is the gradient of f(x) = x³ − mx² + 4 strictly increasing
• 2017 Exam 2 Section A Q10 — 6% — points of inflection when f''(x) = (x + a)²(x − b)/g(x) with g(x) < 0; the repeated factor gives no sign change
• 2019 Exam 2 Section B Q3b.iii — 38% — justify that Q = logₑ(e^t + e − 1) has no point of inflection
• 2021 Exam 2 Section A Q9 — 38% — which derivative has an antiderivative with a turning point but no point of inflection
• 2023 Exam 2 Section B Q4e.i — 21% — express d²Q/dt² in terms of Q for the logistic model; the chain rule when differentiating with respect to t was the sticking point
• 2024 Exam 2 Section B Q1d.i — 27% — condition on a parameter for exactly one stationary point
• 2024 Exam 2 Section B Q1d.ii — 27% — condition for exactly three stationary points; the equality sign was the mark
• 2024 Exam 2 Section B Q1d.iii — 25% — condition for exactly five stationary points
• 2025 Exam 2 Section B Q1d.ii — 45.85% — no stationary points forces an asymptote condition; solve for the asymptote equations

4.4 Related rates (7)

• 2006 Exam 2 Section B Q1e (3 marks) — 22% — dA/dt for the surface of wine in a glass; "hence" from the supplied dy/dt
• 2006 Exam 2 Section B Q1e (1 mark, the final part) — 28% — exact depth at which the surface area is a maximum
• 2009 Exam 2 Section B Q4e — 18% — dy/dt assembled from a three-link chain rule
• 2016 Exam 1 Q4 — 40% — rate of growth of the surface area of a cube whose side is x = arctan(t)
• 2018 Exam 2 Section B Q3c.i — 36% — show dh/dt = (4 − 5√h)/(25π(4h² + 1)) for the fountain
• 2021 Exam 2 Section B Q3b.ii — 22% — maximum rate at which the depth decreases, and the depth at which it occurs; maximise dh/dt from the shown expression
• 2024 Exam 2 Section B Q3b — 37% — related rate for a pond, with a unit conversion from centimetres to metres

4.5 Antidifferentiation by standard forms, and antiderivative graphs (3)

• 2006 Exam 1 Q8 — 31% — antiderivative of (2 + 6x)/√(4 − x²); split into an arcsin plus a substitution
• 2008 Exam 2 Section A Q12 — 45% — value of ∫₋₃⁰ f(x) dx read as a height difference on the antiderivative graph
• 2012 Exam 1 Q1 — 46% — antiderivative of (6 + x)/(x² + 4); split into an arctan plus a log

4.6 Integration by substitution (11)

• 2009 Exam 2 Section A Q10 — 23% — "Using the substitution u = cos(x), the area … could be found by evaluating"; 52% chose option E
• 2010 Exam 1 Q6 — 34% — ∫cos²(2x)sin(2x) dx between π/4 and 3π/4; sign and terminals
• 2011 Exam 1 Q6 — 47% — definite integral needing u = e^x, terminals to be changed
• 2012 Exam 1 Q7 — 41% — area under y = (x − 1)√(2 − x) on [1, 2] via u = 2 − x
• 2014 Exam 1 Q5b — 48% — "Use an appropriate substitution in the form u = g(x) … in terms of u only"
• 2014 Exam 1 Q5c — 45% — "Hence evaluate …, giving your answer in the form kπ"
• 2015 Exam 1 Q8a — 47% — show ∫tan(2x) dx = ½logₑ|sec(2x)| + c
• 2020 Exam 1 Q2 — 28% — ∫₋₁⁰ √((1 + x)/(1 − x)) dx in the form a√b + c
• 2022 Exam 1 Q9 — 34% — definite integral by substitution with a trigonometric integrand
• 2024 Exam 2 Section A Q9 — 45% — (November 2024 paper text unavailable in the corpus; the report gives a one-line solution only)
• 2025 Exam 1 Q6 — 38% — volume of revolution; substitution or integration by parts, π and terminals both at risk

4.7 Partial fractions (8)

• 2007 Exam 1 Q4 — 35% — volume of revolution requiring partial fractions of 1/(1 − x²); modulus signs
• 2009 Exam 1 Q8b — 31% — exact area from the supplied split of (2 + x²)/(4 − x²)
• 2011 Exam 1 Q1 — 39% — partial fractions with modulus signs and a final log-law simplification
• 2013 Exam 1 Q2 — 47% — ∫₀¹ (x − 5)/(x² − 5x + 6) dx; the quadratic must be factorised correctly
• 2014 Exam 1 Q6b — 37% — volume of revolution using the identity verified in part a
• 2017 Exam 1 Q2 — 35% — ∫₁^{√3} 1/(x(1 + x²)) dx in the form logₑ(a/b); an irreducible quadratic factor
• 2020 Exam 1 Q8 — 20% — volume of revolution requiring partial fractions of (x² + x + 1)/((x + 1)(x² + 1))
• 2023 Exam 2 Section B Q4a — 48% — A and B in the partial-fraction split of the logistic equation

4.8 Integrals needing a trigonometric identity first (4)

• 2008 Exam 1 Q9b — 37% — volume about the y-axis needing ∫cos²(y) dy and the double-angle identity
• 2011 Exam 1 Q11 — 46% — volume needing sin²(x) = ½(1 − cos 2x); exact-value evaluation at the terminal
• 2013 Exam 1 Q9 — 29% — volume between y = sin x and y = 3x/π; difference of squares plus double angle
• 2022 Exam 1 Q10b — 25% — volume of revolution via a double-angle formula; the resulting quadratic has two roots

