OxfordAQA International AS and A-level Mathematics: Integration by parts, partial fractions and volumes of revolution (9660) – 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 P2.7 (the second part of Integration) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2: integration by parts, partial fractions and volumes of revolution. This content is part of Unit P2 (Pure Maths) in the International A-level and is not examined at International AS.

Learn the methods first in the study guide and the revision notes. See also the OxfordAQA A-level Mathematics course page and the printable checklist. For other topics, try the free 10-minute diagnostics. Leave answers exact unless the question asks for a decimal.

Questions

1. Find ∫ x e^(4x) dx. [3]

2. (a) Differentiate x cos 4x with respect to x. [2] (b) Hence find ∫ x sin 4x dx. [3]

3. Show that ∫ from 1 to 9 of (ln x)/√x dx = 12 ln 3 − 8. [5]

4. Show that ∫ from 0 to 1 of x² e^(−x) dx = 2 − 5/e. [5]

5. (a) Express (2x − 17)/((2x + 1)(x − 4)) in the form A/(2x + 1) + B/(x − 4), where A and B are constants. [3] (b) Hence show that ∫ from 5 to 8 of (2x − 17)/((2x + 1)(x − 4)) dx = ln(289/484). [4]

6. Find the exact value of ∫ from 5 to 9 of (2x² − 3x + 3)/((x − 3)(x + 1)) dx, writing it as p + q ln 3 + r ln 5 with p, q and r constants to be stated. [7]

7. The region under the curve y = 1 + eˣ, between x = 0 and x = ln 3, is rotated through 360° about the x-axis. Show that the solid formed has volume π(8 + ln 3). [4]

8. The finite region S is enclosed by the curve y = 1 + √x, the y-axis and the line y = 3. S is rotated through 360° about the y-axis. Calculate the exact volume of the solid formed. [4]

9. The curve C has equation y = √(x ln x), for x ≥ 1. The finite region enclosed by C, the x-axis and the line x = e is rotated through 360° about the x-axis. Determine the exact volume of the solid generated. [6]

10. The region under the curve y = √(10/((x + 1)(x + 6))), between x = 0 and x = 4, is rotated through 360° about the x-axis. Show that the volume generated is 2π ln 3. [7]

11. A planter is made by rotating the part of the curve y = x⁴ (for x ≥ 0) with 0 ≤ y ≤ 16 through 360° about the y-axis. Lengths are in decimetres, so a volume in dm³ is a number of litres. (a) Show that the volume of soil in the planter when the depth is h dm is (2π/3) h^(3/2) dm³. [3] (b) Find the capacity of the planter, to the nearest litre. [2] (c) Find the depth of soil when the planter is half full. Give your answer to 3 significant figures. [2]

Answers

1. Take u = x and dv/dx = e^(4x), so v = ¼ e^(4x) [1]. Then ∫ x e^(4x) dx = ¼ x e^(4x) − ¼ ∫ e^(4x) dx [1]. This gives ¼ x e^(4x) − (1/16) e^(4x) + c, which equals (4x − 1) e^(4x)/16 + c [1]. [3] Examiner insight: State your choice of u and dv/dx. If the final line has a slip, a clear first step still shows the method.

2. (a) Product rule: d/dx(x cos 4x) = x(−4 sin 4x) + cos 4x [1]. This is cos 4x − 4x sin 4x [1]. (b) Integrate the result of (a): x cos 4x = ∫ cos 4x dx − 4 ∫ x sin 4x dx [1]. Rearranging, 4 ∫ x sin 4x dx = (1/4) sin 4x − x cos 4x [1]. So ∫ x sin 4x dx = (1/16) sin 4x − (1/4) x cos 4x + c [1]. [5] Examiner insight: When a question says “hence”, use the earlier result. Starting afresh by parts may be valid, but it is slower and carries the risk of sign slips.

3. Write the integrand as ln x × x^(−1/2). Let u = ln x and v′ = x^(−1/2), so that v = 2√x [1]. Then ∫ (ln x)/√x dx = 2√x ln x − ∫ 2√x (1/x) dx [1]. The remaining integral is 2 ∫ x^(−1/2) dx = 4√x [1]. So the antiderivative is 2√x ln x − 4√x. At x = 9: 6 ln 9 − 12 = 12 ln 3 − 12, using ln 9 = 2 ln 3 [1]. At x = 1 the value is −4. Subtract: 12 ln 3 − 12 + 4 = 12 ln 3 − 8 [1]. [5] Examiner insight: Where the result is printed, the working is what earns credit. Include both limit substitutions and the step ln 9 = 2 ln 3.

4. Take u = x² and dv/dx = e^(−x), so v = −e^(−x) [1]. Then ∫ x² e^(−x) dx = −x² e^(−x) + 2 ∫ x e^(−x) dx [1]. Second pass: ∫ x e^(−x) dx = −x e^(−x) − e^(−x) [1]. So the antiderivative is −e^(−x)(x² + 2x + 2) [1]. At 1 the value is −5e^(−1). At 0 it is −2. Subtract: −5/e + 2 = 2 − 5/e [1]. [5] Examiner insight: Differentiate your antiderivative as a check before substituting limits. It takes under a minute and catches sign errors.

5. (a) 2x − 17 = A(x − 4) + B(2x + 1) [1]. Put x = 4: −9 = 9B, so B = −1 [1]. Put x = −½: −18 = −4.5A, so A = 4 [1]. The result is 4/(2x + 1) − 1/(x − 4). (b) ∫ [4/(2x + 1) − 1/(x − 4)] dx = 2 ln|2x + 1| − ln|x − 4| [2]. At 8: 2 ln 17 − ln 4. At 5: 2 ln 11 − ln 1 = 2 ln 11 [1]. Subtract and use the laws of logarithms: 2 ln 17 − ln 4 − 2 ln 11 = ln(289/(4 × 121)) = ln(289/484) [1]. [7] Examiner insight: The 2 in 2 ln|2x + 1| comes from 4 × ½. Writing the integration rule next to your working helps you avoid dropping the factor.

