AQA AS Level Mathematics Paper 1, June 2025: Question 11
8 marks · Medium difficulty · Multi-step Problem
Expand a cubic expression, evaluate a definite integral between -4 and 2, explain why this definite integral does not equal the total shaded area bounded by the curve and the x-axis, and calculate the actual total area.
Practise this questionQuestion
Question text
11 (a) Expand x(x – 2)(x + 4)
[1 mark]
11 (b) Show that
∫ xx x((x–−2)(2)(xx++4)4)ddxx==3636
− 4
Fully justify your answer.
[4 marks]
11 (c) The curve C has equation
y = x(x – 2)(x + 4)
A sketch of C is shown in the diagram.
y
U
(13)
–4 0 2 x
11 (c) (i) Explain why your answer to part (b) will not give the total area of the shaded region
bounded by C and the x‑axis.
[1 mark]
11 (c) (ii) Find the total area of the shaded region bounded by C and the x‑axis.
[2 marks]
(14)
Mark scheme
Show the mark scheme
![Mark scheme for Question 11: 11(a) B1 for x^3 + 2x^2 - 8x. 11(b) M1 for integrating with at least one term correct, A1F for fully correct integration [x^4/4 + 2x^3/3 - 4x^2] from -4 to 2, M1 for substituting limits 2 and -4 and subtracting in correct order, R1 for obtaining 36 with all powers evaluated. 11(c)(i) E1 for explaining that part of the area is above the x-axis and part is below (or below gives negative value). 11(c)(ii) M1 for adding the absolute values 128/3 + 20/3, A1 for 148/3.](/.netlify/images?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fexamquestions-bf477.firebasestorage.app%2Fo%2FmarkSchemes%252FAQA-ALEVEL-MAT-AS.Q25.L1.11.SRC.aqa-7356-1%25257C2025%25257CJUN%25257Cuntiered-1790799356095.png%3Falt%3Dmedia&w=1000)
Q Marking instructions AO Marks Typical solution
11(a) Obtains x3 + 2x2 − 8x 1.1b B1 x3 + 2x2 − 8x
Subtotal 1
11(b) Integrates their answer to (a) 1.1a M1 2
∫ x x( − 2)(x + 4) dx
with at least one term correct −4
Completes integration correctly 1.1b A1F 2
= x3 + 2x2 − 8x dx
FT their answer to (a) provided it ∫−4
is a three-term cubic expression 4 3 2
x 2x 2
Substitutes limits explicitly into 1.1a M1 = + − 4x
their integrated function and 4 3 −4
subtracts in the correct order 4 3
( )2 2 2( ) 2
Completes reasoned argument 2.1 R1 = + − 4 2( )
to show that 4 3
24 3
∫ x x( − 2)(x + 4) dx= 36 (−4) 2(−4) 2
−4 − + − 4(−4)
AG 4 3
Must see a correct expression 20 128
with all powers evaluated prior = − − −
to obtaining 36
= 36
Subtotal 4
11(c)(i) Explains that part of the area is 2.3 E1 The area is above and below the
above and part is below the x- axis, so cannot be evaluated as a
axis single integral
Or
Explains that any part below the
axis will give a negative value
when using integration
Do not accept comments that
reference negative area
Subtotal 1
11(c)(ii) 128 20 2.2a M1 128 20 148
Adds their ± + their ± + =
33 3 3 3
to obtain an answer not equal to
OE could be in integral form
148 1.1b A1
Obtains
OE CAO
Subtotal 2
Question 11 Total 8
How to answer it
Definite Integration & Bounded Area Between Curves
What this question tests
- Polynomial Expansion: Expanding three linear factors into a standard cubic polynomial.
- Definite Integration: Integrating term-by-term using the power rule ∫ xⁿ dx = (xⁿ⁺¹)/(n + 1) + c.
- Rigorous Working ("Show that"): Demonstrating full numerical substitution and evaluating powers explicitly before obtaining the final target value.
- Calculus vs. Geometry: Understanding why an integral over an interval where the curve crosses the x-axis differs from the geometric area.
- Calculating Total Bounded Area: Splitting intervals at roots and taking the absolute magnitude of regions below the x-axis.
Expanding Three Linear Brackets
Expand x(x − 2)(x + 4)
📐 Step-by-Step Expansion
1 Expand the two brackets first:
(x − 2)(x + 4) = x² + 4x − 2x − 8 = x² + 2x − 8
2 Multiply each term by x:
x(x² + 2x − 8) = x³ + 2x² − 8x
✅ Final Answer
x³ + 2x² − 8x
• B1: Obtains x³ + 2x² − 8x (AO 1.1b)
Evaluating a Definite Integral ("Show that")
Show that ∫₋₄² x(x − 2)(x + 4) dx = 36. Fully justify your answer.
