Question Simulator

O Level Additional Mathematics (4037)

Integration Banker

22 questions · about 10% of the marks we indexed · examined in 9 of 9 sessions.

Questions by year

2026

  • 2026 May/June · Paper 12 · Q1211 marksArea under a curve
  • 2026 May/June · Paper 22 · Q3 (b)3 marksIndefinite integrals

2025

  • 2025 Oct/Nov · Paper 12 · Q9 (a)11 marksFinding a curve from its gradient
  • 2025 Oct/Nov · Paper 12 · Q105 marksDefinite integrals
  • 2025 Oct/Nov · Paper 22 · Q7 (b)4 marksIntegration as reverse of differentiation
  • 2025 Oct/Nov · Paper 22 · Q1010 marksArea under a curve
  • 2025 May/June · Paper 12 · Q4 (a)2 marksDefinite integrals
  • 2025 May/June · Paper 12 · Q126 marksFinding a curve from its gradient
  • 2025 May/June · Paper 22 · Q811 marksArea under a curve

2024

  • 2024 Oct/Nov · Paper 12 · Q5 (c)4 marksDefinite integrals
  • 2024 Oct/Nov · Paper 12 · Q98 marksArea under a curve
  • 2024 May/June · Paper 12 · Q11 (b)3 marksIntegration as reverse of differentiation
  • 2024 May/June · Paper 22 · Q2 (a, b)7 marksDefinite integrals · Indefinite integrals
  • 2024 May/June · Paper 22 · Q99 marksArea under a curve

2023

  • 2023 Oct/Nov · Paper 22 · Q66 marksDefinite integrals
  • 2023 Oct/Nov · Paper 22 · Q8 (c)5 marksIntegration as reverse of differentiation
  • 2023 May/June · Paper 12 · Q910 marksArea under a curve
  • 2023 May/June · Paper 22 · Q7 (c(ii))3 marksIntegration as reverse of differentiation

2022

  • 2022 Oct/Nov · Paper 12 · Q9 (a, b)9 marksDefinite integrals
  • 2022 May/June · Paper 12 · Q7 (a, b)8 marksDefinite integrals
  • 2022 May/June · Paper 22 · Q119 marksArea under a curve
  • 2022 May/June · Paper 22 · Q127 marksFinding a curve from its gradient

