Integration methods

Practice

Recall it first, then use it — testing yourself builds the memory, where rereading the notes above only feels easier.

Recall and application shuffled together — you have to work out what each item wants before you can answer it, which is the part the exam actually tests. Both kinds feed the same spacing schedule, so anything you miss comes back sooner.

Say the answer out loud before flipping — that attempt is where the learning happens, even when you get it wrong.

Exam-style only. Instant feedback, shuffled each attempt, best score saved on this device.

What actually comes up

Three years of real questions, part by part, with the setup and values from the paper and the grade band each part sat at according to that year's marking schedule. Start with the pattern table — it is the fastest way to decide what to revise.

Across 3 years

The pattern — what to prepare

The 2025 paper ↗ Marking schedule ↗

Question 1 — standard integrals, separation of variables

Five independent parts, no shared scenario: a standard integral, a separable differential equation with a boundary condition, a pair of definite integrals that pin down two constants, a trig area, and a differential equation you have to rearrange before you can separate it.

• a Find ∫ 6 sec(3x) tan(3x) dx. Achieved
• b Solve the differential equation dy/dx = 3√x + 2/√x, given that y = 10 when x = 4. Merit
• c It is given that ∫₁² (a − 6k/x²) dx = 3 and ∫₁ᵏ 6x dx = a, where a and k are constants. Determine the possible value(s) of k. Excellence
• d The graph shows part of y = 4 sin(5x) cos(3x), shaded from the origin to the first return to the x-axis. Find the shaded area under the curve between x = 0 and x = π/6. Merit
• e Consider the differential equation y − xy − (1 + x)·dy/dx = 0. Given that y = 3 when x = 0, find the value(s) of y when x = 2. Excellence

Question 2 — substitution-type, constants, kinematics

A power-of-a-linear integral, a rate of change integrated back to its function, a constant found from a definite integral, a kinematics chain (acceleration → velocity → displacement), and an odd-power trig area.

• a Find ∫ 10/(2x + 1)⁶ dx. Achieved
• b The rate of change of a particular function, p, is given by dp/dt = 5 cos(4t). Find the function, given that p = 8 when t = π/24. Merit
• c Find the value of the constant k, given that ∫₀ᵏ 1/√(4x + 1) dx = 1. Merit
• d A particle’s acceleration is modelled by a(t) = 2.4e^(−0.3t) − 1, where a(t) is in m s⁻² and t is the time in seconds from the start of timing. Initially, at a fixed point P, the particle had a velocity of 6 m s⁻¹. How far from the point P is the particle 3 seconds after timing started?
  • a(t) = 2.4e^(−0.3t) − 1 m s⁻²
  • v(0) = 6 m s⁻¹
  • t = 3 s
  Merit
• e The graph shows y = sin³x cos³x, shaded from the origin to (π/2, 0). Find the shaded area under the curve between x = 0 and x = π/2. Excellence

Question 3 — trapezium rule, areas, centroid

Opens with a numerical estimate from measured depths, then a log integral left in terms of a limit, a separable differential equation, an area that must be turned into a cubic, and a balance point found from a ratio of two integrals.

• a The diagram shows the cross-section of a hole dug in the ground. The depth is measured every 5 metres across the top of the hole. Using the trapezium rule, find an estimate for the area of the cross-section of the hole.
  • 5 m intervals
  • depths 2.12, 2.32, 2.65, 2.54, 1.88 m
  Achieved
• b Find ∫₁ᵏ 10/(2x − 1) dx, giving your answer in terms of k, where k is a constant and k > 1. Merit
• c Consider the differential equation dy/dx = √(4y + 1)/x². Given that y = 2 when x = ⅔, find the value(s) of y when x = ⅘. Merit
• d The graph shows part of the curve y = (x² + 6)/x⁴, where x > 0, shaded between x = p and x = 2p. The area of the shaded region is 9/4 units². Prove that 9p³ − 2p² − 7 = 0. Excellence
• e Murray is hanging a piece of art whose curved edge is f(x) = ⅓(x² + 3)², from x = 0 to x = 3. The x-value of the balance point is given by ∫₀³ x·f(x) dx ÷ ∫₀³ f(x) dx. Use this formula to find the x-value of the hanging point.
  • f(x) = ⅓(x² + 3)²
  • 0 ≤ x ≤ 3
  Excellence

The 2024 paper ↗ Marking schedule ↗

Question 1 — integrals & areas

A sec–tan integral, an area under a cubic-in-a-bracket, a kinematics chain that starts 360 m from the reference point, a trig differential equation, and an area between two trig curves.

