AQA A-Level Mathematics: H: Integration (7357) – 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 H: Integration (H1 to H8) of the AQA A-level Mathematics (7357) specification, version 1.3 (31 January 2018), for A-level exams from June 2018 onwards. Section H is Paper 1 content and can also be assessed on Papers 2 and 3. A calculator is required in every paper of this specification, but where a question asks for an exact answer or says “show that”, give full working.

Learn the content first with the Integration study guide and the Integration revision notes. Course hub: /boards/aqa/a-level/mathematics/. Printable checklist: /checklists/aqa/a-level/mathematics/. Free 10-minute diagnostics: /diagnostics/.

Questions

1. Find ∫ (8x³ − 6/x² + 1/√x) dx. [3]

2. Find ∫ (4e^(−2x) + 3 cos 3x − 2/x) dx. [3]

3. A curve has gradient dy/dx = 6x − 4/x², for x > 0, and passes through the point (2, 9). Find the equation of the curve. [4]

4. Given that k > 1 and ∫₁ᵏ (2x − 1) dx = 12, find the value of k. [3]

5. Find the exact value of ∫₀^(π/6) (sin 2x + cos 3x) dx. [4]

6. The curve y = x² − 2x and the line y = x + 4 intersect at two points. Find the exact area of the region enclosed between them. [5]

7. Write lim (δx → 0) Σ (3x² + eˣ) δx, summed from x = 0 to x = 2, as a definite integral, and find its exact value. [3]

8. Use the substitution u = 3x + 1 to find the exact value of ∫₀¹ x/√(3x + 1) dx. [5]

9.

(a) Use integration by parts to find ∫ x² eˣ dx. [5] (b) Hence find the exact value of ∫₀¹ x² eˣ dx. [1]

10. Let f(x) = (5x + 5)/((x + 2)(2x − 1)).

(a) Express f(x) in partial fractions. [3] (b) Hence find ∫₁³ f(x) dx, giving your answer in the form p ln 5 + q ln 3, where p and q are rational. [4]

11. The number of fish in a lake, P thousand, t years after the fish were introduced, is modelled by

dP/dt = P(10 − P)/20, with P = 2 when t = 0, and 0 < P < 10.

(a) Show that 20/(P(10 − P)) ≡ 2/P + 2/(10 − P). [2] (b) Solve the differential equation to show that P = 10e^(t/2)/(4 + e^(t/2)). [5] (c) Find the time taken for the number of fish to reach 8000. Give your answer to 3 significant figures. [2] (d) State the long-term number of fish predicted by the model, and give one limitation of the model. [2]

12. A particle moves in a straight line. At time t seconds its velocity is v m s⁻¹, and its motion satisfies dv/dt = −v²/20, with v = 10 when t = 0.

(a) Show that v = 20/(t + 2). [4] (b) Find the time at which the velocity is 4 m s⁻¹. [1] (c) Find the distance travelled in the first 8 seconds, giving your answer to 3 significant figures. [3] (d) Comment on what the model predicts about the motion as t becomes large. [2]

Answers

1. Rewrite: ∫ (8x³ − 6x⁻² + x^(−1/2)) dx. 8x³ → 2x⁴ [1]; −6x⁻² → 6x⁻¹ [1]; x^(−1/2) → 2x^(1/2), with + c [1]. 2x⁴ + 6/x + 2√x + c [3] Examiner insight: the last accuracy mark needs + c; an otherwise correct answer without it usually loses a mark.

2. 4e^(−2x) → −2e^(−2x) [1]; 3 cos 3x → sin 3x [1]; −2/x → −2 ln|x|, with + c [1]. −2e^(−2x) + sin 3x − 2 ln|x| + c [3] Examiner insight: each term is marked separately, so one wrong factor of k costs one mark, not three.

3. y = 3x² [1] + 4/x + c [1]. At (2, 9): 9 = 12 + 2 + c [1], so c = −5. y = 3x² + 4/x − 5 [1] [4] Examiner insight: the method mark for c needs the given point substituted into your integrated expression, not into dy/dx.

4. [x² − x]₁ᵏ = (k² − k) − 0 [1]. k² − k − 12 = 0, so (k − 4)(k + 3) = 0 [1]. Since k > 1, k = 4 [1] [3] Examiner insight: this is the Fundamental Theorem of Calculus in reverse; the final mark needs k = −3 rejected with a reason.

5. ∫ sin 2x dx = −(1/2) cos 2x [1]; ∫ cos 3x dx = (1/3) sin 3x [1]. [−(1/2) cos 2x + (1/3) sin 3x]₀^(π/6) = (−1/4 + 1/3) − (−1/2 + 0) [1] = 7/12 [1] [4] Examiner insight: the lower limit is not zero here (cos 0 = 1), so dropping the lower-limit value loses the accuracy mark.

6. x² − 2x = x + 4 gives x² − 3x − 4 = 0 [1], so x = −1 or x = 4 [1]. Area = ∫₋₁⁴ ((x + 4) − (x² − 2x)) dx = ∫₋₁⁴ (−x² + 3x + 4) dx [1] = [−x³/3 + 3x²/2 + 4x]₋₁⁴ [1] = 56/3 − (−13/6) = 125/6 [1] [5] Examiner insight: “line minus curve” works even though part of the region is below the x-axis; splitting at x = 0 or x = 2 is unnecessary and adds chances to slip.

7. ∫₀² (3x² + eˣ) dx [1] = [x³ + eˣ]₀² [1] = (8 + e²) − (0 + 1) = 7 + e² [1] [3] Examiner insight: the limits come from the range of the sum; writing the integral without limits loses the first mark.