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1Area under a curve · Core · Paper 1 style · 4 marks
  1. (a) Find the coordinates of the points where the curve y = 6x − x² meets the x-axis. [1]
  2. (b) Find the area of the region enclosed by the curve and the x-axis. [3]
Show answer
  1. (a) (0, 0) and (6, 0)
    B1
  2. (b) 36
    M1 integrate: 3x² − x³/3; M1 limits 0 and 6; A1 108 − 72 = 36
Q2Area under a curve · Core · Paper 1 style · 4 marks
  1. (a) Find the exact area of the region bounded by the curve y = 2 + 4/x, the x-axis and the lines x = 1 and x = e. [4]
Show answer
  1. (a) 2e + 2
    M1 ∫(2 + 4/x)dx = 2x + 4 ln x; M1 correct limits; A1 (2e + 4) − 2; A1 2e + 2
Q3Area under a curve · Core · Paper 2 style · 4 marks
  1. (a) The shape of a ramp at a skate park in Lahore is modelled by y = e^(0.5x) for 0 ≤ x ≤ 4, with lengths in metres. Find the exact area of the side face of the ramp, bounded by the curve, the x-axis, the y-axis and the line x = 4. [3]
  2. (b) Give the area correct to 3 significant figures. [1]
Show answer
  1. (a) 2e² − 2 m²
    M1 ∫e^(0.5x)dx = 2e^(0.5x); M1 limits; A1 2e² − 2
  2. (b) 12.8 m²
    B1
Q4Area under a curve · Core · Paper 1 style · 5 marks
  1. (a) Find the area of the region bounded by the curve y = sin 2x and the x-axis for 0 ≤ x ≤ π/2. [3]
  2. (b) Find the area of the region bounded by y = 3 cos x, the x-axis, the y-axis and the line x = π/6. [2]
Show answer
  1. (a) 1
    M1 −½cos 2x; M1 limits; A1 ½ + ½ = 1
  2. (b) 3/2
    M1 3 sin x; A1 3 × ½
Q5Area under a curve · Standard · Paper 1 style · 6 marks
  1. (a) The line y = x + 2 meets the curve y = 4 − x² at the points A and B. Find the x-coordinates of A and B. [2]
  2. (b) Find the area of the region enclosed between the line and the curve. [4]
Show answer
  1. (a) x = −2 and x = 1
    M1 x² + x − 2 = 0; A1
  2. (b) 9/2
    M1 ∫(4 − x² − x − 2)dx; M1 2x − x²/2 − x³/3; M1 limits −2 to 1; A1 7/6 + 10/3 = 9/2
Q6Area under a curve · Standard · Paper 2 style · 5 marks
  1. (a) The curve y = x³ − 4x meets the x-axis at x = −2, x = 0 and x = 2. Explain why ∫ from −2 to 2 of (x³ − 4x) dx = 0. [2]
  2. (b) Find the total area of the two regions enclosed by the curve and the x-axis. [3]
Show answer
  1. (a) The region above the axis (−2 to 0) and the region below (0 to 2) have equal area, and areas below the axis give a negative integral, so they cancel.
    B1 equal areas above (−2 to 0) and below (0 to 2) the axis, by symmetry (x³ − 4x is an odd function: f(−x) = −f(x)); B1 the part below the axis gives a negative integral, so the two contributions cancel
  2. (b) 8
    M1 ∫ from 0 to 2 of (x³ − 4x) dx = −4; M1 use of modulus / symmetry; A1 8
Q7Area under a curve · Standard · Paper 2 style · 4 marks
  1. (a) Find the area of the region bounded by the curve y = √(2x + 1), the x-axis, the y-axis and the line x = 4. Give your answer as an exact fraction and also correct to 3 significant figures. [4]
Show answer
  1. (a) 26/3 ≈ 8.67
    M1 (2x + 1)^(3/2) ÷ (3/2 × 2); A1 ⅓(2x + 1)^(3/2); M1 (27 − 1)/3; A1 26/3 = 8.67
Q8Area under a curve · Standard · Paper 1 style · 7 marks
