IB DP Mathematics: Applications and Interpretation – Differentiation, integration, optimisation and the trapezoidal rule 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 core 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 5.1–5.8, which are common content for SL and HL. 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 SL and HL sessions.

Every paper in this course requires technology (a GDC), so every question here is labelled “(calculator allowed)”. Some still expect you to differentiate or integrate by hand, because that working earns the method marks. Give answers exactly or to 3 significant figures unless the question says otherwise.

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) The table shows values of k(x) = (√(x + 4) − 2)/x.

x	−0.1	−0.01	0.01	0.1
k(x)	0.25158	0.25016	0.24984	0.24846

Estimate the limit of k(x) as x → 0, and explain why k(0) cannot be found by substitution. [2]

2. (calculator allowed) Let f(x) = 4x³ − 7x + 6/x² − 5. Find f′(x) and hence find f′(2). [3]

3. (calculator allowed) Find ∫ (6x² − 3 + 4/x³) dx. [3]

4. (calculator allowed) Let f(x) = 2x³ + 3x² − 36x + 10. Find the interval on which f is decreasing. [4]

5. (calculator allowed) A curve has equation y = x³ − 4x + 8/x.

(a) Find the equation of the tangent to the curve at the point where x = 2. [3] (b) Find the equation of the normal at the same point, in the form ax + by + d = 0, where a, b, d ∈ ℤ. [2]

6. (calculator allowed) The gradient of a curve is given by dy/dx = 3x² − 8x + 1. The curve passes through (2, −3). Find the equation of the curve. [4]

7. (calculator allowed) Let f(x) = −x² + 2x + 8.

(a) Find the x-intercepts of the graph of f. [2] (b) Write down an integral for the area of the region enclosed by the graph of f and the x-axis, and find this area. [2]

8. (calculator allowed) A drainage channel is 12 m wide. Its depth is measured every 2 m across its width.

Distance across (m)	0	2	4	6	8	10	12
Depth (m)	0	1.4	2.3	2.9	2.6	1.5	0

(a) Use the trapezoidal rule to estimate the cross-sectional area of the channel. [3] (b) A 50 m length of the channel has this cross-section throughout and is full. Estimate the volume of water in it. [2]

9. (calculator allowed) Let f(x) = 3 + 12/x², for 1 ≤ x ≤ 3.

(a) Use the trapezoidal rule with 4 intervals of equal width to estimate ∫₁³ f(x) dx. [3] (b) Use your GDC to find the exact value of ∫₁³ f(x) dx. [2] (c) Find the percentage error in your estimate from (a). [2]

10. (calculator allowed) A closed storage box has a square base of side x metres and height h metres. Its volume is 2 m³. Material for the base and the top costs 15 dollars per m². Material for the four sides costs 10 dollars per m².

(a) Show that h = 2/x². [1] (b) Show that the total cost, C dollars, is C = 30x² + 80/x. [2] (c) Find dC/dx. [2] (d) Find the value of x that minimises the cost. [2] (e) Justify that this value gives a minimum. [1] (f) Find the minimum cost. [1]

11. (calculator allowed) Let f(x) = x³ − 9x² + 24x − 10, for 0 ≤ x ≤ 6.

(a) Find f′(x). [2] (b) Find the coordinates of the local maximum and local minimum points of the graph of f, justifying which is which. [4] (c) Find the greatest value of f on the domain 0 ≤ x ≤ 6. [2] (d) Find the equation of the tangent to the graph of f at x = 5. [2]

12. (calculator allowed) A workshop makes x bicycles per month. Its monthly profit is P dollars, where dP/dx = −0.06x² + 2.4x + 30. When no bicycles are made, P = −500.

(a) Find an expression for P in terms of x. [4] (b) Find the number of bicycles that maximises the profit, and justify that it is a maximum. [3] (c) Find the maximum monthly profit. [1] (d) Find the least number of bicycles the workshop must make in a month to make a profit. [2]

Answers

1. Values approach 0.25 from both sides, so limit ≈ 0.25 [1]. At x = 0 the expression gives 0/0, which is undefined [1] [2] Examiner insight: A limit estimate must use values from both sides; quoting only the positive-x column can lose the first mark.

2. 6/x² = 6x⁻² [1]; f′(x) = 12x² − 7 − 12x⁻³ [1]; f′(2) = 48 − 7 − 1.5 = 39.5 [1] [3] Examiner insight: A GDC value of 39.5 alone scores only the final mark; “find f′(x)” means the expression must be written.

3. 2x³ − 3x [1]; 4x⁻³ integrates to −2x⁻² [1]; 2x³ − 3x − 2x⁻² + C [1] [3] Examiner insight: The final accuracy mark needs + C; a correct function without it is not a full anti-derivative.

4. f′(x) = 6x² + 6x − 36 [1]. f′(x) = 0 or f′(x) < 0 considered [1]. 6(x + 3)(x − 2) = 0 gives x = −3, x = 2 [1]. f′ < 0 between the roots: −3 < x < 2 [1] [4] Examiner insight: Giving the roots but not the interval, or writing x < −3, x > 2 (where f increases), loses the final mark.

