IB DP Mathematics: Analysis and Approaches – Integration, areas and kinematics 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 integration unit of IB Diploma Programme Mathematics: Analysis and Approaches. It is aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 5.5, 5.9, 5.10 and 5.11. All questions are SL content, so they suit both 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.

Paper 1 allows no technology; Paper 2 requires it (and HL Paper 3 too). Questions are labelled “(calculator-free)” or “(calculator allowed)” to match. Give calculator answers exactly or to 3 significant figures.

Learn the methods first in the integration study guide and the revision notes. These questions go further than the short set on the AA calculus strand page. The IB DP Maths AA course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator-free) Find ∫ (8x³ − 6/x³ + 5) dx. [3]

2. (calculator-free) f′(x) = 4x − 3x² and f(2) = 7. Find f(x). [3]

3. (calculator-free) Find

(a) ∫ 6/(3x − 2) dx, for x > 2/3 [2] (b) ∫ 6e^(1 − 2x) dx [2] (c) ∫ 8/√(2x + 1) dx [2]

4. (calculator-free)

(a) Find ∫ x²(x³ + 1)⁵ dx. [3] (b) Show that ∫₀^(π/3) (sin x)/(cos² x) dx = 1. [4]

5. (calculator-free) Let f(x) = 3x² − 12.

(a) Find ∫₀³ f(x) dx. [2] (b) Find the total area of the region enclosed by y = f(x), the x-axis, x = 0 and x = 3. [4]

6. (calculator-free) Find the area of the region enclosed by y = x² − 2x and y = 4 − x². [6]

7. (calculator allowed) The curves y = 2 cos x and y = x² − 1 enclose a region R.

(a) Find the x-coordinates of the points where the curves meet. [2] (b) Find the area of R. [3]

8. (calculator-free) A particle moves in a straight line. Its velocity is v(t) = 3t² − 12t + 9 m s⁻¹, for 0 ≤ t ≤ 4. When t = 0, its displacement from O is 2 m.

(a) Find the acceleration when t = 4. [2] (b) Find the times when the particle is at rest. [2] (c) Find an expression for the displacement s(t). [3] (d) Find the total distance travelled from t = 0 to t = 4. [4]

9. (calculator allowed) A particle moves in a line with velocity v(t) = 6e^(−0.4t) cos(1.2t) m s⁻¹, for 0 ≤ t ≤ 5.

(a) Find the first time when the particle is at rest. [2] (b) Find the acceleration when t = 2. [2] (c) Find the displacement from t = 0 to t = 5. [2] (d) Find the total distance travelled from t = 0 to t = 5. [2]

10. (calculator-free) Consider y = cos x and y = sin 2x for 0 ≤ x ≤ π/2.

(a) Show that the curves meet when x = π/6 and x = π/2. [3] (b) Find the area of the region bounded by the two curves and the y-axis. [4] (c) Hence find the total area enclosed between the two curves for 0 ≤ x ≤ π/2. [2]

11. (calculator allowed) A cyclist moves along a straight road. Her acceleration is a(t) = 1.2 − 0.24t m s⁻², t seconds after she passes a marker, and her velocity at the marker is 3 m s⁻¹.

(a) Show that v(t) = 3 + 1.2t − 0.12t². [3] (b) Find her maximum velocity. [2] (c) Find the distance she travels in the first 10 seconds. [2] (d) Find the time when she comes to rest. [2]

12. (calculator-free) Given that ∫₂ᵏ 1/(2x − 3) dx = ln 3, where k > 2, find k. [4]

Answers

1. 8x³ − 6x⁻³ + 5 → 2x⁴ [1], + 3x⁻² [1], + 5x + C [1]. 2x⁴ + 3/x² + 5x + C [3] Examiner insight: A missing +C usually costs the final accuracy mark even when every term is correct.

2. f(x) = 2x² − x³ + C [1]. f(2) = 8 − 8 + C = 7 [1], so C = 7: f(x) = 2x² − x³ + 7 [1] [3] Examiner insight: The last mark needs the whole function written with C replaced, not just “C = 7”.

3. (a) 6 · (1/3) ln(3x − 2) [1] = 2 ln(3x − 2) + C [1] (b) 6 · (1/(−2)) e^(1 − 2x) [1] = −3e^(1 − 2x) + C [1] (c) 8(2x + 1)^(1/2) / ((1/2) · 2) [1] = 8√(2x + 1) + C [1] Examiner insight: The method mark is for dividing by a, the coefficient of x; multiplying by it scores nothing.

4. (a) Inner function x³ + 1 with derivative 3x² [1]. Try (x³ + 1)⁶: derivative 18x²(x³ + 1)⁵ [1]. (1/18)(x³ + 1)⁶ + C [1] (b) Let u = cos x, du = −sin x dx [1]. ∫ −u⁻² du = u⁻¹, so the anti-derivative is 1/cos x [1]. [1/cos x] from 0 to π/3 = 1/(1/2) − 1/1 [1] = 2 − 1 = 1 [1] Examiner insight: In a “show that” the values 1/cos(π/3) = 2 and 1/cos 0 = 1 must both appear; jumping to “= 1” loses the last mark.

