| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Deduce related integral from numerical approximation |
| Difficulty | Moderate -0.3 This is a straightforward trapezium rule application followed by routine integral manipulations using standard properties (linearity and substitution). Part (a) is mechanical calculation, while parts (b)(i) and (b)(ii) test basic understanding of integral properties that are commonly practiced. The question requires no problem-solving insight, just careful application of learned techniques. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.09f Trapezium rule: numerical integration |
| \(x\) | - 1 | - 0.5 | 0 | 0.5 | 1 |
| \(y\) | 2.287 | 4.470 | 6.719 | 7.291 | 2.834 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h = 0.5\) | B1 | oe seen or implied by sight of \(\frac{0.5}{2}\) in front of bracket; \(h=-0.5\) is B0 |
| \(A \approx \frac{1}{2} \times \frac{1}{2}\{2.287 + 2.834 + 2(4.470 + 6.719 + 7.291)\}\) | M1 | Correct application of trapezium rule with their \(h\); correct bracket structure; condone slips copying values or omission of final brackets |
| \(= \text{awrt } 10.52\) | A1 | Correct answer with no working scores B1M1A1; if \(h=-0.5\) used, maximum is B0M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A \approx \text{"10.52"} - 4\) | M1 | Must use answer to part (a); allow for expression \(\text{"10.52"}-[2x]_{-1}^{1}\); do not condone omission of brackets or arithmetical errors; \(\text{"10.52"}-(2-2)\) is M0 |
| \(= 6.52\) | A1ft | awrt 6.52; correct or ft allowing if they round part (a) to 2dp |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A \approx \text{"10.52"}\) | B1ft | Same answer as part (a); condone "same as part (a)"; allow if they round to 2dp |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h = 0.5$ | B1 | oe seen or implied by sight of $\frac{0.5}{2}$ in front of bracket; $h=-0.5$ is B0 |
| $A \approx \frac{1}{2} \times \frac{1}{2}\{2.287 + 2.834 + 2(4.470 + 6.719 + 7.291)\}$ | M1 | Correct application of trapezium rule with their $h$; correct bracket structure; condone slips copying values or omission of final brackets |
| $= \text{awrt } 10.52$ | A1 | Correct answer with no working scores B1M1A1; if $h=-0.5$ used, maximum is B0M1A0 |
## Part (b)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A \approx \text{"10.52"} - 4$ | M1 | Must use answer to part (a); allow for expression $\text{"10.52"}-[2x]_{-1}^{1}$; do not condone omission of brackets or arithmetical errors; $\text{"10.52"}-(2-2)$ is M0 |
| $= 6.52$ | A1ft | awrt 6.52; correct or ft allowing if they round part (a) to 2dp |
## Part (b)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A \approx \text{"10.52"}$ | B1ft | Same answer as part (a); condone "same as part (a)"; allow if they round to 2dp |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f6af51c1-5f85-4952-b3c4-9dca42b2a309-02_614_739_248_664}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation $y = \mathrm { f } ( x )$\\
The table below shows some corresponding values of $x$ and $y$ for this curve.\\
The values of $y$ are given to 3 decimal places.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & - 1 & - 0.5 & 0 & 0.5 & 1 \\
\hline
$y$ & 2.287 & 4.470 & 6.719 & 7.291 & 2.834 \\
\hline
\end{tabular}
\end{center}
Using the trapezium rule with all the values of $y$ in the given table,
\begin{enumerate}[label=(\alph*)]
\item obtain an estimate for
$$\int _ { - 1 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x$$
giving your answer to 2 decimal places.
\item Use your answer to part (a) to estimate
\begin{enumerate}[label=(\roman*)]
\item $\int _ { - 1 } ^ { 1 } ( \mathrm { f } ( x ) - 2 ) \mathrm { d } x$
\item $\int _ { 1 } ^ { 3 } \mathrm { f } ( x - 2 ) \mathrm { d } x$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2023 Q1 [6]}}