  1. (a) The line x + y = 6 meets the curve y = 8/x at two points. Find their coordinates. [3]
  2. (b) Find the exact area of the region enclosed by the line and the curve, giving your answer in the form a − b ln 2. [4]
Show answer
  1. (a) (2, 4) and (4, 2)
    M1 x(6 − x) = 8; M1 x² − 6x + 8 = 0; A1
  2. (b) 6 − 8 ln 2
    M1 ∫(6 − x − 8/x)dx; A1 6x − x²/2 − 8 ln x; M1 limits 2 to 4; A1 6 − 8 ln 2
Q9Area under a curve · Standard · Paper 2 style · 5 marks
  1. (a) Show that the curve y = 2 sin x + 1 lies above the x-axis for 0 ≤ x ≤ π. [1]
  2. (b) Find the exact area of the region bounded by the curve, the x-axis and the lines x = 0 and x = π, and give it also correct to 3 significant figures. [4]
Show answer
  1. (a) sin x ≥ 0 on this interval so y ≥ 1 > 0
    B1
  2. (b) 4 + π ≈ 7.14
    M1 −2 cos x + x; M1 limits; A1 (2 + π) − (−2) = 4 + π; A1 7.14
Q10Area under a curve · Challenge · Paper 1 style · 7 marks
  1. (a) The curve y = e^(2x) meets the line y = 9 at the point P. Find the exact x-coordinate of P. [2]
  2. (b) Find the exact area of the region bounded by the curve, the y-axis and the line y = 9. [5]
Show answer
  1. (a) x = ln 3
    M1 2x = ln 9; A1 x = ln 3
  2. (b) 9 ln 3 − 4
    M1 area = ∫ from 0 to ln 3 of (9 − e^(2x)) dx; A1 9x − ½e^(2x); M1 limits; A1 9 ln 3 − (9/2 − 1/2); A1 9 ln 3 − 4
Q11Area under a curve · Challenge · Paper 1 style · 7 marks
  1. (a) The curve y = 3 + 2x − x² has a tangent at the point where x = 2. Find the equation of this tangent. [3]
  2. (b) Find the area of the region bounded by the curve, the tangent and the y-axis. [4]
Show answer
  1. (a) y = 7 − 2x
    M1 dy/dx = 2 − 2x = −2; B1 point (2, 3); A1
  2. (b) 8/3
    M1 ∫ from 0 to 2 of [(7 − 2x) − (3 + 2x − x²)] dx; A1 integrand x² − 4x + 4; M1 limits; A1 8/3
Q12Area under a curve · Challenge · Paper 2 style · 8 marks
  1. (a) Find the x-coordinates of the points where the curve y = x(x − 3)² meets the line y = x. [3]
  2. (b) Find the total area of the two regions enclosed between the curve and the line. [5]
Show answer
  1. (a) x = 0, 2, 4
    M1 x(x − 3)² − x = 0; M1 x[(x − 3)² − 1] = 0; A1
  2. (b) 8 (4 + 4)
    M1 curve − line = x³ − 6x² + 8x; A1 x⁴/4 − 2x³ + 4x²; M1 from 0 to 2: 4; M1 from 2 to 4: −4, area 4; A1 total 8
Q13Definite integrals · Core · Paper 1 style · 4 marks
  1. (a) Find the exact value of ∫ from 1 to 4 of (3√x − 2/x²) dx. [4]
Show answer
  1. (a) 25/2
    M1 2x^(3/2); M1 + 2/x; M1 limits (16 + ½) − (2 + 2); A1 25/2
Q14Definite integrals · Core · Paper 1 style · 3 marks
  1. (a) Find the exact value of ∫ from 0 to π/4 of (sec² x + 2 cos 2x) dx. [3]
Show answer
  1. (a) 2
    B1 tan x; B1 sin 2x; B1 (1 + 1) − 0 = 2
Q15Definite integrals · Core · Paper 2 style · 3 marks
  1. (a) Find ∫ e^(3x − 1) dx. [1]
  2. (b) Hence evaluate ∫ from 0 to 2 of e^(3x − 1) dx, giving your answer in exact form and correct to 3 significant figures. [2]
Show answer
  1. (a) ⅓e^(3x − 1) + c
    B1
  2. (b) ⅓(e⁵ − e⁻¹) ≈ 49.3
    M1 limits; A1
Q16Definite integrals · Standard · Paper 1 style · 4 marks
