| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area under polynomial curve |
| Difficulty | Moderate -0.3 This is a straightforward integration question testing basic antiderivatives, definite integrals, and understanding of odd functions. Part (a) is routine integration, part (b) exploits symmetry (odd function over symmetric interval gives zero), and part (c) requires recognizing that negative areas cancel positive areas. While it tests conceptual understanding of signed area vs absolute area, the techniques are standard AS-level material with no complex problem-solving required. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| 9(a)(i) | Integrates to obtain terms in x2 and x4 | 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| Condone omission of +c | 1.1b | A1 |
| Subtotal | 2 |
| Answer | Marks |
|---|---|
| 9(a)(ii) | Obtains answer of 0 |
| Answer | Marks | Guidance |
|---|---|---|
| shown then CSO | 1.1b | B1 |
| Subtotal | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 9(b) | Shows curve with three zeros with | |
| correct orientation | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| drawn or omitted) | 2.4 | E1 |
| Subtotal | 2 |
| Answer | Marks |
|---|---|
| 9(c) | Uses 2 × (OE) |
| Answer | Marks | Guidance |
|---|---|---|
| two separate integrals | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains correct area | 1.1b | A1 |
| Subtotal | 2 | |
| Question Total | 7 | |
| Q | Marking Instructions | AO |
Question 9:
--- 9(a)(i) ---
9(a)(i) | Integrates to obtain terms in x2 and x4 | 1.1a | M1 | 4
2 𝑑𝑑
2𝑑𝑑 − +𝑐𝑐
4
Obtains fully correct integral (ISW)
Condone omission of +c | 1.1b | A1
Subtotal | 2
--- 9(a)(ii) ---
9(a)(ii) | Obtains answer of 0
nb correct answer can be obtained
directly from calculator, if working
shown then CSO | 1.1b | B1 | 0
Subtotal | 1
--- 9(b) ---
9(b) | Shows curve with three zeros with
correct orientation | 1.1b | B1 | The area between -2 and 0 lies
below the axis so its integral has a
negative value
Explains that the integral for the area
below the axis has a negative value
(PI)
(Allow even if the graph is incorrectly
drawn or omitted) | 2.4 | E1
Subtotal | 2
--- 9(c) ---
9(c) | Uses 2 × (OE)
2 3
∫0 4𝑑𝑑 − 𝑑𝑑 𝑑𝑑𝑑𝑑
Or
Obtains values of 4, and -4 or 4 from
two separate integrals | 1.1a | M1 | 2 ×
2 3
∫0 4𝑑𝑑 − 𝑑𝑑 𝑑𝑑𝑑𝑑
2
4 2
2 𝑑𝑑
�2𝑑𝑑 − 4�
= 8 0
Obtains correct area | 1.1b | A1
Subtotal | 2
Question Total | 7
Q | Marking Instructions | AO | Marks | Typical Solution\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find
$$\int (4x - x^3) dx$$
[2 marks]
\item Evaluate
$$\int_{-2}^{2} (4x - x^3) dx$$
[1 mark]
\end{enumerate}
\item Using a sketch, explain why the integral in part (a)(ii) does not give the area enclosed between the curve $y = 4x - x^3$ and the $x$-axis.
[2 marks]
\item Find the area enclosed between the curve $y = 4x - x^3$ and the $x$-axis.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2020 Q9 [7]}}