8. x = (u − 1)/3 and dx = (1/3) du [1]. Limits: x = 0 → u = 1, x = 1 → u = 4 [1]. ∫₁⁴ ((u − 1)/3) u^(−1/2) (1/3) du = (1/9) ∫₁⁴ (u^(1/2) − u^(−1/2)) du [1] = (1/9) [(2/3)u^(3/2) − 2u^(1/2)]₁⁴ [1] = (1/9)[(16/3 − 4) − (2/3 − 2)] = 8/27 [1] [5] Examiner insight: using the x-limits 0 and 1 with a u-integral is a common slip; it loses the limits mark and the final accuracy mark.

9. (a) u = x², dv/dx = eˣ: ∫ x² eˣ dx = x² eˣ − ∫ 2x eˣ dx [1][1]. Again, u = 2x, dv/dx = eˣ: ∫ 2x eˣ dx = 2x eˣ − ∫ 2eˣ dx [1] = 2x eˣ − 2eˣ [1]. So ∫ x² eˣ dx = eˣ(x² − 2x + 2) + c [1] (b) [eˣ(x² − 2x + 2)]₀¹ = e − 2, so e − 2 [1] [6] Examiner insight: the sign error on the second application (−2x eˣ + 2eˣ) is where the accuracy mark usually goes; a quick differentiation check catches it.

10. (a) 5x + 5 ≡ A(2x − 1) + B(x + 2) [1]. x = −2: −5 = −5A, so A = 1. x = 1/2: 7.5 = 2.5B, so B = 3 [1]. f(x) = 1/(x + 2) + 3/(2x − 1) [1] (b) ∫ f(x) dx = ln|x + 2| [1] + (3/2) ln|2x − 1| [1]. [ln|x + 2| + (3/2) ln|2x − 1|]₁³ = (ln 5 + (3/2) ln 5) − (ln 3 + 0) [1] = (5/2) ln 5 − ln 3, so p = 5/2, q = −1 [1] [7] Examiner insight: writing 3 ln|2x − 1| (missing the 1/2) loses the second mark in (b), but follow-through usually allows the limits mark.

11. (a) 20 ≡ A(10 − P) + BP [1]. P = 0: A = 2. P = 10: B = 2. So 20/(P(10 − P)) ≡ 2/P + 2/(10 − P) [1] [2] (b) ∫ (2/P + 2/(10 − P)) dP = ∫ dt [1]. 2 ln P − 2 ln(10 − P) = t + c [1]. At t = 0, P = 2: c = 2 ln 2 − 2 ln 8 = −2 ln 4 [1]. 2 ln(P/(10 − P)) = t − 2 ln 4, so P/(10 − P) = e^(t/2)/4 [1]. 4P = 10e^(t/2) − Pe^(t/2), so P(4 + e^(t/2)) = 10e^(t/2), giving P = 10e^(t/2)/(4 + e^(t/2)) [1] [5] (c) P = 8: 8/2 = e^(t/2)/4, so e^(t/2) = 16 [1]. t = 2 ln 16 ≈ 5.55 years [1] [2] (d) As t → ∞, P → 10, so 10 000 fish [1]. Limitation, e.g. the model treats the number of fish as continuous and ignores fishing, disease or changes in food supply [1] [2] Examiner insight: in (b) the integral of 2/(10 − P) is −2 ln(10 − P); losing that minus sign is the usual error, and it prevents the given answer being reached, so later marks depend on correct working.

12. (a) Separate: ∫ v⁻² dv = ∫ −1/20 dt [1]. −1/v = −t/20 + c [1]. At t = 0, v = 10: c = −1/10 [1]. 1/v = t/20 + 1/10 = (t + 2)/20, so v = 20/(t + 2) [1] [4] (b) 20/(t + 2) = 4, so t = 3 s [1] [1] (c) Distance = ∫₀⁸ 20/(t + 2) dt [1] = [20 ln(t + 2)]₀⁸ [1] = 20 ln 10 − 20 ln 2 = 20 ln 5 ≈ 32.2 m [1] [3] (d) v is always positive, so the model says the particle never stops [1]. The distance 20 ln(t + 2) increases without limit, which is unrealistic; other resistance such as friction would bring the particle to rest [1] [2] Examiner insight: “show that” in (a) needs the constant found and the rearrangement shown line by line; jumping from −1/v = −t/20 + c to the given result gets no credit for the final mark.

Where marks are usually lost

• Multiplying by k instead of dividing when integrating e^(kx), sin kx or cos kx.
• Forgetting the minus sign in ∫ sin kx dx = −(1/k) cos kx.
• Missing + c on indefinite integrals, including the final line after parts.
• Not rewriting 6/x² or 1/√x as powers before integrating.
• Treating a negative integral as a negative area, or cancelling regions either side of a root.
• Keeping x-limits after a substitution, or leaving x and u mixed in one integral.
• Dropping the 1/2 in ∫ 1/(2x − 1) dx or the minus sign in ∫ 1/(10 − P) dP.
• Finding the constant in a differential equation after an incorrect rearrangement.
• Giving a decimal where an exact or log form is asked for.
• Stating a long-term value but no limitation, or a limitation that does not refer to the context.

Next steps

• Integration revision notes
• Integration study guide
• Differentiation practice questions, since every integral can be checked by differentiating
• AQA A-level Mathematics course hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA A-level Mathematics (7357) specification, version 1.3, 31 January 2018, A-level exams June 2018 onwards, published by AQA. Section 3.9, H: Integration (H1 to H8), with Appendix A (mathematical notation) and Appendix B (mathematical formulae and identities).