4.9 Areas of regions (10)

• 2008 Exam 1 Q9a — 28% — area bounded by y = cos⁻¹(x), the x-axis and x = −1; symmetry shortcut
• 2008 Exam 2 Section B Q5f — 6% — area of a major segment of a circle; geometric method far shorter than integration
• 2010 Exam 1 Q10 — 23% — area of a region below the x-axis, in the form a√b/c; sign of the integral
• 2011 Exam 2 Section B Q1e.ii — 41% — area as a quarter circle minus a triangle
• 2012 Exam 2 Section B Q2d — 44% — area of a portion of an annulus; the correct portion eluded most
• 2015 Exam 1 Q8d — 23% — area enclosed by f(x) = ½arctan(x), the x-axis and x = √3, using part a
• 2017 Exam 2 Section B Q3d — 26% — acute angle between two edges at the origin, from the derivatives of the branches
• 2018 Exam 2 Section B Q2e — 31% — segment area; standard formula beat the integral
• 2021 Exam 2 Section B Q2c.ii — 22% — segment area; the sector angle was the obstacle
• 2022 Exam 2 Section B Q2d — 32% — segment area; the definite-integral route "usually led to error"

4.10 Volumes of revolution (10)

• 2006 Exam 1 Q6c — 49% — find a given that the volume is 5π/18
• 2008 Exam 2 Section B Q1d.iii — 26% — evaluate the volume after the substitution; π omitted, logₑ(u²) invented
• 2009 Exam 2 Section B Q4d.ii — 45% — evaluate the volume integral; π present in one part and absent in the other
• 2012 Exam 2 Section A Q12 — 48% — which definite integral gives the volume; squaring and terminals
• 2017 Exam 1 Q10c — 14% — volume from y = arccos(x/2) using the derivative shown in part a; terminals −2 to 2
• 2018 Exam 1 Q9c — 19% — volume bounded by x² − 2y² = 1 and y = x − 1
• 2018 Exam 2 Section B Q3a — 32% — show V = (π/4)(4h³/3 + h) for rotation about the y-axis to depth h
• 2019 Exam 2 Section B Q1e — 3% — "write down, but do not evaluate, a definite integral in terms of t" for a volume about the y-axis
• 2022 Exam 2 Section B Q1d.i — 45% — integral for a volume between two curves; square of the difference written instead of difference of the squares
• 2022 Exam 2 Section B Q1d.ii — 37% — evaluate that volume; π dropped at the evaluation stage

4.11 Arc length and surface area (8)

• 2016 Exam 1 Q7 — 43% — Cartesian arc length of y = ⅓(x² + 2)^{3/2} from 0 to 2; the perfect square under the root (content retired in 2023)
• 2017 Exam 1 Q7 — 30% — parametric arc length of r(t) = cos³(t)i + sin³(t)j on [0, π/4]
• 2017 Exam 2 Section B Q3e — 16% — show the border length equals 2∫₀²(a + b/√(4 − x²)) dx and find a, b
• 2020 Exam 1 Q9b — 14% — parametric arc length; the integrand is a perfect square that must be spotted
• 2023 Exam 1 Q7 — 31% — surface area of revolution from parametric equations, in the form π(a√b/c − d)
• 2023 Exam 2 Section A Q11 — 46% — surface area about the y-axis for y = cos⁻¹(x)
• 2023 Exam 2 Section B Q3c — 38% — total surface area: curved surface plus two circular discs
• 2023 Exam 2 Section B Q3d — 24% — efficiency ratio for a second solid with a variable upper terminal

4.12 Differential equations by separation of variables (14)

• 2006 Exam 1 Q2 — 45% — dy/dx = x√(x² − 16), y(5) = 13/3; substitution plus a constant of integration
• 2007 Exam 1 Q7b — 33% — solve dy/dx = 1/x and evaluate at the Euler target; answer logₑ(1.2) + 1
• 2007 Exam 1 Q8b — 46% — solve dy/dx = (1 + y²)/2 with y(0) = −1; arctan(−1) and the algebra of the rearrangement
• 2009 Exam 1 Q9a — 34% — solve dx/dy = (y + 2)² + 4 giving y as a function of x; reciprocal of a sum
• 2013 Exam 1 Q5b — 41% — Newton's law of cooling; temperature after 10 minutes using e^{−5k} = ¾
• 2016 Exam 1 Q10 — 14% — √(2 − x²) dy/dx = 1/(2 − y), y(1) = 0; arcsin, completing the square, and the correct root
• 2016 Exam 2 Section B Q3a — 42% — solve dx/dt + x/(20 + t) = 0 with the constant evaluated
• 2017 Exam 1 Q8b — 36% — solve dy/dx = −x/(1 + y²), y(−1) = 1, in the form ay³ + by + cx² + d = 0
• 2018 Exam 1 Q8b — 23% — salt tank; Q as a function of t in the form a/(16 + 2t)^{b/c}
• 2019 Exam 1 Q1 — 42% — dy/dx = 2ye^{2x}/(1 + e^{2x}), y(0) = π; log and index laws at the end
• 2021 Exam 1 Q7b — 8% — maximum displacement of the particle and the times at which it occurs
• 2022 Exam 2 Section B Q3b.i — 23% — equation of the horizontal asymptote of x = logₑ(tan⁻¹(2t) + 1)
• 2024 Exam 1 Q7 — 27% — separable equation; y as a function of x, with the correct sign chosen
• 2007 Exam 2 Section A Q11 — 50% — value of y at x = π/3 given dy/dx = sin(x) and y(0) = 1, as a definite integral plus the initial value

4.13 Formulating a differential equation (8)

• 2006 Exam 2 Section A Q10 — 47% — "The differential equation which models this process is"; dissolution proportional to the undissolved amount
• 2007 Exam 2 Section A Q14 — 37% — logistic-form equation for the spread of bird flu through a population of 1000
• 2009 Exam 2 Section A Q13 — 38% — differential equation for the concentration, not the amount, of salt
• 2011 Exam 2 Section B Q5b — 45% — rate in minus rate out, rearranged to the stated equation
• 2016 Exam 2 Section B Q3b — 34% — expression for the concentration of salt at time t
• 2016 Exam 2 Section B Q3c — 40% — show dy/dt + y/(10 + t) = 1/3, using the concentration from part b
• 2018 Exam 1 Q8a — 44% — show dQ/dt = −3Q/(16 + 2t); the rate in must be stated as zero
• 2023 Exam 2 Section A Q8 — 37% — which equation models a spa pool with 15 L/min in and 20 L/min out

