Cambridge International AS & A Level Mathematics 9709: Integration (Pure Mathematics 1) – 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 1.8 Integration of the Cambridge International AS & A Level Mathematics 9709 syllabus for exams in 2026 and 2027 (Version 4). The content is Pure Mathematics 1, examined on Paper 1, which is compulsory for AS Level and A Level. A scientific calculator is allowed on every 9709 paper, so all questions here are calculator allowed, but no marks are given for unsupported calculator answers: show each integration and each substitution of limits. Give exact answers where asked; otherwise give 3 significant figures.

Related: the study guide, the revision notes, the A Level Mathematics hub, the printable 9709 checklist and the free AS diagnostic. For mixed Paper 1 questions, see Pure Mathematics 1 mixed practice.

Total: 60 marks. Questions 10 and 11 are extended problems that combine several outcomes.

Questions

1. Find ∫ (4x³ − 6/x⁴ + 5) dx. [3]

2. Find ∫ 12/(5 − 2x)⁴ dx. [2]

3. A curve is such that dy/dx = 3√x − 6/x². The curve passes through the point (1, 3). Find the equation of the curve. [4]

4. Find the exact value of ∫₁⁹ (x − 3)/√x dx. [4]

5. Evaluate each of the following, giving exact answers.

(a) ∫₄^∞ 6/(2x + 1)² dx [3] (b) ∫₀² 1/√(2 − x) dx [3]

6. A curve is such that dy/dx = k/√(x + 3), where k is a constant. The curve passes through the points (1, 5) and (6, 11). Find the value of k and the equation of the curve. [5]

7. The curves y = x² and y = √(8x) meet at the origin O and at the point P.

(a) Find the x-coordinate of P. [2] (b) Find the exact area of the region enclosed between the two curves. [4]

8. The region R is bounded by the curve y = x³ for x ≥ 0, the y-axis and the line y = 8. By integrating with respect to y, find the area of R. [3]

9. The region S is bounded by the curve y = √(9 − 2x) and the coordinate axes.

(a) S is rotated through 360° about the x-axis. Find the exact volume of the solid formed. [3] (b) S is rotated through 360° about the y-axis. Find the exact volume of the solid formed. [4]

10. The curve y = 6x − x² and the line y = 5 meet at the points A and B.

(a) Find the x-coordinates of A and B. [2] (b) Find the exact area of the region R enclosed by the curve and the line. [3] (c) R is rotated through 360° about the x-axis. Find the exact volume of the solid formed. [5]

11. A curve is such that dy/dx = 3x² − 10x + 6. The curve passes through the point (1, 2).

(a) Find the equation of the curve. [3] (b) Show that the curve crosses the x-axis where x = 0, x = 2 and x = 3. [2] (c) Find the total area of the two regions enclosed by the curve and the x-axis. [5]

Answers

1. Write the integrand as 4x³ − 6x⁻⁴ + 5. ∫ 4x³ dx = x⁴ [1]. ∫ −6x⁻⁴ dx = −6x⁻³/(−3) = 2x⁻³ [1]. ∫ 5 dx = 5x, so the answer is x⁴ + 2/x³ + 5x + c [1]. Examiner insight: each correct term usually earns its own mark, but the last mark needs + c; a missing constant on an indefinite integral costs the final accuracy mark even when every term is right.

2. Write as 12(5 − 2x)⁻⁴ and integrate to a multiple of (5 − 2x)⁻³ [1]. Divide by the new power (−3) and by a = −2: 12 ÷ ((−3)(−2)) = 2, giving 2(5 − 2x)⁻³ + c, or 2/(5 − 2x)³ + c [1]. Examiner insight: the method mark is for the correct new power of the bracket; the accuracy mark needs both divisions, so an answer of −2(5 − 2x)⁻³ (sign of a ignored) loses it.

3. Write dy/dx = 3x^(1/2) − 6x⁻². Integrate: 3x^(3/2)/(3/2) = 2x^(3/2) [1], and −6x⁻¹/(−1) = 6x⁻¹, so y = 2x^(3/2) + 6/x + c [1]. Substitute (1, 3): 3 = 2 + 6 + c [1], so c = −5 and y = 2x^(3/2) + 6/x − 5 [1]. Examiner insight: the mark for substituting the point is a method mark, so it can be earned even after an integration slip, but only if + c was included at the integration step; the final mark needs the full equation, not just “c = −5”.

4. Divide through by x^(1/2): x^(1/2) − 3x^(−1/2) [1]. Integrate: [(2/3)x^(3/2) − 6x^(1/2)]₁⁹ [1]. Substitute: (2/3 × 27 − 6 × 3) − (2/3 − 6) = 0 − (−16/3) [1] = 16/3 [1]. Examiner insight: integrating the numerator and denominator separately earns nothing; the first mark is for rewriting as powers of x, and “exact” means 5.33 on its own does not get the final mark.

5. (a) Integrate to a multiple of (2x + 1)⁻¹ [1]: ∫ 6(2x + 1)⁻² dx = 6(2x + 1)⁻¹/(2 × (−1)) = −3/(2x + 1) [1]. As x → ∞, −3/(2x + 1) → 0, so the value is 0 − (−3/9) = 1/3 [1]. (b) Integrate to a multiple of (2 − x)^(1/2) [1]: ∫ (2 − x)^(−1/2) dx = (2 − x)^(1/2)/((−1)(1/2)) = −2√(2 − x) [1]. The value is (−2√0) − (−2√2) = 2√2 [1]. Examiner insight: for an infinite limit, state that the term tends to 0 rather than writing −3/∞; the final mark in each part needs the exact form, and 0.333 or 2.83 alone does not meet an “exact” instruction.

