Cambridge International AS & A Level Mathematics 9709: Pure Mathematics 2 Integration – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 2.5, Integration, of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). Section 2.5 is part of Pure Mathematics 2 and is examined on Paper 2, the AS Level Pure Mathematics route of Paper 1 plus Paper 2. Paper 1 knowledge is assumed, so some questions also use areas and volumes. A scientific calculator is allowed on every 9709 paper, so all questions are calculator allowed, but unsupported calculator answers earn no marks. Where a question asks for an exact answer, the working must be algebraic. Otherwise give answers to 3 significant figures. All angles are in radians.

Related: the study guide, the revision notes, the Pure Mathematics 1 integration practice, the whole-paper Pure Mathematics 2 practice set, the A Level Mathematics hub, the printable 9709 checklist and the free AS diagnostic.

Questions

1. Find ∫ (8e^(2x − 3) − 3 sin 3x + 10 sec²5x) dx. [3]

2. Find the exact value of ∫₂⁷ 6/(3x − 1) dx, giving your answer in the form k ln 2. [4]

3. Show that ∫₀^(π/4) (cos x − sin x)² dx = (π − 2)/4. [4]

4.

(a) Show that 8 sin²x cos²x ≡ 1 − cos 4x. [2] (b) Hence find the exact value of ∫₀^(π/8) 8 sin²x cos²x dx. [3]

5.

(a) Use the trapezium rule with 4 intervals to estimate ∫₀² ln(1 + 2x) dx, giving your answer correct to 3 significant figures. [3] (b) State, with a reason, whether the trapezium rule gives an over-estimate or an under-estimate of the true value. [2]

6. By first using a double-angle formula, find the exact value of ∫₀^(π/3) 2/(1 + cos 2x) dx. [4]

7. The curve y = e^(2x) − 5eˣ + 6 crosses the x-axis at the points A and B.

(a) Find the exact x-coordinates of A and B. [3] (b) Find the exact area of the region enclosed by the curve and the x-axis between A and B, giving your answer in the form p − q ln(3/2). [4]

8. It is given that ∫₀ᵏ 4e^(−2x) dx = 1.5, where k is a positive constant. Find the exact value of k. [4]

9.

(a) Express 3 cos x + √3 sin x in the form R cos(x − α), where R > 0 and 0 < α < ½π, giving the exact values of R and α. [2] (b) Hence find the exact value of ∫ 1/(3 cos x + √3 sin x)² dx between the limits x = π/6 and x = 5π/12. [4]

10.

(a) Use the trapezium rule with 4 intervals to estimate ∫₀^(π/2) sec²(½x) dx, giving your answer correct to 3 significant figures. [3] (b) Find the exact value of ∫₀^(π/2) sec²(½x) dx. [2] (c) Calculate the percentage error in your estimate from part (a). Explain, with reference to the shape of the curve y = sec²(½x), why the estimate is too large. [3]

11. A curve is such that dy/dx = 4/(2x − 1) + 3e^(1 − x). The curve passes through the point (1, 4).

(a) Find the equation of the curve. [4] (b) Find the y-coordinate of the point on the curve where x = 3, giving your answer correct to 3 significant figures. [1]

12. The region R is bounded by the curve y = e^(−x), the x-axis, the y-axis and the line x = ln 3. Find the exact volume of the solid formed when R is rotated through 360° about the x-axis. [4]

Answers

1. ∫ 8e^(2x − 3) dx = 8 × ½ e^(2x − 3) = 4e^(2x − 3) [1]. ∫ −3 sin 3x dx = −3 × (−⅓ cos 3x) = cos 3x [1]. ∫ 10 sec²5x dx = 10 × ⅕ tan 5x = 2 tan 5x, so the answer is 4e^(2x − 3) + cos 3x + 2 tan 5x + c [1]. Examiner insight: each term is usually marked independently, so one wrong factor costs only that term; the final mark here also needs + c, which is a common single-mark loss on indefinite integrals.