4.14 Verifying a given solution (4)

• 2010 Exam 2 Section B Q3a — 5% — verify that P = 20000(4 − 3e^{0.01t}) satisfies both the equation and P(0) = 20000
• 2011 Exam 1 Q2 — 49% — show k = 3 by substituting a given y into a second-order differential equation
• 2013 Exam 2 Section B Q3a — 30% — verify a solution by substitution rather than by solving
• 2016 Exam 2 Section B Q3d — 17% — "verify by differentiation and substitution into the left side", plus the initial condition

4.15 Second antiderivatives and d²y/dx² = f(x) (2)

• 2008 Exam 1 Q6 — 47% — f''(x) = −sec²(2x), gradient −1 at x = π/8; find the gradient at x = π/12
• 2010 Exam 1 Q7 — 33% — d²y/dx² = 4x/(1 − x²)² with a derivative supplied as a hint; two constants, two conditions, partial fractions in between

4.16 Slope and direction fields (7)

• 2006 Exam 2 Section A Q11 — 40% — which differential equation the direction field could represent
• 2007 Exam 1 Q8a — 29% — sketch the slope field of dy/dx = (1 + y²)/2 at 25 grid points
• 2007 Exam 1 Q8c — 13% — sketch the solution curve found in part b on the slope field of part a
• 2008 Exam 2 Section A Q9 — 40% — which differential equation the direction field could represent
• 2010 Exam 2 Section A Q11 — 40% — direction field for reservoir volume; identify dV/dt
• 2015 Exam 2 Section A Q13 — 47% — which point the solution curve through (−2.5, 1.5) could also pass through
• 2017 Exam 1 Q8a — 17% — sketch the solution curve for y(−1) = 1 and hence estimate x when y = 0

4.17 Euler's method (5)

• 2007 Exam 1 Q7a — 36% — y₂ for dy/dx = 1/x, h = 0.1, answer as a fraction
• 2009 Exam 1 Q9b — 44% — y₁ with step size 0.1; f not f'
• 2017 Exam 2 Section A Q9 — 45% — y₂ approximating y(0.8) starting from x = 1; the step is negative
• 2018 Exam 2 Section B Q3e — 25% — depth after 30 s from the depth at 25 s; the method had to be shown
• 2025 Exam 2 Section B Q3c — 38.93% — Q(30) with h = 15; Q(15) and the final answer both had to appear

4.18 Parametric calculus (4)

• 2012 Exam 1 Q9c — 35% — show dy/dx = (1 + √3)/(1 − √3) at t = 1 using part a's derivatives
• 2012 Exam 2 Section B Q1d — 47% — dy/dx on a parametrised hyperbola; implicit or parametric route
• 2013 Exam 2 Section B Q1b — 41% — values of t in a restricted domain for a given dy/dx
• 2016 Exam 2 Section A Q7 — 37% — dy/dx in terms of t for x = sin t − cos t, y = ½sin 2t

4.19 Integration by parts (2)

• 2014 Exam 1 Q7c — 26% — "hence evaluate the area enclosed by g(x) = arctan(2x)"; by parts in disguise via the supplied derivative
• 2023 Exam 2 Section A Q10 — 33% — reduction formula Iₙ = −1 + nIₙ₋₁; option E traps the unevaluated boundary term

4.20 Kinematics tagged as Calculus (1)

• 2024 Exam 1 Q9b — 35% — acceleration via d/dx(½v²); the chain-rule alternative was error-prone and the negative sign was often dropped

4.21 Modelling, interpretation and graphing of solutions (20)

• 2008 Exam 2 Section B Q1b.i — 37% — write the polynomial equation whose solutions give the stationary points
• 2009 Exam 2 Section B Q4c — 29% — reject the negative square root with a stated reason
• 2010 Exam 2 Section B Q3d.i–iii — 28% — sketch the solution curve with correct concavity, within the specified domain and to scale
• 2010 Exam 2 Section B Q3e.ii — 17% — interpret the value of k in context (arrivals versus departures)
• 2011 Exam 2 Section A Q16 — 25% — Calculus multiple-choice item; the 2011 paper text extracted with a CID-shifted font and is unreadable, so the stem cannot be quoted
• 2011 Exam 2 Section B Q5c.i — 11% — four-mark part of the salt-tank question; the report carries no commentary on it
• 2011 Exam 2 Section B Q5d (3 marks) — 46% — sketch x against t with the turning point (8.3, 35.3) correctly placed
• 2011 Exam 2 Section B Q5d (2 marks, final part) — 1% — total mass of salt as a definite integral of the flow rate, not of x(t)
• 2014 Exam 2 Section B Q4c — 32% — time to fill the tank by a definite integral in h
• 2014 Exam 2 Section B Q4d — 12% — depth x as a function of time from a cubic V(x); only a minority inverted it
• 2016 Exam 2 Section B Q3e — 23% — time at which the concentration reaches 0.095 kg/L, not dy/dt
• 2018 Exam 2 Section B Q1e.ii — 49% — piecewise rule for a derivative graph; several responses were not functions
• 2018 Exam 2 Section B Q3f — 24% — distance from the top of the fountain at which the level stabilises; dh/dt = 0 then subtract
• 2021 Exam 2 Section B Q3b.iii — 23% — maximum rate at which water can be added without the vessel overflowing
• 2021 Exam 2 Section B Q3c — 12% — time to refill from a depth of 25 cm under a net inflow, as a definite integral in h
• 2019 Exam 2 Section B Q3a.ii — 20% — both pairs of conditions under which a constant is defined; most gave only one
• 2022 Exam 2 Section B Q3e — 50% — ratio of the speeds of two particles at the moment they are equidistant from O
• 2023 Exam 2 Section B Q4g — 40% — maximum sustainable population under a harvested logistic model dQ/dt = (11/10)Q(1 − Q/1000) − 0.055Q; the non-zero equilibrium
• 2024 Exam 2 Section B Q3d — 17% — surface area obtained by dividing the volume by the constant depth
• 2024 Exam 2 Section B Q3e — 4% — the five-day delay was not taken into account by most who attempted it

4.22 Which types separate most

Ranking by the median pct of the separators within each group, and by how much of the group is separating:

Two readings of that table matter.