6. Integrate: ∫ k(x + 3)^(−1/2) dx = 2k√(x + 3) [1], so y = 2k√(x + 3) + c [1]. Substitute both points: at (1, 5), 4k + c = 5; at (6, 11), 6k + c = 11 [1]. Subtract: 2k = 6, so k = 3 [1], then c = −7 and y = 6√(x + 3) − 7 [1]. Examiner insight: there are two unknowns, so you need two equations; a common loss is using only one point, then guessing c = 0, which scores the integration marks but nothing after.

7. (a) Set x² = √(8x) and square: x⁴ = 8x [1]. Then x(x³ − 8) = 0, so P has x = 2 [1]. (b) For 0 < x < 2, √(8x) is above x² (at x = 1, 2.83 > 1). Area = ∫₀² (2√2 x^(1/2) − x²) dx [1]. Integrate: [(4√2/3)x^(3/2) − x³/3]₀² [1]. At x = 2: (4√2/3)(2√2) − 8/3 = 16/3 − 8/3 [1] = 8/3 [1]. Examiner insight: squaring an equation can introduce extra roots, so check each one; in (b), subtracting in the wrong order gives −8/3, and a negative value quoted as an area loses the final mark.

8. Make x the subject: x = y^(1/3) [1]. Area = ∫₀⁸ y^(1/3) dy = [(3/4)y^(4/3)]₀⁸ [1] = (3/4)(16) − 0 = 12 [1]. Examiner insight: the question says “by integrating with respect to y”, so a correct answer from another method (such as a rectangle minus ∫ x³ dx) scores nothing; the limits must be y-values, 0 and 8, not x-values.

9. (a) The curve meets the x-axis at x = 9/2. V = π ∫₀^(9/2) (9 − 2x) dx [1] = π[9x − x²]₀^(9/2) [1] = π(81/2 − 81/4) = 81π/4 [1]. (b) Make x the subject: x = (9 − y²)/2, with y from 0 to 3 [1]. V = π ∫₀³ (9 − y²)²/4 dy = (π/4) ∫₀³ (81 − 18y² + y⁴) dy [1] = (π/4)[81y − 6y³ + y⁵/5]₀³ [1] = (π/4)(243 − 162 + 243/5) = 162π/5 [1]. Examiner insight: a volume without π, or with y instead of y², loses the method mark; in (b) the limits must change to y-values, and using 0 and 9/2 with dy is a method error.

10. (a) 6x − x² = 5, so x² − 6x + 5 = 0 [1], giving x = 1 and x = 5 [1]. (b) Area = ∫₁⁵ (6x − x² − 5) dx [1] = [3x² − x³/3 − 5x]₁⁵ [1] = (50 − 125/3) − (−7/3) = 32/3 [1]. (c) The outer radius is 6x − x² and the inner radius is 5, so V = π ∫₁⁵ ((6x − x²)² − 5²) dx [1]. Expand: x⁴ − 12x³ + 36x² − 25 [1]. Integrate: [x⁵/5 − 3x⁴ + 12x³ − 25x]₁⁵ [1]. Substitute: 125 − (−79/5) [1], so V = 704π/5 [1]. Examiner insight: in (c) the method mark is for outer² − inner²; π ∫ (6x − x² − 5)² dx is a wrong method and loses every later mark, even though it uses the same limits.

11. (a) Integrate term by term [1]: y = x³ − 5x² + 6x + c [1]. At (1, 2): 2 = 1 − 5 + 6 + c, so c = 0 and y = x³ − 5x² + 6x [1]. (b) y = x(x² − 5x + 6) [1] = x(x − 2)(x − 3), which is zero when x = 0, 2 or 3 [1]. (c) Let F(x) = x⁴/4 − 5x³/3 + 3x² [1]. ∫₀² y dx = F(2) − F(0) = 8/3 [1]. ∫₂³ y dx = F(3) − F(2) = 9/4 − 8/3 = −5/12 [1]. This region is below the x-axis, so its area is 5/12 [1]. Total area = 8/3 + 5/12 = 37/12 [1]. Examiner insight: (b) is a “show that”, so a bare list of roots earns nothing: the factorisation must be seen; in (c) a single integral from 0 to 3 gives 9/4 and loses the last three marks.

Where marks are usually lost

• Not rewriting 6/x⁴ or 3√x as powers of x before integrating.
• Integrating a quotient such as (x − 3)/√x as “integral of top over integral of bottom”.
• Forgetting to divide by a in ∫ (ax + b)ⁿ dx, or losing the sign when a is negative, as in (5 − 2x) and (2 − x).
• No + c, so the given point cannot be used and the equation of the curve cannot be found.
• Using one point when two unknowns (k and c) need two equations.
• Substituting only the upper limit, or not showing the substitution needed for the method mark.
• Treating ∞ as a number instead of saying the term tends to 0.
• Integrating straight across a root, so areas above and below the axis cancel.
• Using x-limits in a ∫ x dy integral, or y² instead of x² for rotation about the y-axis.
• Squaring the difference of radii for a solid with a hole, or leaving out π.

Next steps

• Integration revision notes for the formula table and a quick self-test.
• Integration study guide for full explanations and worked examples.
• Cambridge A Level Mathematics hub and the printable 9709 checklist.
• The 9709 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 2026 and 2027 (Version 4, published December 2025), Cambridge Assessment International Education. Section 1.8 Integration, Pure Mathematics 1 (Paper 1).