5. (a) At x = 2, y = 8 − 8 + 4 = 4 [1]. dy/dx = 3x² − 4 − 8x⁻², which is 12 − 4 − 2 = 6 at x = 2 [1]. y − 4 = 6(x − 2), so y = 6x − 8 [1] (b) Normal gradient −1/6 [1]; y − 4 = −(1/6)(x − 2), so x + 6y − 26 = 0 [1] Examiner insight: In (b), a correct normal left as y = −x/6 + 13/3 loses the final mark because the form with integer coefficients was asked for.

6. y = x³ − 4x² + x + C [1][1]. Substitute (2, −3): 8 − 16 + 2 + C = −3 [1]. C = 3, so y = x³ − 4x² + x + 3 [1] [4] Examiner insight: The two integration marks are for the terms; forgetting C removes the substitution and final marks together.

7. (a) −x² + 2x + 8 = 0 [1] gives x = −2 and x = 4 [1] (b) Area = ∫₋₂⁴ (−x² + 2x + 8) dx [1] = 36 [1] Examiner insight: The guide expects the integral written before the GDC value; a bare 36 with wrong limits in the calculator earns nothing.

8. (a) h = 2 and 6 strips [1]. Area ≈ (2/2)[0 + 0 + 2(1.4 + 2.3 + 2.9 + 2.6 + 1.5)] [1] = 21.4 m² [1] (b) Volume ≈ 21.4 × 50 [1] = 1070 m³ [1] Examiner insight: In (b), follow-through applies: a wrong area from (a) multiplied correctly by 50 still earns both marks.

9. (a) y-values 15, 8.3333, 6, 4.92, 4.3333 with h = 0.5 [1]. Estimate = 0.25[15 + 4.3333 + 2(8.3333 + 6 + 4.92)] [1] = 14.46 ≈ 14.5 [1] (b) ∫₁³ (3 + 12/x²) dx written [1] = 14 [1] (c) (14.46 − 14)/14 × 100 [1] = 3.29% [1] Examiner insight: Using the rounded 14.5 in (c) gives 3.57%, which loses the accuracy mark; carry unrounded values forward.

10. (a) x²h = 2, so h = 2/x² [1] (b) Base and top: 2x² × 15 = 30x² [1]. Sides: 4xh × 10 = 40x(2/x²) = 80/x, so C = 30x² + 80/x [1] (c) C = 30x² + 80x⁻¹ [1], dC/dx = 60x − 80x⁻² [1] (d) 60x − 80/x² = 0 [1], x³ = 4/3, x = 1.10 m (1.1006…) [1] (e) dC/dx at x = 1 is −20 (< 0) and at x = 1.2 is 16.4 (> 0), so a minimum [1] (f) C = 30(1.1006…)² + 80/1.1006… = $109 [1] Examiner insight: In a “show that” such as (b), every step must be written; stating the result without showing where 80/x comes from earns 0.

11. (a) f′(x) = 3x² − 18x + 24 [1][1] (b) 3(x − 2)(x − 4) = 0 gives x = 2, x = 4 [1]. f(2) = 10, so local maximum (2, 10) [1]; f(4) = 6, so local minimum (4, 6) [1]. f′(1) = 9 > 0, f′(3) = −3 < 0, f′(5) = 9 > 0 [1] (c) f(0) = −10 and f(6) = 26 compared with the local maximum 10 [1]; greatest value 26 (at x = 6) [1] (d) f(5) = 10 and f′(5) = 9 [1]; y − 10 = 9(x − 5), so y = 9x − 35 [1] Examiner insight: In (c), answering 10 (the local maximum) scores 0; the guide expects you to know a local maximum need not be the greatest value on a domain.

12. (a) P = −0.02x³ + 1.2x² + 30x + C [1][1]. P(0) = −500 gives C = −500 [1]. P = −0.02x³ + 1.2x² + 30x − 500 [1] (b) −0.06x² + 2.4x + 30 = 0, i.e. x² − 40x − 500 = 0 [1]. x = 50 (x = −10 rejected) [1]. dP/dx = 30 > 0 at x = 40 and −42 < 0 at x = 60, so a maximum at 50 bicycles [1] (c) P(50) = $1500 [1] (d) P(x) = 0 on the GDC gives x = 12.04 (3 s.f.: 12.0) [1]. P(12) < 0, so at least 13 bicycles [1] Examiner insight: In (d), the context needs a whole number; giving 12.0 or rounding down to 12 loses the final mark.

Where marks are usually lost

• Not rewriting 6/x² or 8/x as negative powers before differentiating (questions 2 and 5).
• Missing + C, or finding C and not stating the final equation (questions 3, 6 and 12).
• Giving the stationary x-values instead of the interval where f is decreasing (question 4).
• Normal equations not in the form requested (question 5).
• Areas given as GDC values with no integral written (questions 7 and 9).
• Trapezoidal rule with h wrong, or the end values doubled (questions 8 and 9).
• Percentage error worked from a rounded estimate (question 9).
• “Show that” steps skipped in optimisation set-up (question 10).
• No justification that a stationary point is a maximum or minimum (questions 10, 11 and 12).
• Answers not interpreted in context, such as a non-integer number of bicycles (question 12).

Next steps

• Recap the key facts in the revision notes.
• Re-learn any weak topic in the study guide.
• See the rest of the course on the IB DP Maths AI course hub and tick off the printable syllabus checklist.
• Plan your final weeks with the AI exam preparation guide.
• 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.