Question 11 - A-Level Maths

AQA Paper 1 2020 June — Question 11 9 marks

Exam Board	AQA
Module	Paper 1 (Paper 1)
Year	2020
Session	June
Marks	9
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Numerical integration
Type	Trapezium rule with stated number of strips
Difficulty	Moderate -0.3 This is a straightforward application of the trapezium rule with standard function evaluation (ln), requiring basic calculation skills and understanding of density/volume. Part (a) uses a single trapezium (very routine), part (b) requires six ordinates with scaling for mass calculation (mechanical but more steps), and part (c) tests conceptual understanding of trapezium rule approximation. While multi-part with several marks, each component is standard A-level fare with no novel problem-solving required.
Spec	1.08e Area between curve and x-axis: using definite integrals 1.09f Trapezium rule: numerical integration

11 The region $R$ enclosed by the lines $x = 1 , x = 6 , y = 0$ and the curve $$y = \ln ( 8 - x )$$ is shown shaded in Figure 3 below. \begin{figure}[h]

\captionsetup{labelformat=empty} \caption{Figure 3} \includegraphics[alt={},max width=\textwidth]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-17_419_869_598_587}

\end{figure} All distances are measured in centimetres.
11

Use a single trapezium to find an approximate value of the area of the shaded region, giving your answer in $\mathrm { cm } ^ { 2 }$ to two decimal places.
[0pt] [2 marks]
\section*{Question 11 continues on the next page} 11
Shape $B$ is made from four copies of region $R$ as shown in Figure 4 below. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 4} \includegraphics[alt={},max width=\textwidth]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-18_707_711_438_667}
\end{figure} Shape $B$ is cut from metal of thickness 2 mm
The metal has a density of $10.5 \mathrm {~g} / \mathrm { cm } ^ { 3 }$ Use the trapezium rule with six ordinates to calculate an approximate value of the mass of Shape B. Give your answer to the nearest gram.
11
Without further calculation, give one reason why the mass found in part (b) may be:
11 (c) (i) an underestimate.
11 (c) (ii) an overestimate.

Show mark scheme Show mark scheme source

Question 11(a):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$f(1) = 1.945910149\ldots$, $f(6) = 0.69314718\ldots$, $A = \frac{5}{2}(1.9459 + 0.6931) = 6.5976\ldots$	M1	Evaluates $f(1)$ and $f(6)$ using exact logs or decimals; award if seen embedded in calculations using more than one trapezium
$\approx 6.60 \text{ cm}^2$	A1	Evaluates approximate value of area of R; AWRT 6.60; condone omission of units

Question 11(b):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$x = 1,2,3,4,5,6$ giving $f(x) = 1.9459, 1.7918, 1.6094, 1.3863, 1.0986, 0.6931$	B1	Writes or uses the six ordinates as $\ln 7, \ln 6, \ln 5, \ln 4, \ln 3, \ln 2$; or obtains values of correct six ordinates in decimal form
Area $= \frac{1}{2} \times 1 \times (1.9459 + 0.6931 + 2(1.7918 + 1.6094 + 1.3863 + 1.0986))$	M1	Uses correct formula for trapezium rule with their six ordinates and $h = 1$; award if seven ordinates used with $h = \frac{5}{6}$; answer for seven $= 7.2145648\ldots$
Area $= 7.205633 \text{ cm}^2$	A1	Must have used six ordinates; AWRT 7.2; PI by correct final answer
Volume of Shape B $= 4 \times 7.205633 \times 0.2 = 5.7645\ldots \text{ cm}^3$	M1	Forms expression for mass of either one section or all four sections using their area and consistent units; PI by correct final answer
Mass of Shape B $= 5.7645 \text{ cm}^3 \times 10.5 \text{ g/cm}^3 = 60.52731\text{ g} \approx 61\text{ g}$	A1	Obtains approximate value for correct mass of Shape B; must state units; if seven ordinates used this mark can be awarded as answer would be 61g; CAO