First, the top of it is not about technique. Verifying a solution, interpreting a model, and reading a graph are not hard mathematics; they are hard instructions. A candidate who can integrate flawlessly will still score 5% on 2010 Exam 2 Section B Q3a if they solve the equation instead of verifying it. This is the highest-value correction available in the whole area of study.

Second, the bottom of it is not safe. Even the "easiest" separating group — slope fields, median 40% — contains 2007 Exam 1 Q8c at 13% and 2017 Exam 1 Q8a at 17%, both of which are two-mark sketching parts.

What the reports say went wrong, in order of how often they say it.

1. The constant of integration — named in at least nine distinct reports ([RPT06 E1], [RPT07 E1], [RPT09 E1], [RPT12 E2], [RPT13 E1], [RPT16 E1], [RPT16 E2], [RPT18 E1], [RPT25 E2]). [RPT12 E2] on Q3d.i: "a surprising number of students omitted the constant of integration."
2. Terminals not changed at a substitution — [RPT10 E1], [RPT11 E1], [RPT12 E1], [RPT14 E1], [RPT17 E1], [RPT25 E1]. [RPT25 E1] on Q6 (38%): "neglecting to adjust the terminals for the definite integral if a substitution was used."
3. π omitted from a volume — [RPT07 E1], [RPT08 E2], [RPT11 E1], [RPT11 E2], [RPT15 E2], [RPT22 E2], [RPT23 E2].
4. Modulus signs omitted from logarithmic antiderivatives — [RPT07 E1], [RPT09 E1], [RPT11 E1], [RPT13 E1].
5. Answer not given in the demanded form — [RPT09 E1], [RPT16 E1], [RPT17 E1], [RPT18 E1], [RPT23 E2], [RPT24 E1], [RPT24 E2].
6. "Show that" with insufficient working — [RPT08 E2], [RPT11 E1], [RPT12 E1], [RPT13 E1], [RPT16 E2], [RPT18 E1], [RPT18 E2], [RPT25 E2].
7. Algebra and arithmetic — [RPT16 E1], [RPT17 E1], [RPT19 E1] and [RPT25 E1] all name it as a top-level weakness independent of the mathematics. [RPT17 E1]: "algebraic skills. The inability to simplify expressions often prevented students from completing the question. Incorrect attempts to factorise, expand and simplify were common. Poor use of brackets was also common."
8. Notation — the missing dx / du / dt — [RPT16 E1] and [RPT17 E1]: "notation, especially the omission of the dx or equivalent in integration".
9. Doing more than was asked — [RPT08 E1], [RPT10 E1], [RPT16 E1], [RPT17 E1].
10. Non-attempts. [RPT11 E2] on Q5d (1%): "Only a small number of students attempted this part, with very few getting it correct." [RPT14 E2] on Q4c (32%), [RPT16 E2] on Q3e (23%), [RPT17 E2] on Q3d (26%) and [RPT24 E2] on Q3d (17%) and Q3e (4%) all begin with a variant of "Many students did not attempt this question." On the last parts of Section B, a substantial fraction of the separation is simply time.

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5. What makes a hard one hard

Seven mechanisms account for nearly all of the 158 separators. None of them is "the calculus is too advanced". Every one is a decision point where a correct method still produces a wrong mark.

5.1 Choosing the substitution

The substitution is the only step in VCE integration where the paper gives you nothing and the answer depends entirely on a choice. Everything downstream is mechanical; the choice is not.

What a correct choice looks like. The derivative of the inner function is already present, up to a constant. ∫cos²(2x)sin(2x) dx: the derivative of cos(2x) is −2sin(2x), and sin(2x) is sitting there — so u = cos(2x). ∫(x − 1)√(2 − x) dx: nothing differentiates to anything, so the substitution is the simplifying kind, u = 2 − x, which turns (x − 1) into (1 − u).

What the archive shows about wrong choices. They rarely fail outright; they fail slowly, and the candidate runs out of time. [RPT10 E1] on Q6 (34%): "Other students tried to use u = sin(2x) or u = cos²(2x), or expanded using double angle formulas and then used u = sin(x) or u = cos(x). Some of these are possible but are very time-consuming, and students who ventured down these paths rarely obtained the correct answer." [RPT15 E1] on Q8a (47%): "There were many instances of poor choices of substitution, such as u = sin(2x), u = tan(2x), u = sec(2x) or u = cos(x) (after the use of double-angle formulas) rather than u = cos(2x). These attempts led to a more complicated solution and were rarely successful." [RPT14 E1] on Q5b (48%) says the same thing about u = sin(6x), u = sin(3x), u = cos(3x), u = cos²(6x).

The decision rule. Before writing anything: (i) is the integrand already a standard form on [FS]? If yes, no substitution. (ii) Is there an inner function whose derivative is a factor? If yes, that is u. (iii) Is there an awkward surd or a linear denominator? If yes, substitute it. (iv) Only then consider anything else — and if the answer is not visible within two lines, the choice was wrong.

The 45+ discipline. [RPT06 E1] on Q8 (31%) describes the discrimination this creates exactly: "Overall this question was not well done and there was a clear difference between the students who knew how to approach this type of question and those who did not."

5.2 Changing the limits

This is the most-repeated correction in twenty years of Specialist Mathematics reports, and it is worth quoting in full because VCAA states the principle, not just the error. [RPT10 E1]:

"Too many students changed the variable correctly but left the terminals unchanged; it should be emphasised that this is not logically correct, even if changing back to the original variable later enables them to obtain a correct answer."

[RPT11 E1] reprints the same sentence. [RPT12 E1] spells out the consequence for marking:

"Several of the students who used the appropriate substitution but kept the x terminals achieved the correct answer by substituting back for x before using the terminals. They could not be awarded full marks due to the inconsistency in their working."

Why it is genuinely hard, not merely careless. A substituted definite integral has four things to change — integrand, differential, lower limit, upper limit — and only three of them appear on the same line. The limits sit in a different visual position. VCAA knows this, which is why it sets question parts whose entire content is the limits (Type 13, Type 14, Type 34).