5. (a) ∫ (3x² − 12) dx = x³ − 12x [1]; (27 − 36) − 0 = −9 [1] (b) Root in [0, 3] at x = 2 [1]. ∫₀² f dx = 8 − 24 = −16 [1]; ∫₂³ f dx = −9 − (−16) = 7 [1]. Area = 16 + 7 = 23 [1] Examiner insight: Giving −9 or 9 in part (b) shows you have not split at the root and usually scores only the first mark.

6. x² − 2x = 4 − x² → 2x² − 2x − 4 = 0 [1] → x = −1, x = 2 [1]. Top curve is 4 − x² [1]. Area = ∫₋₁² (4 + 2x − 2x²) dx [1] = [4x + x² − (2/3)x³] from −1 to 2 = 20/3 − (−7/3) [1] = 9 [1] [6] Examiner insight: A correct integral with limits earns method marks even if the arithmetic slips, so write it before evaluating.

7. (a) Using the GDC to solve 2 cos x = x² − 1 [1]: x = −1.27 and x = 1.27 (x = ±1.2654…) [1] (b) Area = ∫ from −1.2654… to 1.2654… of (2 cos x − x² + 1) dx [1] [1] = 4.99 [1] Examiner insight: Both marks for the expression need the correct integrand with top minus bottom and the limits; a bare GDC value can lose them.

8. (a) a(t) = 6t − 12 [1]; a(4) = 12 m s⁻² [1] (b) 3(t − 1)(t − 3) = 0 [1], t = 1 and t = 3 [1] (c) s(t) = t³ − 6t² + 9t + C [1]; s(0) = 2 gives C = 2 [1]; s(t) = t³ − 6t² + 9t + 2 [1] (d) s(0) = 2, s(1) = 6, s(3) = 2, s(4) = 6 [1]. Distances 4, 4 and 4 over the three intervals [1] [1]. Total 12 m [1] Examiner insight: s(4) − s(0) = 4 m is the displacement; the particle turns at t = 1 and t = 3, so the distance must be split there.

9. (a) v = 0 when cos(1.2t) = 0 [1]: t = 1.31 s (π/2.4) [1] (b) a = dv/dt at t = 2 using the GDC [1] = −1.39 m s⁻² [1] (c) ∫₀⁵ v(t) dt [1] = 1.13 m [1] (d) ∫₀⁵ |v(t)| dt [1] = 8.34 m [1] Examiner insight: The method mark in (d) needs the modulus written inside the integral; the same value without |…| reads as displacement.

10. (a) sin 2x = 2 sin x cos x, so cos x − 2 sin x cos x = 0 [1]. cos x(1 − 2 sin x) = 0 [1]. cos x = 0 gives x = π/2; sin x = 1/2 gives x = π/6 [1] (b) On [0, π/6], cos x ≥ sin 2x [1]. Area = ∫₀^(π/6) (cos x − sin 2x) dx [1] = [sin x + (1/2) cos 2x] from 0 to π/6 [1] = (1/2 + 1/4) − (0 + 1/2) = 1/4 [1] (c) ∫ from π/6 to π/2 of (sin 2x − cos x) dx = [−(1/2) cos 2x − sin x] = (1/2 − 1) − (−1/4 − 1/2) = 1/4 [1]. Total 1/2 [1] Examiner insight: Factorise rather than divide by cos x in (a); dividing loses x = π/2 and the final mark.

11. (a) v = ∫ (1.2 − 0.24t) dt = 1.2t − 0.12t² + C [1]. v(0) = 3 [1] gives C = 3 [1], so v(t) = 3 + 1.2t − 0.12t². (b) a = 0 when t = 5 [1]; v(5) = 6 m s⁻¹ [1] (c) ∫₀¹⁰ v(t) dt [1] = 50 m [1] (v > 0 on this interval) (d) 0.12t² − 1.2t − 3 = 0 [1], t > 0: t = 12.1 s (5 + 5√2) [1] Examiner insight: Reject the negative root in (d) with a reason (t ≥ 0); leaving both roots can cost the accuracy mark.

12. ∫ 1/(2x − 3) dx = (1/2) ln(2x − 3) [1]. (1/2) ln(2k − 3) − (1/2) ln 1 = ln 3 [1]. ln(2k − 3) = 2 ln 3 = ln 9 [1], so 2k − 3 = 9, k = 6 [1] [4] Examiner insight: Forgetting the factor 1/2 gives 2k − 3 = 3 and k = 3; this loses the anti-derivative mark and the accuracy mark that follows.

Where marks are usually lost

• A GDC value given with no integral expression, so the method mark cannot be awarded.
• +C left off an indefinite integral, or kept in a definite one.
• The 1/a factor missing for f(ax + b), especially with ln and e.
• Integrating across a root and calling the signed value an area.
• Area written as bottom minus top, giving a negative answer.
• Displacement given when total distance was asked for.
• Rounded intersection points used as limits, changing the third significant figure.
• Answers not given to 3 s.f., or exact answers turned into decimals on Paper 1.
• Units left off in kinematics answers.
• Dividing by a factor such as cos x and losing a solution.

Next steps

• Integration and kinematics revision notes
• Integration and kinematics study guide
• AA exam preparation guide
• IB DP Maths AA course hub
• Printable syllabus checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020). Sections SL 5.5, 5.9, 5.10 and 5.11.