• a Find ∫ 6 sec(2x) tan(2x) dx. Achieved
• b The graph shows the function y = 40x(5x² − 3)³, shaded from x = −0.4 to the origin. Find the shaded area under y = 40x(5x² − 3)³ between x = −0.4 and x = 0. Merit
• c An object’s velocity is modelled by v(t) = 26.4·∛t, where v is in m s⁻¹ and t is the time in seconds since the start of timing. Initially, the object was 360 metres from a point P. Calculate the distance that this object has travelled from the point P when it has reached a velocity of 264 m s⁻¹.
  • v(t) = 26.4·∛t
  • initial displacement 360 m from P
  • v = 264 m s⁻¹
  Merit
• d Consider the differential equation dy/dx = 24cos³(x)·sin(x). Given that y = 6 when x = π/3, find the value(s) of y when x = π/2. Excellence
• e The graph shows the curves y = 3sec²x and y = 2tan²x. Find the area of the shaded region enclosed by the two curves, x = 1, and the y-axis. Excellence

Question 2 — expansion, constants, DE, mass

Expand-then-integrate, a lower limit recovered from a definite integral, a separable equation in y², an area between a curve and a horizontal line, and a physical integral left entirely in symbols.

• a Find ∫ (3x⁴ + 4)² dx. Achieved
• b Find the value of k, given that ∫ₖ¹⁶ 3√x dx = 112. Merit
• c Consider the differential equation dy/dx = 12y²e^(3x). Given that y = 0.5 when x = 0, find the value of y when x = ⅓. Merit
• d The graph shows part of the function y = sin²x together with the line y = 1. Find the shaded area enclosed between y = sin²x, y = 1, x = −π/2 and x = π/2. Excellence
• e The mass M of a spherical object of radius p can be approximated by M = ∫₀ᵖ 4πr²·a/(1 + br³) dr, where a, b and p are all positive constants. Using this formula, find an expression for the mass M of a spherical object, giving your answer in terms of a, b, p and π.
  • M = ∫₀ᵖ 4πr²·a/(1 + br³) dr
  • a, b, p > 0
  Excellence

Question 3 — e^ integrals, a coffee, equal areas

Two exponential terms to integrate, a log-form differential equation, an upper limit that appears on both sides, a bisected cosine area, and Newton’s law of cooling built from scratch.

• a Find ∫ (e^(2x) + 3/e^(4x)) dx. Achieved
• b Solve the differential equation dy/dx = 5/(4x − 3), given that y = 10 when x = 1. Merit
• c Find the value of m, given that ∫₋₁^m (4x + 5)/(2x + 3) dx = 2m. Merit
• d The graph shows y = 2cos(x/2) from x = 0 to x = π, divided at x = k into Area A (left) and Area B (right). Find the value of k so that the shaded Area A will be equal to the shaded Area B. Excellence
• e A teacher makes a cup of coffee at the start of interval and leaves it in the staff room, where the temperature is 18 °C. After 30 minutes the coffee is 50 °C. After a further one hour it has cooled to 30 °C. The rate at which the temperature changes at any instant is proportional to the difference between the temperature of the coffee, N, and the temperature of the room. Write a differential equation that models this situation, and then solve it to calculate the temperature of the cup of coffee when it was made.
  • room 18 °C
  • 50 °C after 30 min
  • 30 °C after a further 60 min
  Excellence

The 2023 paper ↗ Marking schedule ↗

Question 1 — a rational integral, kinematics, two bounded areas

A linear-plus-reciprocal integral, a kinematics chain in km hr⁻¹ starting 3 km from the reference point, an area between a root and a parabola, a separable equation, and an area bounded by three different curves at once.