The compounding trap. Changing the limits and then changing back to x without changing the limits back. [RPT10 E1]: "Some changed to a new variable u including changing the terminals, integrated, and then changed back to the original variable but without changing the terminals back."

The mechanical fix. Write the substituted integral with its new limits on a fresh line, as a complete object, before integrating: ∫_{u=a'}^{u=b'} h(u) du. Never carry x-limits past the substitution line.

5.3 Absolute values inside logarithmic antiderivatives

[FS] gives ∫1/(ax + b) dx = (1/a)logₑ|ax + b| + c. The modulus is printed on the sheet. The state still drops it, and the reports are explicit that dropping it is not cosmetic.

[RPT07 E1] on Q4 (35%):

"Some reversed the order in the denominator and then continued with ∫(1/(x+1) − 1/(x−1)) dx = logₑ(x + 1) − logₑ(x − 1). The lack of modulus signs in this case leads to logarithms of negative numbers, so a correct answer cannot be properly obtained."

[RPT13 E1] on Q2 (47%): "The most common error was the lack of modulus signs leading to the logarithms of negative numbers."

The two situations that differ. 1. The interval lies wholly on one side of the singularity. Then |ax + b| can legitimately be replaced by ±(ax + b), and a candidate who says so is safe. [RPT11 E1] records this being done correctly: "A few students justified removing modulus signs due to x ∈ R \ {−3, 3}" — though note VCAA is reporting, not endorsing, and the safe route is to keep the modulus and let it resolve at evaluation. 2. The integral is indefinite, or the interval straddles nothing in particular. Then the modulus must stay.

Where it bites hardest. Solving a differential equation by separation. [RPT10 E2] on Q3c: "Some students introduced modulus signs that caused confusion when they tried to reconcile their solution with the given result." The correct discipline is: keep |·| through the integration, apply the initial condition, determine the sign of the bracket from the initial condition, then drop the modulus with that sign fixed.

The associated fake rule. ∫1/f(x) dx ≠ logₑ(f(x)). Named in [RPT07 E1] ("As in the past, too many students used the incorrect 'log rule'"), [RPT09 E1] ("Some students did not attempt partial fractions and gave the common logarithm error ∫6/(4 − x²) dx = 6logₑ(4 − x²)"), [RPT07 E1] again on ∫1/((y+2)²+4) dy = logₑ((y+2)²+4), and [RPT17 E1] ("Students gave answers such as logₑ(x(1 + x²))").

5.4 Constants of integration and initial conditions

Nine separate reports name the missing constant. It is the most banal failure in the area and among the most expensive.

Three distinct failures hide under one name.

1. Omission. [RPT12 E2]: "a surprising number of students omitted the constant of integration." [RPT18 E1] on Q8b (23%): "the arbitrary constant of integration frequently missing."
2. Adding a constant to both sides. [RPT16 E2] on Q3a (42%): "Errors with constants were common among students who added a constant to both sides of the expression before attempting to find its value." One constant, one side.
3. Having it and failing to evaluate it. [RPT06 E1] on Q2 (45%): "Some students omitted the constant of integration and of those who did include it, a surprisingly large number were not able to evaluate it due either to arithmetic errors or an inability to simplify 9^{3/2}. Some attempted to multiply both sides of the equation by 3 to find the constant of integration but multiplied all terms other than the c."

The second-order case doubles the exposure. d²y/dx² = f(x) needs two constants and two conditions, and the conditions arrive in different forms — one is a gradient, one is a point. [RPT10 E1] on Q7 (33%): "A large number of students omitted the x term from their final answer. Some of these errors were due to the earlier omission of the constant of integration, and others were due to students simply forgetting to incorporate it back into their answer after working on the latter part of the question."

Reading the initial condition. y(1) = 0 means y = 0 when x = 1. [RPT16 E1] on Q10 (14%): "A number of students interpreted y(1) = 0 as x = 0 when y = 1." [RPT07 E1] records candidates using the Euler point (1.1, 1.1) as the initial condition rather than the given (1, 1).

The branch. Once the constant is found, an implicit solution often has to be made explicit, and there are two roots. Only one satisfies the initial condition. [RPT16 E1] on Q10 (14%): "A large number of students, when confronted with a square equals a constant, gave only the positive root. Many gave both roots but did not realise that only the negative root satisfied the initial conditions." [RPT24 E1] on Q7 (27%): "Some students failed to choose the correct sign."

5.5 The sign of an area

An area is positive. A definite integral is not. The gap between those two sentences is worth four marks in 2010 Exam 1 Q10, which 77% of the state did not earn in full.

[RPT10 E1] is the definitive statement:

"a large number did not recognise that area is a positive quantity, and did not place a negative sign in front of their integral, use absolute values, or reverse the terminals. As the area was below the x-axis, it was essential that one of these techniques be applied — this is assumed knowledge from Mathematical Methods CAS Units 3 and 4. Some students presented a negative answer as the area. Some simply dropped the negative sign or wrote A = −32√2/35 = 32√2/35, which is not a reasonable statement."

Note the last clause. Writing a false equation to repair a sign is marked as an error in its own right. [RPT12 E1] generalises it: "Equals signs must not be placed between quantities that are not equal."

The mirror error. Assuming a definite integral must be positive because it looks like an area. [RPT10 E1] on Q6 (34%): "A very large proportion of students gave 1/6 as their final answer. There were several reasons for this: many students simply made a sign error due to the proliferation of negatives involved, and quite a few students got the correct answer but then dropped the negative sign, seemingly assuming that a definite integral had to represent an area."

The discipline. Decide, before integrating, whether the question asks for a signed quantity (evaluate this integral) or an unsigned one (find the area). If unsigned and the region is below the axis, commit to one repair — negate, reverse, or modulus — and show it.

5.6 Setting up rather than evaluating

VCAA has built a whole question-part genre around the fact that most of the difficulty in an applied integral is in writing it down (§3.1, §3.5). The genre exists because the two skills separate cleanly, and the reports confirm they do.