2. ∫ 6/(3x − 1) dx = k ln(3x − 1), a log of the linear function [1], with k = 6 × ⅓ = 2 [1]. Limits: 2 ln 20 − 2 ln 5 [1] = 2 ln 4 = 4 ln 2 [1]. Examiner insight: the question fixes the form k ln 2, so 2 ln 4 or 2 ln 20 − 2 ln 5 left unsimplified loses the final accuracy mark even though the value is correct.

3. (cos x − sin x)² = cos²x − 2 sin x cos x + sin²x [1] = 1 − sin 2x, using sin²x + cos²x = 1 and sin 2x = 2 sin x cos x [1]. ∫ (1 − sin 2x) dx = x + ½ cos 2x [1]. Limits: (π/4 + ½ cos(π/2)) − (0 + ½ cos 0) = π/4 − ½ = (π − 2)/4 [1]. Examiner insight: on a “show that” with a given answer, the final mark needs both limit substitutions written out; jumping from the integral straight to (π − 2)/4 earns no credit for that step.

4. (a) 8 sin²x cos²x = 2(2 sin x cos x)² = 2 sin²2x [1]. Using cos 4x = 1 − 2 sin²2x, 2 sin²2x = 1 − cos 4x, as required [1]. (b) ∫ (1 − cos 4x) dx = x − ¼ sin 4x [1]. Limits: (π/8 − ¼ sin(π/2)) − (0 − 0) [1] = π/8 − ¼ [1]. Examiner insight: “hence” means the result from (a) must be used; integrating 8 sin²x cos²x by any other route that is not shown in full gains little, and the factor ¼ from the 4x is where the accuracy mark is usually lost.

5. (a) h = (2 − 0)/4 = 0.5 [1]. Ordinates at x = 0, 0.5, 1, 1.5, 2: 0, 0.6931, 1.0986, 1.3863, 1.6094 [1]. Estimate = ½ × 0.5 × {0 + 2(0.6931 + 1.0986 + 1.3863) + 1.6094} = 0.25 × 7.9655 = 1.99 (3 s.f.) [1]. (b) The graph of y = ln(1 + 2x) bends downward on 0 ≤ x ≤ 2, so each chord lies below the curve [1]. The trapezium rule therefore gives an under-estimate [1]. Examiner insight: in (b) the word “under-estimate” alone usually earns nothing; the mark scheme wants a reason tied to the shape of this curve, such as a sketch or a statement that the chords lie below it.

6. cos 2x = 2 cos²x − 1, so 1 + cos 2x = 2 cos²x [1]. The integrand becomes 2/(2 cos²x) = sec²x [1]. ∫ sec²x dx = tan x [1]. [tan x]₀^(π/3) = √3 − 0 = √3 [1]. Examiner insight: choosing cos 2x = 1 − 2 sin²x here leads nowhere; the method mark is for picking the form of the identity that cancels the 1, so write that identity down explicitly.

7. (a) Let u = eˣ: u² − 5u + 6 = 0 [1]. (u − 2)(u − 3) = 0, so eˣ = 2 or eˣ = 3 [1]. x = ln 2 and x = ln 3 [1]. (b) ∫ (e^(2x) − 5eˣ + 6) dx = ½e^(2x) [1] − 5eˣ + 6x [1]. Limits ln 2 to ln 3: (9/2 − 15 + 6 ln 3) − (2 − 10 + 6 ln 2) = −5/2 + 6 ln(3/2) [1]. This is negative because the region is below the x-axis, so the area is 5/2 − 6 ln(3/2) [1] (about 0.0672). Examiner insight: an area cannot be negative; leaving −5/2 + 6 ln(3/2) as the final answer usually loses the last mark, and so does using e^(2 ln 3) = 6 instead of 9.

8. ∫ 4e^(−2x) dx = −2e^(−2x) [1]. Limits: −2e^(−2k) − (−2) = 1.5 [1]. So 2e^(−2k) = 0.5, giving e^(−2k) = ¼ [1]. −2k = ln ¼, so k = ½ ln 4 = ln 2 [1]. Examiner insight: the lower limit gives e⁰ = 1, not 0; forgetting it is the usual error, and a decimal k = 0.693 does not earn the final mark when the exact value is asked for.