  1. (a) Show that ∫ from 2 to 5 of 6/(2x − 1) dx = a ln 3, stating the value of a. [4]
Show answer
  1. (a) a = 3 (= ln 27)
    M1 k ln(2x − 1); A1 3 ln(2x − 1); M1 3 ln 9 − 3 ln 3; A1 3 ln 3
Q17Definite integrals · Standard · Paper 2 style · 5 marks
  1. (a) Given that k > 1 and ∫ from 1 to k of (2x + 3) dx = 18, show that k² + 3k − 22 = 0. [3]
  2. (b) Hence find k correct to 3 significant figures. [2]
Show answer
  1. (a) k² + 3k − (1 + 3) = 18 ⇒ k² + 3k − 22 = 0
    M1 x² + 3x; M1 limits; A1 shown
  2. (b) k = 3.42
    M1 formula; A1 reject negative root
Q18Definite integrals · Challenge · Paper 1 style · 4 marks
  1. (a) Find the exact value of ∫ from 0 to π/6 of (cos 3x − 2 sin 2x) dx. [4]
Show answer
  1. (a) −1/6
    B1 ⅓ sin 3x; B1 + cos 2x; M1 (⅓ + ½) − (0 + 1); A1 −1/6
Q19Definite integrals · Challenge · Paper 1 style · 4 marks
  1. (a) Given that a > 0 and ∫ from 0 to a of e^(x/2) dx = 4, find the exact value of a. [4]
Show answer
  1. (a) a = 2 ln 3 (= ln 9)
    M1 2e^(x/2); M1 2e^(a/2) − 2 = 4; M1 e^(a/2) = 3; A1 a = 2 ln 3
Q20Finding a curve from its gradient · Core · Paper 1 style · 4 marks
  1. (a) A curve has gradient function dy/dx = 6x² − 4x + 1 and passes through the point (1, 5). Find the equation of the curve. [4]
Show answer
  1. (a) y = 2x³ − 2x² + x + 4
    M1 integrate; A1 2x³ − 2x² + x + c; M1 substitute (1, 5); A1 c = 4
Q21Finding a curve from its gradient · Core · Paper 2 style · 4 marks
  1. (a) A curve is such that dy/dx = 4e^(2x). It passes through the point (0, 7). Find the equation of the curve. [3]
  2. (b) Find the value of y when x = 1, correct to 3 significant figures. [1]
Show answer
  1. (a) y = 2e^(2x) + 5
    M1 2e^(2x) + c; M1 7 = 2 + c; A1
  2. (b) 19.8
    B1 2e² + 5
Q22Finding a curve from its gradient · Standard · Paper 1 style · 4 marks
  1. (a) The gradient of a curve is given by dy/dx = 6/√(4x + 1). The curve passes through the point (2, 10). Find the equation of the curve. [4]
Show answer
  1. (a) y = 3√(4x + 1) + 1
    M1 k(4x + 1)^(1/2); A1 k = 6 ÷ (½ × 4) = 3; M1 10 = 9 + c; A1
Q23Finding a curve from its gradient · Standard · Paper 1 style · 6 marks
  1. (a) A curve has dy/dx = 4 cos 2x − 3 sin x and passes through the point (π/2, 2). Find the equation of the curve. [4]
  2. (b) Find the exact value of y when x = π/6. [2]
Show answer
  1. (a) y = 2 sin 2x + 3 cos x + 2
    B1 2 sin 2x; B1 + 3 cos x; M1 substitute x = π/2; A1 c = 2
  2. (b) √3 + (3√3)/2 + 2 = (5√3)/2 + 2
    M1; A1
Q24Finding a curve from its gradient · Challenge · Paper 2 style · 10 marks
  1. (a) A curve is such that d²y/dx² = 6x − 4. The curve has a stationary point at (2, 1). Find dy/dx in terms of x. [3]
  2. (b) Find the equation of the curve. [3]
  3. (c) Find the coordinates of the other stationary point and determine the nature of both stationary points. [4]
Show answer
  1. (a) dy/dx = 3x² − 4x − 4
    M1 3x² − 4x + c; M1 0 = 12 − 8 + c; A1 c = −4
  2. (b) y = x³ − 2x² − 4x + 9
    M1 integrate again; M1 1 = 8 − 8 − 8 + k; A1 k = 9
  3. (c) (−2/3, 283/27) ≈ (−0.667, 10.5) maximum; (2, 1) minimum
