IB DP Mathematics: Applications and Interpretation – Further differentiation and integration, volumes of revolution and kinematics (HL) Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

This practice set covers the HL further calculus unit of IB Diploma Programme Mathematics: Applications and Interpretation. It is aligned to the IB Mathematics: applications and interpretation guide, first assessment 2021, syllabus sections 5.9–5.13, and every question is HL only (AHL). It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 HL sessions.

All three AI HL papers require technology, so every question here is labelled “(calculator allowed)”. Where a question says “show that” or asks for an exact answer, write the analytic working; a GDC value alone will not earn the method marks. Give other answers exactly or to 3 significant figures.

Learn the methods first in the study guide and the revision notes. The IB DP Maths AI course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator allowed) Let f(x) = 4√x − 3 ln x + tan x, for x > 0. Find f′(x). [3]

2. (calculator allowed)

(a) Differentiate y = x³e⁻²ˣ, giving your answer in factorised form. [2] (b) Show that d/dx(sin x/(1 + cos x)) = 1/(1 + cos x). [3]

3. (calculator allowed) A drone rises vertically at 3 m s⁻¹ from a point on level ground 40 m from an observer. Let θ be the angle of elevation of the drone from the observer, in radians. Find the rate at which θ is increasing when the drone is 30 m high. [5]

4. (calculator allowed) Let f(x) = xe⁻ˣ, for x ≥ 0.

(a) Find f′(x). [2] (b) Find f′′(x) and use it to show that the stationary point is a maximum. [2] (c) Find the coordinates of the point of inflexion, justifying your answer. [2]

5. (calculator allowed)

(a) Find ∫ (6√x + 3/x − 2/cos²x) dx. [3] (b) Find ∫ e^(1 − 4x) dx. [2]

6. (calculator allowed)

(a) Use the substitution u = sin x to find ∫ cos x e^(sin x) dx. [3] (b) Show that ∫₁² x/(x² + 1) dx = (1/2) ln(5/2). [3]

7. (calculator allowed) Let f(x) = x³ − 4x.

(a) Find ∫₀³ f(x) dx. [2] (b) Find the area of the region enclosed by the curve y = f(x), the x-axis and the line x = 3, for 0 ≤ x ≤ 3. [3]

8. (calculator allowed) The inside of a bowl is modelled by rotating the curve y = x²/4, 0 ≤ y ≤ 9, through 2π about the y-axis. All lengths are in cm.

(a) Show that the volume of water in the bowl when the depth is h cm is V = 2πh² cm³. [3] (b) Find the capacity of the bowl. [1] (c) Water is poured in at 20 cm³ s⁻¹. Find the rate at which the depth is increasing when h = 5. [3] (d) Find the depth of water when the bowl is half full. [2]

9. (calculator allowed) The region enclosed by y = √x + 1, the x-axis, the y-axis and the line x = 4 is rotated through 2π about the x-axis. Find the exact volume of the solid. [4]

10. (calculator allowed) A particle moves in a straight line. Its velocity is v(t) = t² − 6t + 8 m s⁻¹ for 0 ≤ t ≤ 6 seconds, and its displacement from O at t = 0 is 1 m.

(a) Find the times when the particle is at rest. [2] (b) Find the acceleration when t = 5. [2] (c) Find the displacement of the particle from O when t = 6. [3] (d) Find the total distance travelled for 0 ≤ t ≤ 6. [3] (e) Determine whether the particle is speeding up or slowing down at t = 3.5. [2]

11. (calculator allowed)

(a) The displacement of a particle is x = 4 sin 2t + 3 cos 2t. Show that ẍ = −4x. [3] (b) A second particle has velocity v = s² + 2s m s⁻¹, where s is its displacement in metres. Find its acceleration when s = 1. [3]

12. (calculator allowed) A small boat moves in a straight line with velocity v(t) = 20e^(−0.3t) − 8 m s⁻¹ for 0 ≤ t ≤ 8 seconds.

(a) Find the time at which the boat is at rest. [2] (b) Find the acceleration at t = 2. [2] (c) Find the displacement of the boat over the 8 seconds. [2] (d) Find the total distance travelled over the 8 seconds. [2]

Answers

1. f′(x) = 2x^(−1/2) [1] − 3/x [1] + 1/cos²x [1], so f′(x) = 2/√x − 3/x + 1/cos²x [3] Examiner insight: Each term is a separate accuracy mark, so one wrong term costs only that mark if the others are right.

2. (a) 3x²e⁻²ˣ + x³(−2e⁻²ˣ) [1] = x²e⁻²ˣ(3 − 2x) [1] (b) ((1 + cos x)cos x − sin x(−sin x))/(1 + cos x)² [1] = (cos x + cos²x + sin²x)/(1 + cos x)² [1]. Using cos²x + sin²x = 1, this is (1 + cos x)/(1 + cos x)² = 1/(1 + cos x) [1] Examiner insight: In a “show that”, the line using cos²x + sin²x = 1 must be written; jumping to the given result loses the final mark.

3. tan θ = h/40 [1]. Differentiate with respect to t: (1/cos²θ)(dθ/dt) = (1/40)(dh/dt) [1]. At h = 30 the hypotenuse is 50, so cos θ = 40/50 = 0.8 [1]. dθ/dt = 0.8² × 3/40 [1] = 0.048 rad s⁻¹ [1] Examiner insight: Substituting h = 30 before differentiating makes h a constant and scores only the first mark.

