| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Apply trapezium rule to given table |
| Difficulty | Moderate -0.8 This is a straightforward application of the trapezium rule with equally-spaced ordinates provided in a table, requiring only substitution into the standard formula. Part (a) involves sketching a simple phase-shifted sine curve and finding x-intercepts using basic trigonometry. Both parts are routine procedural questions with no problem-solving or conceptual challenges, making this easier than average for A-level. |
| Spec | 1.05f Trigonometric function graphs: symmetries and periodicities1.09f Trapezium rule: numerical integration |
| \(x\) | 0 | \(\frac { \pi } { 8 }\) | \(\frac { \pi } { 4 }\) | \(\frac { 3 \pi } { 8 }\) | \(\frac { \pi } { 2 }\) |
| \(y\) | 0.5 | 0.793 | 0.966 | 0.991 | 0.866 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Correct shape: one complete cycle of translated \(\sin x\) between 0 and \(2\pi\), positive \(y\)-intercept, finishing above \(x\)-axis at \(2\pi\) | B1 | Correct shape and position for \(\sin x\) translated horizontally only; above and below \(x\)-axis; ignore graphs to left of \(y\)-axis; if more than one cycle then \(x=2\pi\) must be labelled |
| Either \(\left(\frac{5\pi}{6}, 0\right)\) or \(\left(\frac{11\pi}{6}, 0\right)\) marked | B1 | Allow in degrees: \((150°,0)\) or \((330°,0)\); may be indicated as just \(\frac{5\pi}{6}\) or \(\frac{11\pi}{6}\) on \(x\)-axis; condone awrt 2.62 or awrt 5.76 |
| Both \(\left(\frac{5\pi}{6}, 0\right)\) and \(\left(\frac{11\pi}{6}, 0\right)\) marked, no others between 0 and \(2\pi\) | B1 | Must be in radians; \(\left(1\frac{5\pi}{6}, 0\right)\) acceptable; no decimals accepted |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| States \(h = \frac{\pi}{8}\) or uses in trapezium rule | B1 | |
| \(\left\{0.5 + 0.866 + 2(0.793 + 0.966 + 0.991)\right\}\) | M1A1 | |
| \(\frac{1}{2} \times \frac{\pi}{8}\{6.866\} = 1.35\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h = \frac{\pi}{8}\) or strip width of \(\frac{\pi}{8}\) or awrt 0.39 | B1 | |
| Correct \(\{\ldots\}\) outer bracket structure: first + last \(y\) values, inner bracket = 2 × remaining \(y\) values | M1 | Allow one copying error/omission from inner bracket as slip. Extra repeated term forfeits M mark. M0 if \(x\) values used instead of \(y\) values. Allow recovery of invisible brackets. |
| Correct bracket \(\{\ldots\}\) | A1 | |
| \(\frac{1}{2} \times \frac{\pi}{8} \times 0.5 + 0.866 + 2(0.793 + 0.966 + 0.991) = \) awrt 1.35 | B1M1A1A1 | |
| \(\frac{1}{2} \times \frac{\pi}{8} \times 0.5 + 0.866 + 2(0.793 + 0.966 + 0.991) = \) awrt 6.46 | B1M1A0A0 | |
| \(\frac{1}{2} \times \frac{\pi}{8} \times (0.5 + 0.866) + 2(0.793 + 0.966 + 0.991) = \) awrt 5.77 | B1M1A0A0 | |
| 1.35 only | A1 | Beware: integrating between 0 and \(\frac{\pi}{2}\) gives 1.37 → A0 (possibly 0 marks if no trapezium rule shown) |
## Question 7:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Correct shape: one complete cycle of translated $\sin x$ between 0 and $2\pi$, positive $y$-intercept, finishing above $x$-axis at $2\pi$ | B1 | Correct shape and position for $\sin x$ translated horizontally only; above and below $x$-axis; ignore graphs to left of $y$-axis; if more than one cycle then $x=2\pi$ must be labelled |
| Either $\left(\frac{5\pi}{6}, 0\right)$ or $\left(\frac{11\pi}{6}, 0\right)$ marked | B1 | Allow in degrees: $(150°,0)$ or $(330°,0)$; may be indicated as just $\frac{5\pi}{6}$ or $\frac{11\pi}{6}$ on $x$-axis; condone awrt 2.62 or awrt 5.76 |
| Both $\left(\frac{5\pi}{6}, 0\right)$ and $\left(\frac{11\pi}{6}, 0\right)$ marked, no others between 0 and $2\pi$ | B1 | Must be in radians; $\left(1\frac{5\pi}{6}, 0\right)$ acceptable; no decimals accepted |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| States $h = \frac{\pi}{8}$ or uses in trapezium rule | B1 | |
| $\left\{0.5 + 0.866 + 2(0.793 + 0.966 + 0.991)\right\}$ | M1A1 | |
| $\frac{1}{2} \times \frac{\pi}{8}\{6.866\} = 1.35$ | A1 | |
# Trapezium Rule Question:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h = \frac{\pi}{8}$ or strip width of $\frac{\pi}{8}$ or awrt 0.39 | B1 | |
| Correct $\{\ldots\}$ outer bracket structure: first + last $y$ values, inner bracket = 2 × remaining $y$ values | M1 | Allow one copying error/omission from inner bracket as slip. Extra repeated term forfeits M mark. M0 if $x$ values used instead of $y$ values. Allow recovery of invisible brackets. |
| Correct bracket $\{\ldots\}$ | A1 | |
| $\frac{1}{2} \times \frac{\pi}{8} \times 0.5 + 0.866 + 2(0.793 + 0.966 + 0.991) = $ awrt 1.35 | B1M1A1A1 | |
| $\frac{1}{2} \times \frac{\pi}{8} \times 0.5 + 0.866 + 2(0.793 + 0.966 + 0.991) = $ awrt 6.46 | B1M1A0A0 | |
| $\frac{1}{2} \times \frac{\pi}{8} \times (0.5 + 0.866) + 2(0.793 + 0.966 + 0.991) = $ awrt 5.77 | B1M1A0A0 | |
| 1.35 only | A1 | Beware: integrating between 0 and $\frac{\pi}{2}$ gives 1.37 → A0 (possibly 0 marks if no trapezium rule shown) |
---7. (a) Sketch the graph of $y = \sin \left( x + \frac { \pi } { 6 } \right) , \quad 0 \leqslant x \leqslant 2 \pi$
Show the coordinates of the points where the graph crosses the $x$-axis.
The table below gives corresponding values of $x$ and $y$ for $y = \sin \left( x + \frac { \pi } { 6 } \right)$.\\
The values of $y$ are rounded to 3 decimal places where necessary.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0 & $\frac { \pi } { 8 }$ & $\frac { \pi } { 4 }$ & $\frac { 3 \pi } { 8 }$ & $\frac { \pi } { 2 }$ \\
\hline
$y$ & 0.5 & 0.793 & 0.966 & 0.991 & 0.866 \\
\hline
\end{tabular}
\end{center}
(b) Use the trapezium rule with all the values of $y$ from the table to find an approximate value for
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } \sin \left( x + \frac { \pi } { 6 } \right) \mathrm { d } x$$
Give your answer to 2 decimal places.
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel C12 2019 Q7 [7]}}