    M1 (3x + 2)(x − 2) = 0; A1 y = 283/27; M1 d²y/dx² at each; A1 −8 < 0 max, 8 > 0 min
Q25Integration as reverse of differentiation · Core · Paper 1 style · 6 marks
  1. (a) Find d/dx (x ln x). [2]
  2. (b) Hence find ∫ ln x dx. [2]
  3. (c) Hence find the exact value of ∫ from 1 to e of ln x dx. [2]
Show answer
  1. (a) ln x + 1
    M1 product rule; A1
  2. (b) x ln x − x + c
    M1 ∫(ln x + 1)dx = x ln x; A1
  3. (c) 1
    M1 (e − e) − (0 − 1); A1
Q26Integration as reverse of differentiation · Standard · Paper 1 style · 5 marks
  1. (a) Show that d/dx [(x² + 5)^(3/2)] = kx√(x² + 5), stating the value of k. [2]
  2. (b) Hence find the exact value of ∫ from 0 to 2 of x√(x² + 5) dx. [3]
Show answer
  1. (a) k = 3
    M1 chain rule; A1 3x(x² + 5)^(1/2)
  2. (b) 9 − (5√5)/3
    M1 ⅓(x² + 5)^(3/2); M1 ⅓(27 − 5√5); A1
Q27Integration as reverse of differentiation · Standard · Paper 2 style · 7 marks
  1. (a) Find d/dx (x e^(2x)). [2]
  2. (b) Hence find ∫ x e^(2x) dx. [3]
  3. (c) Evaluate ∫ from 0 to 1 of x e^(2x) dx, giving your answer in exact form and correct to 3 significant figures. [2]
Show answer
  1. (a) e^(2x)(1 + 2x)
    M1 product rule; A1
  2. (b) ½x e^(2x) − ¼e^(2x) + c
    M1 ∫(1 + 2x)e^(2x)dx = x e^(2x); M1 ∫2x e^(2x)dx = x e^(2x) − ½e^(2x); A1 ÷ 2, + c
  3. (c) (e² + 1)/4 ≈ 2.10
    M1; A1
Q28Integration as reverse of differentiation · Challenge · Paper 1 style · 6 marks
  1. (a) Find d/dx (x sin 2x). [2]
  2. (b) Hence find the exact value of ∫ from 0 to π/4 of x cos 2x dx. [4]
Show answer
  1. (a) sin 2x + 2x cos 2x
    M1 product rule; A1
  2. (b) π/8 − ¼
    M1 ∫2x cos 2x dx = x sin 2x − ∫sin 2x dx; A1 x sin 2x + ½cos 2x; M1 halve and use limits; A1 (π/8 + 0) − (0 + ¼)
Q29Indefinite integrals · Core · Paper 1 style · 5 marks
  1. (a) Find ∫ (4x³ − 6/x³ + 2) dx. [3]
  2. (b) Find ∫ (3x − 2)⁵ dx. [2]
Show answer
  1. (a) x⁴ + 3/x² + 2x + c
    B1 x⁴; B1 + 3x⁻²; B1 2x + c
  2. (b) (3x − 2)⁶/18 + c
    M1 (3x − 2)⁶/6; A1 ÷ 3
Q30Indefinite integrals · Standard · Paper 1 style · 6 marks
  1. (a) Find ∫ (5e^(−2x) + 4 sec² 2x) dx. [3]
  2. (b) By first expanding and simplifying, find ∫ (x² + 1)²/x² dx. [3]
Show answer
  1. (a) −(5/2)e^(−2x) + 2 tan 2x + c
    B1 −(5/2)e^(−2x); B1 2 tan 2x; B1 + c
  2. (b) x³/3 + 2x − 1/x + c
    M1 x² + 2 + x⁻²; A1 x³/3 + 2x; A1 − 1/x + c

Similar questions

Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.

References only: find each question in its paper (official link where available, otherwise through your school).

Area under a curve 7 questions · 7/9 sessions
Definite integrals 7 questions · 7/9 sessions
Finding a curve from its gradient 3 questions · 3/9 sessions
Integration as reverse of differentiation 4 questions · 4/9 sessions
Indefinite integrals Rare2 questions · 2/9 sessions

Subtopics

Our analysis of how often each subtopic was examined and how many marks it carried, 2022 to May/June 2026.

SubtopicMarks indexedSessions
Area under a curve 687/9
Definite integrals 387/9
Finding a curve from its gradient 243/9
Integration as reverse of differentiation 154/9
Indefinite integrals Rare62/9

← All Additional Mathematics topics