The evidence that set-up is the harder half. On 2015 Exam 2 Section B Q1f, set-up scored 79% and evaluation 71%. On 2023 Exam 2 Section B Q3a, set-up 89%, evaluation 85%. On 2009 Exam 2 Section B Q4d, set-up 57%, evaluation 45%. On 2022 Exam 2 Section B Q1d, set-up 45%, evaluation 37%. In every case the evaluation part scores lower than the set-up part, because the evaluation inherits the set-up's errors and then adds its own. The two-part structure is not a kindness; it is a trap with a visible tripwire.

What a complete set-up contains. [RPT15 E2] lists what was missing: "not squaring f(x), incorrect terminals and the occasional omission of π and/or dx". Four separate things, each independently losable, in a one-mark part.

"Do not evaluate" means do not evaluate. But it also means the expression alone must be perfect, because there is no second chance and no partial credit from a downstream number. 2019 Exam 2 Section B Q1e is a two-mark "write down but do not evaluate" and 79% of the state scored zero [RPT19 E2].

On Examination 2, write the instantiated integral. [RPT24 E2]: "Students were expected to write an expression for [the function] within the definite integral instead of stating the generic formula. Many students made transcription errors when transferring their answer from their CAS to the script."

The related set-up-only genre is the "time as an integral" part (§3.1), where the whole question is recognising that t = ∫1/F(h) dh.

5.7 Exactness in Examination 1

The front page says it: "Unless otherwise specified, an exact answer is required for each question." Three consequences, all visible in the percentages.

First, exact values of circular functions are load-bearing. [RPT11 E1] on Q11 (46%): "Exact values were again a problem." [RPT14 E1] on Q5b (48%): "Many students made simplification errors and errors in exact values for circular functions." [RPT13 E1] on Q9 (29%): "Too many students were unable to evaluate sin²(π/3)." [RPT15 E1] on Q8c (86%): "The main errors seen were π/3 and π/12, with some students not knowing the exact values." [RPT07 E1] on Q8b (46%): "Typical errors involved taking tan⁻¹(−1) to be 0, −π/2, or π/4."

Second, surd and fraction manipulation is a separate failure mode from the calculus. [RPT10 E1] on Q10 (23%): "Several students found the area correctly but were not able to express it in the required form, often not able to manipulate the surds or simplify their answer." [RPT16 E1] and [RPT17 E1] both list "arithmetic skills. The inability to evaluate expressions, especially those involving fractions or surds, was common" as a top-level weakness of the whole paper.

Third, over-simplification destroys correct answers. This is the most self-inflicted loss in the area. [RPT11 E1] on Q11 (46%): "Several students simplified their final answer poorly at the end … Other examples involved erroneously attempting to put the answer on a common denominator or to factorise it." [RPT06 E1] on Q5b (28%): "Others, after arriving at a correct answer, tried to go further and incorrectly rationalised the denominator of the fraction." [RPT13 E1] on Q9 (29%): "Some students arrived at the correct answer but then made errors when attempting to use a common denominator (which was unnecessary)." [RPT15 E2] on Q3a: "Some students did unnecessary further working out, attempted to simplify a correct answer and changed it to an incorrect answer."

The rule. Simplify only as far as the demanded form requires. If the question says "in the form a√b/c", stop when you are in that form. If it says nothing, stop when the expression is a single clean exact value.

5.8 A note on time

Several of the deepest separators are not conceptually hard; they are simply last. 2024 Exam 2 Section B Q3e (4%), 2011 Exam 2 Section B Q5d (1%), 2016 Exam 2 Section B Q3e (23%) and 2017 Exam 2 Section B Q3d (26%) are all final or near-final parts of Section B questions, and the reports on all four begin with a variant of "Many students did not attempt this question." For a 45+ candidate the implication is a pacing rule, not a mathematics rule: Section B questions are worth 10–13 marks each and the last two marks of each are the cheapest marks on the paper if you reach them.

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6. A worked method sheet — the ten highest-yield types

Chosen by (marks in the archive) × (probability of appearing) × (how much the reports say is recoverable). For each: the method in the order you should write it, then the technology-free (Examination 1) version called out separately.

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Method 1 — Implicit differentiation for a tangent or normal

Frequency. Examination 1, nineteen years out of twenty. Typically 3–4 marks, often Question 1.

Method. 1. Differentiate every term with respect to x. A term in y alone contributes (d/dy)·dy/dx. A term in xy needs the product rule. A constant contributes 0. 2. Substitute the given point now. Do not solve for dy/dx symbolically first. 3. Collect the dy/dx terms; divide once. 4. Tangent: y − y₁ = m(x − x₁). Normal: replace m with −1/m. 5. Re-read the question. Was it the gradient, the equation, or the normal?

Technology-free version. This is the technology-free version — [SD] says "apply implicit differentiation, by hand in simple cases". Two extra by-hand disciplines: keep every negative inside a bracket ([RPT10 E1]: "The most frequent error involved not distributing the minus sign"), and do not expand a squared bracket before differentiating.

Worked shape (2025 Exam 1 Q1, 48%). xe^{2y} + y²e^x = 8e⁴ at (4, −2): e^{2y} + 2xe^{2y}(dy/dx) + 2ye^x(dy/dx) + y²e^x = 0. Substitute x = 4, y = −2: e^{−4} + 8e^{−4}(dy/dx) − 4e⁴(dy/dx) + 4e⁴ = 0, then solve the single linear equation and write the tangent.

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Method 2 — Separation of variables with an initial condition

Frequency. The most common Calculus type on Examination 1: sixteen instances in the archive, typically 3–5 marks.

Method. 1. Write g(y) dy = f(x) dx. If the equation is given as dx/dy, do not reciprocate a sum term by term — reciprocate the whole side. 2. Integrate both sides. One constant, on the right. 3. Keep modulus signs on every logₑ. 4. Substitute the initial condition immediately, before rearranging. Solve for c. 5. Rearrange to the demanded form. If a square root appears, choose the branch that satisfies the initial condition and say so. 6. Check the demanded form: "y as a function of x"? "with integer coefficients"? "in the form …"?

Technology-free version. Exam 2 permits deSolve — unless the question says "Use calculus to solve", which 2025 Exam 2 Section B Q3d does and which [RPT25 E2] glosses as "students are required to show the steps needed to find the solution to gain all 3 marks". So the by-hand method is required on both papers; the only difference is that on Examination 2 you may check with CAS.