6. Numerator and denominator are both quadratic, so start by dividing. The denominator expands to x² − 2x − 3 [1]. 2x² − 3x + 3 = 2(x² − 2x − 3) + (x + 9) [1]. Write (x + 9)/((x − 3)(x + 1)) = A/(x − 3) + B/(x + 1), so x + 9 = A(x + 1) + B(x − 3) [1]. Put x = 3: 12 = 4A, A = 3. Put x = −1: 8 = −4B, B = −2 [1]. Integrate: 2x + 3 ln|x − 3| − 2 ln|x + 1| [1]. At 9: 18 + 3 ln 6 − 2 ln 10. At 5: 10 + 3 ln 2 − 2 ln 6 [1]. Subtract: 8 + 5 ln 6 − 3 ln 2 − 2 ln 10. With ln 6 = ln 2 + ln 3 and ln 10 = ln 2 + ln 5, the result is 8 + 5 ln 3 − 2 ln 5, so p = 8, q = 5, r = −2 (about 10.3) [1]. [7] Examiner insight: Convert to ln 3 and ln 5 only at the end. The ln 2 terms should cancel, which gives you a check on the arithmetic.

7. V = π ∫ (1 + eˣ)² dx, with (1 + eˣ)² = 1 + 2eˣ + e^(2x) [1]. The integral is x + 2eˣ + ½ e^(2x) [1]. At ln 3: ln 3 + 6 + 9/2 = ln 3 + 21/2. At 0: 0 + 2 + ½ = 5/2 [1]. Difference: ln 3 + 8, so V = π(8 + ln 3) [1]. [4] Examiner insight: Expand the square before integrating. Writing y² as 1 + e^(2x) is the most common slip, and it gives a different answer.

8. Rearrange y = 1 + √x to give x = (y − 1)², so x² = (y − 1)⁴ [1]. The curve meets the y-axis at y = 1, so the limits are 1 and 3 [1]. V = π ∫ from 1 to 3 of (y − 1)⁴ dy = π [(y − 1)⁵/5] from 1 to 3 [1]. This is π(32/5) − 0, so V = 32π/5 [1]. [4] Examiner insight: For a rotation about the y-axis, use y-limits and write x² in terms of y before integrating. An integral in x would answer a different problem.

9. y² = x ln x. The curve meets the x-axis where ln x = 0, at x = 1, so the limits are 1 and e [1]. V = π ∫ from 1 to e of x ln x dx [1]. Take u = ln x, dv/dx = x, v = x²/2 [1]. Then ∫ x ln x dx = (x²/2) ln x − x²/4 [1]. At e: e²/2 − e²/4 = e²/4. At 1: −¼ [1]. V = π(e² + 1)/4 [1]. [6] Examiner insight: Squaring a square root removes it completely. Take the lower limit from the equation y = 0 and show that you have done so.

10. V = π ∫ y² dx, with y² = 10/((x + 1)(x + 6)) [1]. Write 10 = A(x + 6) + B(x + 1) [1]. Put x = −1: A = 2. Put x = −6: B = −2 [1]. So V = π ∫ from 0 to 4 of [2/(x + 1) − 2/(x + 6)] dx [1]. The integral is 2 ln|x + 1| − 2 ln|x + 6| = 2 ln|(x + 1)/(x + 6)| [1]. At 4: 2 ln(5/10) = 2 ln(½). At 0: 2 ln(1/6). Subtract: 2 ln(½ × 6) = 2 ln 3 [1]. So V = 2π ln 3 [1]. [7] Examiner insight: This question joins two skills. Show the squaring of y, the partial fractions and the limits as separate lines so each can be followed.

11. (a) y = x⁴ gives x² = y^(1/2). The volume to depth h is V = π ∫ from 0 to h of x² dy = π ∫ y^(1/2) dy [1] = π [(2/3) y^(3/2)] from 0 to h [1] = (2π/3) h^(3/2) [1]. (b) Put h = 16: 16^(3/2) = 64, so V = 128π/3 [1], which is about 134 litres [1]. (c) Half the capacity is 64π/3. Then (2π/3) h^(3/2) = 64π/3, so h^(3/2) = 32 [1], and h = 32^(2/3) ≈ 10.1 dm [1]. [7] Examiner insight: In part (c), work from the exact capacity rather than the rounded 134, so that the third significant figure is not disturbed.

Where marks are usually lost

• Forgetting the minus sign in uv − ∫ v (du/dx) dx.
• Choosing u = eˣ or u = sin x in a product with x, which makes the new integral harder.
• Stopping after one pass when the integrand contains x².
• Omitting the factor 1/a when integrating 1/(ax + b).
• Skipping the division step when the numerator’s degree equals the denominator’s.
• Squaring only part of y, or forgetting to expand (1 + eˣ)² as a trinomial.
• Using x-limits for a rotation about the y-axis, or leaving the curve in terms of x.
• Substituting limits into only one term of the parts result.
• Rounding before the last line, and then losing the exact form that was asked for.

Next steps

• Re-read the revision notes and redo any question you lost marks on.
• Return to the study guide for the worked examples.
• Visit the course page and the printable checklist.
• Try all the free 10-minute diagnostics at /diagnostics/.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, published by OxfordAQA, for International AS exams from May/June 2018 and International A-level exams from May/June 2019. See section P2.7 (Integration).