📐 Full Step-by-Step Working
1 Write in integrable polynomial form:
∫₋₄² (x³ + 2x² − 8x) dx
2 Integrate term-by-term:
= [ x⁴/4 + 2x³/3 − 4x² ]₋₄²
3 Substitute upper limit (x = 2):
= (2)⁴/4 + 2(2)³/3 − 4(2)²
= 16/4 + 16/3 − 16 = 4 + 16/3 − 16 = −20/3
4 Substitute lower limit (x = −4):
= (−4)⁴/4 + 2(−4)³/3 − 4(−4)²
= 256/4 − 128/3 − 64 = 64 − 128/3 − 64 = −128/3
5 Subtract limits (F(b) − F(a)):
= (−20/3) − (−128/3) = (−20/3) + (128/3) = 108/3 = 36
🧠 Exam Technique: "Fully Justify Your Answer"
- Modern calculators can compute definite integrals instantly with the press of a button. For a "Show that" question, examiners award marks solely for the intermediate working.
- You must write out the explicit substitution with limits and evaluate the powers prior to reaching 36 (e.g. showing −20/3 and −128/3).
❌ Common Pitfalls
- Sign errors with negative powers: Writing (−4)⁴ as −256 instead of +256, or (−4)² as −16.
- Double negative oversight: Subtracting a negative: −20/3 − (−128/3) becomes addition.
- Skipping limits: Jumping from the integrated square brackets directly to 36 loses the final R1 reasoning mark completely.
• M1: Integrates answer to (a) with at least one term correct.
• A1F: Fully correct integration (follow-through from 3-term cubic).
• M1: Explicitly substitutes limits 2 and −4 into integrated function in correct order.
• R1: Fully reasoned argument with all powers evaluated before obtaining 36.
Interpreting Integrals vs. Geometric Area
Explain why your answer to part (b) will not give the total area of the shaded region bounded by C and the x-axis.
💡 Key Knowledge
The integral ∫ₐᵇ y dx calculates the net signed area:
- Regions above the x-axis yield a positive integral value.
- Regions below the x-axis yield a negative integral value.
- Evaluating across both regions in a single integral causes the negative region to cancel out part of the positive region.
✅ Model Answer
"Part of the region is above the x-axis and part is below the x-axis, so integrating in a single step causes the section below the axis to give a negative value which cancels out part of the area."
❌ Examiner Warning
Do NOT write: "Because area cannot be negative" or refer to the region as having "negative area". Area is strictly positive. You must state that the integral/value below the axis is negative or that part is above and part is below the x-axis.
• E1: Explains that part of the area is above and part below the x-axis OR explains that any part below the axis will give a negative value when integrating (AO 2.3).
Finding Total Shaded Area
Find the total area of the shaded region bounded by C and the x-axis.
📐 Step-by-Step Calculation
1 Identify the separate regions from the roots:
Roots are at x = −4, x = 0, and x = 2.
• Region 1 (above x-axis): between x = −4 and x = 0
• Region 2 (below x-axis): between x = 0 and x = 2
2 Evaluate each region using our antiderivative F(x):
F(x) = x⁴/4 + 2x³/3 − 4x²
Notice that F(0) = 0.
From part (b):
• F(2) = −20/3
• F(−4) = −128/3
3 Calculate individual areas:
• Area 1 = ∫₋₄⁰ y dx = F(0) − F(−4) = 0 − (−128/3) = 128/3
• Area 2 = |∫₀² y dx| = |F(2) − F(0)| = |−20/3 − 0| = 20/3
4 Sum the positive areas:
Total Area = 128/3 + 20/3 = 148/3 (or 49⅓)
✅ Final Answer
Total Area = 148/3 (or 49⅓ / 49.33)
🧠 Time-Saving Exam Tip
You do not need to re-integrate! Since F(0) = 0, the two component areas are simply the magnitudes of the values found in part (b):
|-128/3| + |-20/3| = 128/3 + 20/3 = 148/3
• M1: Adds their ±(128/3) + their ±(20/3) to obtain an answer not equal to 36 (AO 2.2a)
• A1: Obtains 148/3 (CAO) (AO 1.1b)
Topics
Pure Mathematics · B: Algebra and functions · H: Integration
Question and mark scheme from the AQA AS Level Mathematics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.