Worked shape (2019 Exam 1 Q1, 42%). dy/dx = 2ye^{2x}/(1 + e^{2x}), y(0) = π: ∫dy/y = ∫2e^{2x}/(1 + e^{2x}) dx → logₑ|y| = logₑ(1 + e^{2x}) + c. At (0, π): logₑ π = logₑ 2 + c. So y = (π/2)(1 + e^{2x}). Note that no substitution is needed on the right — the numerator is already the derivative of the denominator, which is exactly the recognition [RPT19 E1] says the state failed.

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Method 3 — Integration using partial fractions

Frequency. Ten instances, almost all Examination 1, 3–5 marks.

Method. 1. If deg(num) ≥ deg(den), divide first. (2026 NHT Exam 1 Q5: (x² + 2x)/(x + 1) = x + 1 − 1/(x + 1).) 2. Factorise the denominator completely. 3. Choose the form by the factor type: A/(x − a) for each distinct linear factor; (Bx + C)/(x² + k) for each irreducible quadratic; A/(x − a) + B/(x − a)² for a repeated factor. 4. Multiply up and substitute the roots to get A, B, C in one line each. 5. Integrate. Every linear factor gives logₑ| |. Watch the sign when the factor is (a − x): ∫1/(a − x) dx = −logₑ|a − x| + c. 6. Combine with log laws — correctly. p logₑ m − q logₑ n = logₑ(m^p/n^q).

Technology-free version. This is an Examination 1 type almost exclusively. The by-hand accelerator is the cover-up rule: to find A in A/(x − a), substitute x = a into the original expression with (x − a) deleted.

Worked shape (2013 Exam 1 Q2, 47%). ∫₀¹(x − 5)/(x² − 5x + 6) dx. Factorise (x − 2)(x − 3). Then (x − 5)/((x−2)(x−3)) = 3/(x−2) − 2/(x−3). So the integral is [3logₑ|x − 2| − 2logₑ|x − 3|]₀¹ = 2logₑ3 − 5logₑ2 = logₑ(9/32).

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Method 4 — Volume of revolution

Frequency. Nineteen instances, both papers, 1–5 marks.

Method. 1. Identify the axis. About x: V = π∫ᵃᵇ y² dx, terminals in x. About y: V = π∫_{c}^{d} x² dy, terminals in y, and you must first solve for x² in terms of y. 2. Between two curves about x: V = π∫(y_outer² − y_inner²) dx. Difference of squares. 3. Write the constant π at the front before anything else, and never remove it. 4. Square the function. If the function was given with a square root, the square root disappears — that is the design. 5. Classify the resulting integrand and integrate. 6. Evaluate exactly (Examination 1) or to the stated precision (Examination 2).

Technology-free version. V = π∫y² dx is not on [FS]; memorise both forms. On Examination 1, expect the squared integrand to be exactly one of: a partial-fractions rational, a sin²/cos² needing a double angle, or a split standard form.

Worked shape (2020 Exam 1 Q8, 20%). y = 2√((x² + x + 1)/((x + 1)(x² + 1))) on [0, √3]: V = 4π∫₀^{√3} (x² + x + 1)/((x + 1)(x² + 1)) dx. Decompose into A/(x + 1) + (Bx + C)/(x² + 1); the result splits into a logₑ|x + 1| term, a ½logₑ(x² + 1) term and an arctan term. [RPT20 E1]: "Many students identified the correct form of the partial fraction decomposition for the integrand."

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Method 5 — Integration by substitution with a change of limits

Frequency. Eleven separators alone; appears somewhere in essentially every paper.

Method. 1. Choose u = g(x) by the rules in §5.1. 2. Differentiate: du = g'(x) dx. Rearrange so that the exact expression appearing in the integrand is isolated. 3. On a new line, write the whole integral in u, with both limits converted: u_lower = g(a), u_upper = g(b). 4. If an x survives, eliminate it via x = g⁻¹(u). 5. Integrate in u and evaluate at the u-limits. Do not return to x. 6. Sign check: if the substitution introduced a negative, either carry it or swap the limits — one or the other, not both.

Technology-free version. The same, with exact values at the limits. The extra by-hand risk is a trigonometric limit: u = cos(6x) at x = π/12 gives cos(π/2) = 0, and a candidate who writes 1 there loses everything downstream.

Worked shape (2010 Exam 1 Q6, 34%). ∫_{π/2}^{3π/4} cos²(2x)sin(2x) dx. Set u = cos(2x), so du = −2sin(2x) dx and sin(2x) dx = −½ du. Limits: x = π/2 ⇒ u = cos(π) = −1; x = 3π/4 ⇒ u = cos(3π/2) = 0. The integral becomes −½∫_{−1}^{0} u² du = −½[u³/3]_{−1}^{0} = −½(0 + ⅓) = −⅙. [RPT10 E1] records that a very large proportion of the state gave +⅙.

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Method 6 — Integration by parts

Frequency. New in 2023; four instances in three years, 3–4 marks. Growing.

Method. 1. Split the integrand into u and dv/dx. Choose u to be whichever factor becomes simpler on differentiation. Priority order that works for every VCE case: logₑ → inverse circular → power of x → exponential/trigonometric. 2. Write u, du/dx, dv/dx, v in a two-by-two block before substituting. This alone prevents most errors. 3. Apply [FS]: ∫u(dv/dx) dx = uv − ∫v(du/dx) dx. 4. If the remaining integral is still a product of the same type, apply parts again (∫x²cos(2x) dx needs two rounds). 5. For a definite integral, evaluate [uv]ₐᵇ at once and keep the remaining integral definite with the same limits.

Technology-free version. This is the Examination 1 form. [RPT23 E1] reports it as an area of strength in its first year — which makes any mark lost here expensive relative to the field.

Worked shape (2023 Exam 1 Q5, 53%). ∫₁² x² logₑ(x) dx: u = logₑ x, dv/dx = x², so du/dx = 1/x, v = x³/3. = [x³logₑ(x)/3]₁² − ∫₁²(x³/3)(1/x) dx = (8logₑ2)/3 − [x³/9]₁² = (8logₑ2)/3 − 7/9.

