| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area under curve with fractional/negative powers or roots |
| Difficulty | Standard +0.3 Part (a) is a routine integration with standard power rule requiring rewriting 1/√x as x^(-1/2), worth 3 marks. Part (b) requires setting up an equation where two areas are equal, involving finding where the curve crosses the x-axis and solving a resulting equation with surds, but follows a standard 'equal areas' template common in AS-level integration questions. The algebraic manipulation is straightforward once the setup is recognized. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a) | Integrates with one term correct | 1.1a |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains fully correct integral | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | 2.1 | R1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 8(b) | Explains or recognises that area | |
| is linked to integration | 2.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| NB Correct x value is 4 | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6a – 24 a + 24 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| deduce a =9 | 2.2a | R1 |
| Subtotal | 4 | |
| Question 8 Total | 7 | |
| Q | Marking instructions | AO |
Question 8:
--- 8(a) ---
8(a) | Integrates with one term correct | 1.1a | M1 | a1 1 2 a
6 − d x = 6 x − 2 4 x
x 1
= 6a−24 a−6+24
= 6 a − 2 4 a + 1 8
Obtains fully correct integral | 1.1b | A1
Substitutes limits and obtains
the given answer.
AG | 2.1 | R1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 8(b) ---
8(b) | Explains or recognises that area
is linked to integration | 2.4 | M1 | Equal areas, positive and negative,
so integral 1 to a = 0
6a – 24 a + 18 = 0
a – 4 a + 3 = 0
We need a = 9
Equates the answer to part (a)
to 0
or
Finds intersection point with the
x axis and evaluates an integral
between 1 and their x value.
NB Correct x value is 4 | 3.1a | M1
Solves 6a – 24 a + 18 = 0
to obtain a value for a or √a
or
Equates their area of R to their
1
integrated expression in term of
a for R
2
Must have used a positive value
for the area of R
1
NB Correct expression for R is
2
6a – 24 a + 24 | 1.1a | M1
Completes a reasoned
argument with no errors to
deduce a =9 | 2.2a | R1
Subtotal | 4
Question 8 Total | 7
Q | Marking instructions | AO | Marks | Typical solution\begin{enumerate}[label=(\alph*)]
\item Show that
$$\int_1^a \left(6 - \frac{12}{\sqrt{x}}\right) dx = 6a - 24\sqrt{a} + 18$$
[3 marks]
\item The curve $y = 6 - \frac{12}{\sqrt{x}}$, the line $x = 1$ and the line $x = a$ are shown in the diagram below.
The shaded region $R_1$ is bounded by the curve, the line $x = 1$ and the $x$-axis.
The shaded region $R_2$ is bounded by the curve, the line $x = a$ and the $x$-axis.
\includegraphics{figure_8}
It is given that the areas of $R_1$ and $R_2$ are equal.
Find the value of $a$
Fully justify your answer.
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2023 Q8 [7]}}