9. (a) R = √(3² + (√3)²) = √12 = 2√3 [1]. tan α = √3/3 = 1/√3, so α = π/6 [1]. So 3 cos x + √3 sin x = 2√3 cos(x − π/6). (b) The integrand is 1/(12 cos²(x − π/6)) = (1/12) sec²(x − π/6) [1]. ∫ (1/12) sec²(x − π/6) dx = (1/12) tan(x − π/6) [1]. Limits: (1/12)[tan(π/4) − tan 0] [1] = 1/12 [1]. Examiner insight: α must be exact and in radians; writing α = 30° or 0.524 loses the accuracy mark in (a) and makes the exact answer in (b) impossible to reach.

10. (a) h = (π/2)/4 = π/8 [1]. Ordinates at x = 0, π/8, π/4, 3π/8, π/2: 1, 1.0396, 1.1716, 1.4465, 2 [1]. Estimate = ½ × π/8 × {1 + 2(1.0396 + 1.1716 + 1.4465) + 2} = (π/16) × 10.3154 = 2.03 (3 s.f.) [1]. (b) ∫ sec²(½x) dx = 2 tan(½x) [1]. [2 tan(½x)]₀^(π/2) = 2 tan(π/4) − 0 = 2 [1]. (c) Percentage error = (2.0254 − 2)/2 × 100 = 1.27% [1]. The estimate is an over-estimate [1] because y = sec²(½x) bends upward on 0 ≤ x ≤ π/2, so each chord lies above the curve [1]. Examiner insight: ordinates for sec² must be found with the calculator in radians; degree-mode values are all close to 1 and the estimate collapses to about 1.57, losing every accuracy mark in (a).

11. (a) ∫ 4/(2x − 1) dx = 2 ln(2x − 1) [1]. ∫ 3e^(1 − x) dx = −3e^(1 − x) [1]. Substitute (1, 4): 4 = 2 ln 1 − 3e⁰ + c = −3 + c, so c = 7 [1]. y = 2 ln(2x − 1) − 3e^(1 − x) + 7 [1]. (b) y = 2 ln 5 − 3e^(−2) + 7 = 9.81 (3 s.f.) [1]. Examiner insight: the constant must be found after integrating, not before; substituting (1, 4) into dy/dx instead of into y earns no method mark for c.

12. V = π ∫₀^(ln 3) (e^(−x))² dx = π ∫₀^(ln 3) e^(−2x) dx [1]. = π[−½ e^(−2x)]₀^(ln 3) [1]. e^(−2 ln 3) = 1/9, so V = π(−½ × 1/9 + ½) [1] = 4π/9 [1]. Examiner insight: the square must be applied before integrating; π ∫ e^(−x) dx gives a different value and loses all but possibly the limits mark.

Where marks are usually lost

• Multiplying by a instead of dividing by it: ∫ e^(2x − 3) dx given as 2e^(2x − 3).
• Losing the minus sign when integrating sin(ax + b), or when a is negative as in e^(1 − x).
• Using the wrong double-angle form, so the constant does not cancel and the integrand stays unintegrable.
• Omitting + c on indefinite integrals, or finding c from dy/dx rather than from y.
• Evaluating trig ordinates or limits in degree mode.
• Rounding trapezium ordinates to 2 d.p. before adding, so the 3 s.f. answer is wrong.
• Counting ordinates as strips, which gives the wrong h.
• Giving “over-estimate” or “under-estimate” with no reason about how the curve bends.
• Reporting a negative area, or leaving exact answers such as 2 ln 20 − 2 ln 5 unsimplified.
• Forgetting e⁰ = 1 at a lower limit of 0.

Next steps

• Integration revision notes
• Integration study guide
• Pure Mathematics 2 guide
• A Level Mathematics hub
• Printable 9709 checklist
• AS diagnostic, the 9709 self-check bank, or all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Mathematics 9709 syllabus, for exams in 2026 and 2027 (Version 4), Cambridge University Press & Assessment. Topic 2, Pure Mathematics 2 (for Paper 2): section 2.5 Integration.