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Method 7 — Related rates

Frequency. Eleven instances, predominantly Examination 2 Section B, 2–5 marks; also a standing Examination 1 type (2016 Q4).

Method. 1. List every variable with its symbol and unit. Convert units now. 2. Write the geometric or modelling relation connecting them. If two variables are geometrically linked (cone radius and height), eliminate one by similar triangles first. 3. Differentiate the relation with respect to the variable that makes the chain close. 4. Write the chain explicitly: dA/dt = (dA/dx)·(dx/dt), or dh/dt = (dh/dV)·(dV/dt) with dh/dV = 1/(dV/dh). 5. Only now substitute the instantaneous values. 6. If the rate is a net rate, dV/dt = (in) − (out), and say so. 7. If the part says "show that", write the product of the derivatives and then the algebra that turns it into the stated expression.

Technology-free version. 2016 Exam 1 Q4 (40%) is the model: x = arctan(t), A = 6x², so dA/dt = 12x·1/(1 + t²), and at t = 1, x = π/4, giving dA/dt = 12(π/4)/2 = 3π/2. The by-hand risks are the surface-area formula and the confusion between 1 + t² and 1 + x² — both named in [RPT16 E1].

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Method 8 — Euler's method

Frequency. Six instances, both papers, 1–2 marks. Always cheap, frequently dropped.

Method. 1. Read off x₀, y₀, h, and the number of steps. If the target is to the left of the start, h is negative. 2. Draw a four-column table: n, xₙ, yₙ, f(xₙ, yₙ). 3. Apply yₙ₊₁ = yₙ + h·f(xₙ, yₙ) one row at a time. Use f, never f'. 4. Write every intermediate yₙ in the answer space. On Examination 2 this is the difference between one mark and two. 5. Round only at the end.

Technology-free version. Keep exact fractions. [RPT07 E1] on Q7a (36%) shows the failure precisely: the state could not add 1 + 1/10 + 1/11, "Very often the lowest common denominator chosen for 10 and 11 was 121."

Worked shape (2025 Exam 2 Section B Q3c, 38.93%). dQ/dt = (300 − Q)/150, Q(0) = 5, h = 15: Q(15) = 5 + 15(300 − 5)/150 = 5 + 29.5 = 34.5; Q(30) = 34.5 + 15(300 − 34.5)/150 = 34.5 + 26.55 = 61.05. [RPT25 E2]: "A tabulated approach was acceptable as long as both Q(15) and the final answer were shown."

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Method 9 — Setting up a definite integral without evaluating it

Frequency. Eighteen papers contain the instruction. Always 1–2 marks. The highest marks-per-second in the subject.

Method. 1. Identify the quantity: area, volume, arc length, surface area, or time. 2. Select the template and write the constant first: - area = ∫(upper − lower) dx - volume about x = π∫y² dx; about y = π∫x² dy - arc length (parametric) = ∫√((dx/dt)² + (dy/dt)²) dt [FS] - surface area = 2π∫y·√(…) dx about x, = 2π∫x·√(…) dy about y [FS] - time from dh/dt = F(h) = ∫1/F(h) dh 3. Substitute the instantiated function, not the generic symbol. 4. Write both terminals, in the demanded variable. 5. Write the differential. 6. Stop. Do not evaluate if told not to.

Checklist to run on the finished expression: constant present? function squared where required? terminals in the right variable? terminals in the right order? differential present?

Technology-free version. On Examination 1 the instruction usually reads "Express … as a definite integral" (2006 Exam 1 Q6a, 70%) or "Find a, b and c" where the integral form is supplied (2018 Exam 1 Q10). The work is identical; the exactness rules then apply to the terminals.

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Method 10 — Reading the second derivative

Frequency. Fifteen instances across both papers, 1–3 marks, and rising under the 2023 design because the logistic equation and the counter-example items both feed it.

Method. 1. To get d²y/dx² from dy/dx = f(x, y): differentiate f with respect to x, using the product rule as needed; every ∂/∂y term produces a factor dy/dx; substitute f(x, y) back in for that factor. The result should contain no dy/dx. 2. For an inflection: solve d²y/dx² = 0, then verify a sign change. A squared factor gives no sign change and therefore no inflection. 3. For concavity: f'' > 0 is concave up; report an interval. 4. To show that there is no inflection: show f'' is never zero, or that its only zero is at a repeated factor, or that every factor has constant sign — and say which. 5. Give coordinates if asked: substitute back into f, not f' or f''.

Technology-free version. Examination 1 examples are 2020 Exam 1 Q6b ("Hence, show that the graph of f has a point of inflection at x = 2"), 2024 NHT Exam 1 Q3 and 2026 NHT Exam 1 Q3. The by-hand risk is differentiating a quotient twice; converting to a negative index first usually halves the work.

Worked shape (2013 Exam 2 Section B Q3d.i, 9%). dN/dt = 0.4N(6 − logₑ N): d²N/dt² = 0.4(dN/dt)(6 − logₑ N) + 0.4N(−1/N)(dN/dt) = (dN/dt)(0.4(6 − logₑ N) − 0.4) = 0.4N(6 − logₑ N)·(2 − 0.4logₑ N)/… — collapsing to an expression in N alone, which is what makes d²N/dt² = 0 ⇒ logₑ N = 5 ⇒ N = 148 immediate in the next part.

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Appendix — the five questions to work first

If time is short, these five carry the most transferable value, chosen because each is a separator whose report identifies a single recoverable cause.

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Sources: corpus/sm/questions.json (387 Calculus parts, 623 published marks, 158 separators, as at 15 September 2026); corpus/sm/text/*.txt (papers and reports 2006–2026); corpus/sm/text/Documents_exams_mathematics_specmaths1-formula-w.txt (current formula sheet); research/sm/01-study-design.md. Every percentage is VCAA's published full-marks rate. Quotations from papers and reports are verbatim from the plain-text extractions, with obvious extraction artefacts silently corrected only where the reading is unambiguous; where it is not, the gap is marked rather than filled.