AQA A-Level Mathematics Paper 1, June 2025: Question 14

8 marks · Medium difficulty · Multi-step Problem

Show that the shaded area enclosed by the curve y = 4x sin 2x and the x-axis for 0 ≤ x ≤ π is equal to kπ, finding the integer k.

Practise this question

Question

A graph showing the curve y = 4x sin(2x) for 0 ≤ x ≤ π. The curve starts at the origin O, forms a positive loop above the x-axis, intersects the x-axis at an intermediate point, and dips into a deeper negative loop below the x-axis before meeting the x-axis again at x = π. Both regions bounded by the curve and the x-axis are shaded.
Question text

14 The graph of y = 4xsin 2x for 0 ≤ x ≤ π is shown below.

y

O x

Show that the shaded area enclosed by the x‑axis and the curve is kπ, where k is an

integer to be found.

Fully justify your answer.

[8 marks]

… 19

(18)

Mark scheme

Show the mark scheme Mark scheme for Question 14 worth 8 marks: Method marks for integration by parts on 4x sin(2x), giving -2x cos(2x) + sin(2x). Finding root x = π/2 for the intersection with the x-axis. Evaluating the definite integrals over [0, π/2] which gives π, and [π/2, π] which gives -3π, then summing the absolute areas π + 3π to conclude an area of 4π.

Q Marking instructions AO Marks Typical solution

14 Begins integration by parts by 4x sin 2xdx

writing

u = 4x v = sin 2x u = 4x u = 4

u = 4 v = Acos 2x 1

PI by v = sin 2x v = − cos2x

4Axcos 2x− 4A (cos 2x)dx

Or 4xsin 2xdx =

−2xcos 2x+ sin 2x(+c) 4x 4

3.1a M1 − cos 2x − − cos 2xdx

Or 2 2

u = sin 2x v = 4x = −2xcos 2x + sin 2x + c

u = Acos 2x v = 2x2

PI by area above

2x2 sin 2x −2A ( x2cos 2x) dx

π

−2x cos 2x + sin 2x 2 = π

In either approach the constant 0

4 may be contained within the u

or v terms or taken out as a area below

factor. π

−2xcos 2x + sin 2x π =

Substitutes their u u v v,, , of

either of the above forms into

the integration by parts formula 1.1a M1 −2π − π = −3π

PI by −2xcos 2x+ sin 2x(+c)

Applies IBP correctly to obtain

π + 3π = 4π

4xsin 2xdx =

4x 4 1.1b A1

− cos 2x − − cos 2x(dx)

PI by −2 cos 2xx + sin 2x(+c)

Completes integration to obtain

−2 cos 2xx + sin 2x(+c) 1.1b A1

π

Deduces x = at intersection

with x-axis.

π

Accept labelled on the 2.2a B1

diagram, written as a limit in

their integral or seen as the

solution to the equation

4x sin 2x= 0

Evaluates their integral over two

3.1a M1

separate intervals. – A-LEVEL MATHEMATICS – 7357/1 –

Question 14 mark scheme

14 cont Demonstrates

π

−2 cos 2xx+ sin 2x 2 = π

18 0

Or 1.1a M1

π

−2xcos 2x + sin 2x π =

−2π − π = −3π

Completes reasoned argument

to show the area is 4π 2.1 R1

Question 14 Total 8

How to answer it

Integration by Parts: Finding Enclosed Area

📋 WHAT THIS QUESTION TESTS

This 8-mark question assesses proficiency in applying integration by parts to a mixed algebraic-trigonometric function, solving simple trigonometric equations to identify curve intercepts, and correctly calculating total geometric area when a curve lies both above and below the x-axis.

Question 14 • 8 Marks

Enclosed Area for y = 4x sin(2x)

Domain: 0 ≤ x ≤ π

📐 Step-by-Step Solution

1 Find the Intercepts on the x-axis

Set y = 0:

4x sin(2x) = 0
For 0 ≤ x ≤ π, either 4x = 0 ⇒ x = 0
or sin(2x) = 0 ⇒ 2x = 0, π, 2π ⇒ x = 0, π/2, π.
Therefore, the curve crosses the x-axis between the endpoints at x = π/2.

2 Perform Indefinite Integration by Parts

We need to find ∫ 4x sin(2x) dx using ∫ u v′ dx = uv − ∫ u′v dx:

Let u = 4x ⇒ u′ = 4
Let v′ = sin(2x) ⇒ v = −½ cos(2x)

Substitute into the formula:

∫ 4x sin(2x) dx = (4x)(−½ cos 2x) − ∫ 4(−½ cos 2x) dx
= −2x cos(2x) + 2 ∫ cos(2x) dx
= −2x cos(2x) + sin(2x) (+ c)

3 Evaluate the Area Above the x-axis (0 to π/2)

A1 = [−2x cos(2x) + sin(2x)]0π/2
= [ −2(π/2) cos(π) + sin(π) ] − [ −2(0) cos(0) + sin(0) ]
= [ −π(−1) + 0 ] − [ 0 + 0 ]
= π

4 Evaluate the Integral Below the x-axis (π/2 to π)

∫π/2π 4x sin(2x) dx = [−2x cos(2x) + sin(2x)]π/2π
= [ −2π cos(2π) + sin(2π) ] − [ −2(π/2) cos(π) + sin(π) ]
= [ −2π(1) + 0 ] − [ π ]
= −2π − π = −3π
Since this region is below the x-axis, its geometric area is |−3π| = 3π.

5 Calculate Total Shaded Area

Total Area = A1 + |A2| = π + 3π = 4π
Hence, k = 4.

✅ Final Answer

  • Total Area = 4π
  • k = 4
  • Integral splits into:
    • [−2x cos 2x + sin 2x]0π/2 = π
    • [−2x cos 2x + sin 2x]π/2π = −3π

💡 Key Knowledge

  • Parts choice: Set u to the polynomial (4x) so its derivative becomes a constant (4), reducing the difficulty of the remaining integral.
  • Trig integration signs:
    ∫ sin(ax) dx = −1/a cos(ax)
    ∫ cos(ax) dx = 1/a sin(ax)
  • Geometric area: Area = ∫above y dx + |∫below y dx|.

🧠 Exam Technique & Mark Scheme

  • M1 Setup of integration by parts with correct forms for u, u′, v, v′.
  • M1 Correct substitution into parts formula.
  • A1 Correct unsimplified expression.
  • A1 Fully simplified antiderivative: −2x cos(2x) + sin(2x).
  • B1 Stating/identifying the root at x = π/2.
  • M1 Splitting into two separate intervals.
  • M1 Correctly evaluating limits on at least one interval.
  • R1 Fully reasoned argument showing Area = π + 3π = 4π.

❌ Common Errors & Traps

  • Single integral trap: Integrating directly from 0 to π gives π + (−3π) = −2π. This scores 0 marks for the area calculation because areas below the axis cancel out areas above.
  • Sign slip with cosine: Forgetting that integrating sin(2x) introduces a negative sign, leading to incorrect signs inside the parts formula.
  • Missing root justification: Not clearly writing or showing x = π/2 either on the diagram or in the working loses the B1 mark.
  • Unfinished conclusion: Writing −3π and not clearly stating that area is positive or omitting the explicit value k = 4.
Examiner Insight: The command words are "Show that..." and "Fully justify your answer". This means every stage of working must be explicit: identifying π/2, showing substitution of limits into both integrals, handling the negative sign for the region below the axis, and explicitly adding π + 3π = 4π.

Topics

Pure Mathematics · H: Integration · E: Trigonometry

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.