• a Find ∫ (3x + 2 + 1/(3x + 2)) dx. Achieved
• b An object’s velocity is modelled by v(t) = sec²t, where v is in km hr⁻¹ and t is the time in hours from the start of timing. Initially the object was 3 km from a point P. Find the distance of this object from the point P after π/4 hours.
  • v(t) = sec²t
  • initial displacement 3 km from P
  • t = π/4 hours
  Merit
• c The graph shows the functions y = √x and 8y = x², shaded between their intersections. Find the shaded area enclosed between y = √x and 8y = x². Merit
• d Consider the differential equation dy/dx = y(2x − 3x²). Given that y = 1 when x = 2, find the value(s) of y when x = 1. Merit
• e The shaded region is bounded by three graphs: y² = 10 − x, y = 3x, and y = −sin(πx/10). Find the area of the shaded region.
  • y² = 10 − x
  • y = 3x
  • y = −sin(πx/10)
  Excellence

Question 2 — an exponential integral, a rate constant, a chocolate factory

An exponential integral, a negative-power differential equation, an upper limit that appears on both sides, a trig quotient that needs rewriting before it will integrate, and a proportional-to-the-square model built from scratch.

• a Find ∫ 4e^(2x−1) dx. Achieved
• b Solve the differential equation dy/dx = (4x + 1)^(−1/2), where x ≥ 0, given that when x = 6, y = 7.5. Merit
• c Find the value of k, given that ∫₂ᵏ (6x − 3)/(2x − 3) dx = 3k. Merit
• d Find ∫ (cos 2x + sin 2x)/(cos 2x − sin 2x) dx. Merit
• e A cake factory pumps liquid chocolate out of a container so that the rate of change of the volume remaining is proportional to the square of the volume remaining. After one hour of use, p litres remain, where p is a positive constant. After a further one hour, only ⅘p litres remain. Write a differential equation modelling this, and solve it to calculate how much liquid chocolate was in the container at the start of the day, giving your answer in terms of p.
  • rate ∝ (volume remaining)²
  • p litres after 1 hour
  • ⅘p litres after 2 hours
  Excellence

Question 3 — Simpson’s rule, a trig integral, a bounded region

Opens with Simpson’s Rule on five measured offsets from a garden path, then a root quotient, a product-of-sines area, an exponential-quotient kinematics chain, and a differential equation that has to be factorised before it will separate.

• a A garden designer measures a section of garden from a path: offsets of 10, 13, 16, 15 and 14 metres, taken at 3-metre intervals. Using these measurements, and Simpson’s Rule, find an approximation of the area of the garden section.
  • offsets 10, 13, 16, 15, 14 m
  • 3 m intervals
  Achieved
• b Find ∫ ((√x − 3)/√x) dx. Achieved
• c The graph shows the function y = 5 sin(3x) sin(x). Find the shaded area under y = 5 sin(3x) sin(x). Merit
• d An object’s acceleration is modelled by a = e^(2t)/(4e^(2t) − 3), where t ≥ 0, a is in m s⁻² and t is the time in seconds from the start of timing. At t = 0 the object had a velocity of 5 m s⁻¹. Find the object’s velocity when t = 4 seconds.
  • a = e^(2t)/(4e^(2t) − 3)
  • v(0) = 5 m s⁻¹
  • t = 4 s
  Merit
• e Consider the differential equation (1 − x²)(1 + y)·dy/dx + (1 − x)(1 − y²) = 0. Given that y = 0 when x = 2, find the value(s) of y when x = 6. Excellence

Read the last column as a to-do list. Every paper opens with a bare integral, then a differential equation with an initial condition, then climbs to an Excellence area or rate-of-change problem wrapped in a short story — a factory, a garden, a moving object.

Past exams & resources

⏱ Timed opens the paper and starts a countdown matching the exam length. Official NZQA PDFs. If a link 404s, that year’s paper may sit under a slightly different code: browse the NZQA subject page.