4. (a) Product rule: e⁻ˣ + x(−e⁻ˣ) [1] = (1 − x)e⁻ˣ [1] (b) f′′(x) = −e⁻ˣ − (1 − x)e⁻ˣ = (x − 2)e⁻ˣ [1]. f′(x) = 0 at x = 1 and f′′(1) = −e⁻¹ < 0, so a maximum at (1, e⁻¹) [1] (c) f′′(x) = 0 at x = 2, and f′′ changes from negative to positive there (concave-down to concave-up) [1], so the point of inflexion is (2, 2e⁻²) ≈ (2, 0.271) [1] Examiner insight: For (c), stating f′′(2) = 0 without the sign change is not a justification and loses the reasoning mark.

5. (a) 6x^(1/2) → 4x^(3/2) [1]; 3/x → 3 ln|x| [1]; −2/cos²x → −2 tan x, so 4x^(3/2) + 3 ln|x| − 2 tan x + C [1] (b) Dividing by −4 [1]: −(1/4)e^(1 − 4x) + C [1] Examiner insight: A missing “+ C” on an indefinite integral usually costs the final accuracy mark in that part.

6. (a) u = sin x [1], du = cos x dx [1], so ∫ eᵘ du = e^(sin x) + C [1] (b) u = x² + 1, so x dx = (1/2) du [1]; limits x = 1 → u = 2, x = 2 → u = 5 [1]; (1/2)[ln u]₂⁵ = (1/2)(ln 5 − ln 2) = (1/2) ln(5/2) [1] Examiner insight: A GDC value of 0.458 does not show the exact result; the substitution and the changed limits must appear.

7. (a) [x⁴/4 − 2x²]₀³ [1] = 81/4 − 18 = 9/4 [1] (b) The curve crosses the x-axis at x = 2 inside the interval [1]. ∫₀² f(x) dx = −4 and ∫₂³ f(x) dx = 25/4 [1]. Area = 4 + 25/4 = 41/4 = 10.25 [1] Examiner insight: Giving 9/4 again in (b) scores nothing, because the net integral is not the area.

8. (a) x² = 4y [1]. V = π∫₀ʰ 4y dy [1] = π[2y²]₀ʰ = 2πh² [1] (b) 2π(9²) = 162π ≈ 509 cm³ [1] (c) dV/dt = 4πh (dh/dt) [1]. At h = 5: 20 = 20π (dh/dt) [1], so dh/dt = 1/π ≈ 0.318 cm s⁻¹ [1] (d) 2πh² = 81π [1], h = √40.5 ≈ 6.36 cm [1] Examiner insight: Using πy² dx or x-limits in (a) is the wrong method for a y-axis rotation and loses every mark in that part.

9. V = π∫₀⁴ (√x + 1)² dx [1] = π∫₀⁴ (x + 2√x + 1) dx [1] = π[x²/2 + (4/3)x^(3/2) + x]₀⁴ = π(8 + 32/3 + 4) [1] = 68π/3 [1] Examiner insight: The question asks for an exact volume, so the GDC value 71.2 alone earns no final accuracy mark.

10. (a) (t − 2)(t − 4) = 0 [1], so t = 2 s and t = 4 s [1] (b) a = 2t − 6 [1], a(5) = 4 m s⁻² [1] (c) ∫₀⁶ (t² − 6t + 8) dt = [t³/3 − 3t² + 8t]₀⁶ [1] = 12 m [1]. Displacement from O = 1 + 12 = 13 m [1] (d) ∫₀² v dt = 20/3, ∫₂⁴ v dt = −4/3, ∫₄⁶ v dt = 20/3 [1]. Add the absolute values [1]: 44/3 ≈ 14.7 m [1] (e) v(3.5) = −0.75 and a(3.5) = 1 [1]. Opposite signs, so the particle is slowing down [1] Examiner insight: In (c), leaving out the initial displacement of 1 m loses the final mark even with correct integration.

11. (a) ẋ = 8 cos 2t − 6 sin 2t [1]; ẍ = −16 sin 2t − 12 cos 2t [1] = −4(4 sin 2t + 3 cos 2t) = −4x [1] (b) dv/ds = 2s + 2 [1]. a = v(dv/ds) = (1 + 2)(2 + 2) [1] = 12 m s⁻² [1] Examiner insight: In (b), differentiating v with respect to s and stopping at dv/ds = 4 omits the factor v and scores only the first mark.

12. (a) 20e^(−0.3t) = 8 [1], t = ln(2.5)/0.3 ≈ 3.05 s [1] (b) a = −6e^(−0.3t) [1], a(2) = −6e^(−0.6) ≈ −3.29 m s⁻² [1] (c) ∫₀⁸ (20e^(−0.3t) − 8) dt [1] ≈ −3.38 m [1] (d) ∫₀⁸ |20e^(−0.3t) − 8| dt [1] ≈ 34.5 m [1] Examiner insight: Writing the integral before the GDC value earns the method mark; a bare number that is wrong scores zero.

Where marks are usually lost

• Missing the inner derivative, such as writing d/dx(e⁻²ˣ) = e⁻²ˣ.
• Reversing the quotient rule numerator, which changes the sign.
• Substituting values before differentiating in related rates.
• Claiming a point of inflexion from f′′(x) = 0 alone.
• Keeping x-limits after changing the variable to u.
• Using the net integral as an area when part of the region is below the axis.
• Using x-limits or πy² for a rotation about the y-axis.
• Ignoring the starting displacement when asked for position at a later time.
• Confusing displacement with total distance, or giving a negative speed.
• Rounding intermediate GDC values and losing 3 s.f. accuracy in the final answer.

Next steps

• Review the revision notes
• Reread the study guide
• Plan revision with AI exam preparation
• See the IB DP Maths AI course hub
• Tick off topics on the printable syllabus checklist
• Try all free 10-minute diagnostics
• Book a free trial class

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: applications and interpretation guide, first assessment 2021.