Question 11(c)(i):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
The trapezia are all below the curve	E1	Explains trapezia are all below the curve; or explains the curve is concave; or draws a diagram and indicates the gaps

Question 11(c)(ii):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
Numbers in the calculation have been rounded	E1	Explains that numbers have been rounded

## Question 11(a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(1) = 1.945910149\ldots$, $f(6) = 0.69314718\ldots$, $A = \frac{5}{2}(1.9459 + 0.6931) = 6.5976\ldots$ | M1 | Evaluates $f(1)$ and $f(6)$ using exact logs or decimals; award if seen embedded in calculations using more than one trapezium |
| $\approx 6.60 \text{ cm}^2$ | A1 | Evaluates approximate value of area of R; AWRT 6.60; condone omission of units |

---

## Question 11(b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 1,2,3,4,5,6$ giving $f(x) = 1.9459, 1.7918, 1.6094, 1.3863, 1.0986, 0.6931$ | B1 | Writes or uses the six ordinates as $\ln 7, \ln 6, \ln 5, \ln 4, \ln 3, \ln 2$; or obtains values of correct six ordinates in decimal form |
| Area $= \frac{1}{2} \times 1 \times (1.9459 + 0.6931 + 2(1.7918 + 1.6094 + 1.3863 + 1.0986))$ | M1 | Uses correct formula for trapezium rule with their six ordinates and $h = 1$; award if seven ordinates used with $h = \frac{5}{6}$; answer for seven $= 7.2145648\ldots$ |
| Area $= 7.205633 \text{ cm}^2$ | A1 | Must have used six ordinates; AWRT 7.2; PI by correct final answer |
| Volume of Shape B $= 4 \times 7.205633 \times 0.2 = 5.7645\ldots \text{ cm}^3$ | M1 | Forms expression for mass of either one section or all four sections using their area and consistent units; PI by correct final answer |
| Mass of Shape B $= 5.7645 \text{ cm}^3 \times 10.5 \text{ g/cm}^3 = 60.52731\text{ g} \approx 61\text{ g}$ | A1 | Obtains approximate value for correct mass of Shape B; must state units; if seven ordinates used this mark can be awarded as answer would be 61g; CAO |

---

## Question 11(c)(i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| The trapezia are all below the curve | E1 | Explains trapezia are all below the curve; or explains the curve is concave; or draws a diagram and indicates the gaps |

---

## Question 11(c)(ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Numbers in the calculation have been rounded | E1 | Explains that numbers have been rounded |

---

Show LaTeX source

11 The region $R$ enclosed by the lines $x = 1 , x = 6 , y = 0$ and the curve

$$y = \ln ( 8 - x )$$

is shown shaded in Figure 3 below.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
  \includegraphics[alt={},max width=\textwidth]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-17_419_869_598_587}
\end{center}
\end{figure}

All distances are measured in centimetres.\\
11
\begin{enumerate}[label=(\alph*)]
\item Use a single trapezium to find an approximate value of the area of the shaded region, giving your answer in $\mathrm { cm } ^ { 2 }$ to two decimal places.\\[0pt]
[2 marks]\\

\section*{Question 11 continues on the next page}
11
\item Shape $B$ is made from four copies of region $R$ as shown in Figure 4 below.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 4}
  \includegraphics[alt={},max width=\textwidth]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-18_707_711_438_667}
\end{center}
\end{figure}

Shape $B$ is cut from metal of thickness 2 mm\\
The metal has a density of $10.5 \mathrm {~g} / \mathrm { cm } ^ { 3 }$\\
Use the trapezium rule with six ordinates to calculate an approximate value of the mass of Shape B.

Give your answer to the nearest gram.\\

11
\item Without further calculation, give one reason why the mass found in part (b) may be:\\
11 (c) (i) an underestimate.\\

11 (c) (ii) an overestimate.
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 1 2020 Q11 [9]}}

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