AQA AS Paper 1 2022 June — Question 7 6 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
Year2022
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAreas by integration
TypeArea under polynomial curve
DifficultyStandard +0.3 This is a straightforward integration problem requiring students to set up and evaluate a definite integral for area, then solve a simple equation. While it involves multiple steps (finding x-intercepts, setting up the integral, evaluating, and solving for a), each step uses standard AS-level techniques with no conceptual surprises. The 6 marks reflect the working required rather than genuine difficulty, making this slightly easier than average.
Spec1.08e Area between curve and x-axis: using definite integrals

A curve has equation \(y = a^2 - x^2\), where \(a > 0\) The area enclosed between the curve and the \(x\)-axis is 36 units. Find the value of \(a\). Fully justify your answer. [6 marks]

Question 7:
AnswerMarks Guidance
7Integrates with at least one term
correct.3.1a M1
π‘Žπ‘Ž βˆ’π‘₯π‘₯ = 0
(π‘Žπ‘Ž+xπ‘₯π‘₯ = ) (π‘Žπ‘Žβˆ’ orπ‘₯π‘₯ ) = 0
βˆ’π‘Žπ‘Ž π‘Žπ‘Ž
π‘Žπ‘Ž
2 2
οΏ½ (π‘Žπ‘Ž βˆ’π‘₯π‘₯ )𝑑𝑑π‘₯π‘₯ = 36
βˆ’π‘Žπ‘Ž
= 36
3 π‘Žπ‘Ž
2 π‘₯π‘₯
οΏ½π‘Žπ‘Ž π‘₯π‘₯βˆ’ 3οΏ½
βˆ’π‘Žπ‘Ž
3 3
3 π‘Žπ‘Ž 3 π‘Žπ‘Ž
π‘Žπ‘Ž βˆ’ +π‘Žπ‘Ž βˆ’ = 36
3 3
3
4π‘Žπ‘Ž
= 36
3
3
π‘Žπ‘Ž =27
π‘Žπ‘Ž = 3
AnswerMarks Guidance
Obtains correct integral.1.1b A1
Obtains x = or1.1b B1
βˆ’π‘Žπ‘Ž π‘Žπ‘Ž
Substitutes their limits into their
two-term integrated expression
Do not allow any limits involving
AnswerMarks Guidance
x1.1a M1
Equates their expression in
terms of
With limits to to 36
Or π‘Žπ‘Ž
With limits βˆ’ π‘Žπ‘Žto π‘Žπ‘Žor, to to
18
AnswerMarks Guidance
0 π‘Žπ‘Ž βˆ’π‘Žπ‘Ž 0,3.1a M1
Completes a reasoned
AnswerMarks Guidance
argument to obtain a = 32.1 R1
Question 7 Total6
QMarking instructions AO
Question 7:
7 | Integrates with at least one term
correct. | 3.1a | M1 | 2 2
π‘Žπ‘Ž βˆ’π‘₯π‘₯ = 0
(π‘Žπ‘Ž+xπ‘₯π‘₯ = ) (π‘Žπ‘Žβˆ’ orπ‘₯π‘₯ ) = 0
βˆ’π‘Žπ‘Ž π‘Žπ‘Ž
π‘Žπ‘Ž
2 2
οΏ½ (π‘Žπ‘Ž βˆ’π‘₯π‘₯ )𝑑𝑑π‘₯π‘₯ = 36
βˆ’π‘Žπ‘Ž
= 36
3 π‘Žπ‘Ž
2 π‘₯π‘₯
οΏ½π‘Žπ‘Ž π‘₯π‘₯βˆ’ 3οΏ½
βˆ’π‘Žπ‘Ž
3 3
3 π‘Žπ‘Ž 3 π‘Žπ‘Ž
π‘Žπ‘Ž βˆ’ +π‘Žπ‘Ž βˆ’ = 36
3 3
3
4π‘Žπ‘Ž
= 36
3
3
π‘Žπ‘Ž =27
π‘Žπ‘Ž = 3
Obtains correct integral. | 1.1b | A1
Obtains x = or | 1.1b | B1
βˆ’π‘Žπ‘Ž π‘Žπ‘Ž
Substitutes their limits into their
two-term integrated expression
Do not allow any limits involving
x | 1.1a | M1
Equates their expression in
terms of
With limits to to 36
Or π‘Žπ‘Ž
With limits βˆ’ π‘Žπ‘Žto π‘Žπ‘Žor, to to
18
0 π‘Žπ‘Ž βˆ’π‘Žπ‘Ž 0, | 3.1a | M1
Completes a reasoned
argument to obtain a = 3 | 2.1 | R1
Question 7 Total | 6
Q | Marking instructions | AO | Marks | Typical solution
A curve has equation $y = a^2 - x^2$, where $a > 0$

The area enclosed between the curve and the $x$-axis is 36 units.

Find the value of $a$.

Fully justify your answer.

[6 marks]

\hfill \mbox{\textit{AQA AS Paper 1 2022 Q7 [6]}}