| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Deduce related integral from numerical approximation |
| Difficulty | Standard +0.3 Part (a) is routine sketching of an exponential function. Part (b) is standard trapezium rule application with given values. Part (c) requires algebraic manipulation to relate new integrals to the computed value, but the transformations are straightforward: (i) recognizing that changing -2x to +2x affects only the linear term, (ii) factoring out constants. This is slightly above average due to the deduction element in part (c), but the algebraic insights required are accessible to typical P2 students. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.09f Trapezium rule: numerical integration |
| \(x\) | 2 | 2.3 | 2.6 | 2.9 | 3.2 | 3.5 |
| \(y\) | 0 | 0.3246 | 0.8629 | 1.6643 | 2.7896 | 4.3137 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape or asymptote | M1 | Increasing curve in quadrants 1 and 2, or correct asymptote labelled at height 4 with curve approaching it |
| Intercept \((0,5)\) | B1 | Accept 5 on axis or stated as \((0,5)\); condone \((5,0)\) if in right place |
| Fully correct: shape in quadrants 1 and 2, \(y\)-intercept at 5, asymptote \(y=4\) clearly drawn and labelled | A1 | If conflicting information, what is on graph takes precedence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h = 0.3\) | B1 | Implied by sight of \(\frac{0.3}{2}\) in front of bracket |
| Area \(\approx \frac{0.3}{2}\{0+4.3137+2\times(0.3246+0.8629+1.6643+2.7896)\}\) | M1 | Applies trapezium rule with correct bracket; condone slips in copying values |
| \(=\) awrt \(2.34\) | A1 | Calculator answer is 2.30 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_2^{3.5}(2^x+2x)\,dx = \int_2^{3.5}(2^x-2x+4x)\,dx = 2.34+\left[2x^2\right]_2^{3.5}\) | M1 | Realises integral splits; \(4x\) seen with attempt to integrate; could use answer to (b) plus area under trapezium \(\frac{1.5}{2}(4\times2+4\times3.5)\) |
| \(= 2.34+16.5 = 18.84\) | A1ft | Follow through on their 2.34; answer via repeated trapezium rule scores M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int_2^{3.5}(2^{x+1}-4x)\,dx = \int_2^{3.5}2(2^x-2x)\,dx = 2\times2.34=4.68\) | B1ft | \(2\times\) their "2.34" \(=4.68\) (awrt 3 s.f.); answer via repeated trapezium rule permitted if answer is twice their (b) |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape or asymptote | M1 | Increasing curve in quadrants 1 and 2, or correct asymptote labelled at height 4 with curve approaching it |
| Intercept $(0,5)$ | B1 | Accept 5 on axis or stated as $(0,5)$; condone $(5,0)$ if in right place |
| Fully correct: shape in quadrants 1 and 2, $y$-intercept at 5, asymptote $y=4$ clearly drawn and labelled | A1 | If conflicting information, what is on graph takes precedence |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h = 0.3$ | B1 | Implied by sight of $\frac{0.3}{2}$ in front of bracket |
| Area $\approx \frac{0.3}{2}\{0+4.3137+2\times(0.3246+0.8629+1.6643+2.7896)\}$ | M1 | Applies trapezium rule with correct bracket; condone slips in copying values |
| $=$ awrt $2.34$ | A1 | Calculator answer is 2.30 |
## Part (c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_2^{3.5}(2^x+2x)\,dx = \int_2^{3.5}(2^x-2x+4x)\,dx = 2.34+\left[2x^2\right]_2^{3.5}$ | M1 | Realises integral splits; $4x$ seen with attempt to integrate; could use answer to (b) plus area under trapezium $\frac{1.5}{2}(4\times2+4\times3.5)$ |
| $= 2.34+16.5 = 18.84$ | A1ft | Follow through on their 2.34; answer via repeated trapezium rule scores M0A0 |
## Part (c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_2^{3.5}(2^{x+1}-4x)\,dx = \int_2^{3.5}2(2^x-2x)\,dx = 2\times2.34=4.68$ | B1ft | $2\times$ their "2.34" $=4.68$ (awrt 3 s.f.); answer via repeated trapezium rule permitted if answer is twice their (b) |
---\begin{enumerate}
\item (a) Sketch the curve with equation
\end{enumerate}
$$y = a ^ { x } + 4$$
where $a$ is a positive constant greater than 1\\
On your sketch, show
\begin{itemize}
\item the coordinates of the point of intersection of the curve with the $y$-axis
\item the equation of the asymptote of the curve
\end{itemize}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$x$ & 2 & 2.3 & 2.6 & 2.9 & 3.2 & 3.5 \\
\hline
$y$ & 0 & 0.3246 & 0.8629 & 1.6643 & 2.7896 & 4.3137 \\
\hline
\end{tabular}
\end{center}
The table shows corresponding values of $x$ and $y$ for
$$y = 2 ^ { x } - 2 x$$
with the values of $y$ given to 4 decimal places as appropriate.\\
Using the trapezium rule with all the values of $y$ in the given table,\\
(b) obtain an estimate for $\int _ { 2 } ^ { 3.5 } \left( 2 ^ { x } - 2 x \right) \mathrm { d } x$, giving your answer to 2 decimal places.\\
(c) Using your answer to part (b) and making your method clear, estimate\\
(i) $\int _ { 2 } ^ { 3.5 } \left( 2 ^ { x } + 2 x \right) \mathrm { d } x$\\
(ii) $\int _ { 2 } ^ { 3.5 } \left( 2 ^ { x + 1 } - 4 x \right) \mathrm { d } x$
\hfill \mbox{\textit{Edexcel P2 2024 Q6 [9]}}