| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area between curve and line |
| Difficulty | Standard +0.3 This is a straightforward AS-level calculus question requiring substitution to find constants, then integration to find an area. Part (a) involves simple simultaneous equations with square roots. Part (b) requires setting up and evaluating a definite integral of a polynomial (after substituting u=√x), which is standard technique. The 'show that' format and 'fully justify' instruction add slight rigour but the mathematical content is routine for AS Paper 2. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks |
|---|---|
| 9(a) | Substitutes the coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Allow one slip. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ACF | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| b = 5 scores M1 A0 R0 | 1.1b | R1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 9(b) | Integrates x – 4√x + 5, at least |
| Answer | Marks | Guidance |
|---|---|---|
| Condone non-substitution of a | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| FT their b value | 1.1b | A1F |
| Answer | Marks | Guidance |
|---|---|---|
| including subtraction | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| OE including 10.6 recurring | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0 < their integrated area < 16 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | 2.1 | R1 |
| Subtotal | 6 | |
| Question 9 Total | 9 | |
| Q | Marking instructions | AO |
Question 9:
--- 9(a) ---
9(a) | Substitutes the coordinates
(1, 2) and (9, 2) to generate two
equations.
Allow one slip. | 1.1a | M1 | 2 = 1 – a + b
1 = – a + b
2 = 9 – 3a + b
–7 = – 3a + b
a = 4
b = 5
Obtains two correct equations
ACF | 1.1b | A1
Obtains convincingly
a = 4 and b = 5
If a = 4 is assumed to obtain
b = 5 scores M1 A0 R0 | 1.1b | R1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 9(b) ---
9(b) | Integrates x – 4√x + 5, at least
one term correct, using their b
value
Condone non-substitution of a | 3.1a | M1 | 91
Area under curve = (x – 4 x + 5) dx
3
1 8 9
x 2 − x 2 + 5 x
2 3 1
27 17 32
– =
2 6 3
Area of rectangle is 2 × 8 = 16
32 1 6
Shaded area is 16 – =
3 3
3
1 8
Obtains x 2 − x 2 + 5 x
2 3
FT their b value | 1.1b | A1F
Substitutes correct limits into
their integrated expression,
including subtraction | 1.1a | M1
3 2
Obtains area of
3
OE including 10.6 recurring | 1.1b | A1
Finds their integrated area −16
Or
16 − their integrated area
Allow area obtained from
calculator
Provided
0 < their integrated area < 16 | 1.1a | M1
Completes reasoned argument
1 6
to obtain correct area of
3
AG | 2.1 | R1
Subtotal | 6
Question 9 Total | 9
Q | Marking instructions | AO | Marks | Typical solutionA curve has equation
$$y = x - a\sqrt{x} + b$$
where $a$ and $b$ are constants.
The curve intersects the line $y = 2$ at points with coordinates $(1, 2)$ and $(9, 2)$, as shown in the diagram below.
\includegraphics{figure_1}
\begin{enumerate}[label=(\alph*)]
\item Show that $a$ has the value 4 and find the value of $b$
[3 marks]
\item On the diagram, the region enclosed between the curve and the line $y = 2$ is shaded.
Show that the area of this shaded region is $\frac{16}{3}$
Fully justify your answer.
[6 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2024 Q9 [9]}}