| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Apply trapezium rule to given table |
| Difficulty | Easy -1.2 This is a straightforward application of the trapezium rule with values already provided in a table. Students only need to recall and apply the trapezium rule formula with no problem-solving required. The sketch in part (a) is trivial (basic reciprocal graph), and part (b) is pure procedural calculation with equally-spaced ordinates, making this easier than average for A-level. |
| Spec | 1.09f Trapezium rule: numerical integration |
| \(x\) | 1 | 1.5 | 2 | 2.5 | 3 |
| \(y\) | 1 | 0.667 | 0.5 | 0.4 | 0.333 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| General shape correct showing part of a rectangular hyperbola in the first quadrant | M1 | Condone curve meeting/intersecting either axis; condone incorrect asymptotes e.g. at \(y=1\) |
| Accuracy on asymptotes and no crossing of axes | A1 | Do not allow intersections with axes; curve must appear asymptotic to both \(x\) and \(y\) axes; ignore sections where \(x < 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State \(h = 0.5\), or use of \(\frac{1}{2} \times 0.5\) | B1 aef | Strip width of 0.5; may appear as \(\frac{1}{2} \times 0.5\) or \(0.25\) or equivalent |
| \(\left\{1 + 0.333 + 2(0.667 + 0.5 + 0.4)\right\}\) | M1A1 | Correct \(\{\ldots\}\) bracket structure; first \(y\) value + last \(y\) value + 2 times sum of remaining \(y\) values; accept rounding to 2dp |
| \(\frac{1}{2} \times 0.5 \times \{4.467\} = \text{awrt } 1.12\) | A1cao | — |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| General shape correct showing part of a rectangular hyperbola in the first quadrant | M1 | Condone curve meeting/intersecting either axis; condone incorrect asymptotes e.g. at $y=1$ |
| Accuracy on asymptotes and no crossing of axes | A1 | Do not allow intersections with axes; curve must appear asymptotic to both $x$ and $y$ axes; ignore sections where $x < 0$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State $h = 0.5$, or use of $\frac{1}{2} \times 0.5$ | B1 aef | Strip width of 0.5; may appear as $\frac{1}{2} \times 0.5$ or $0.25$ or equivalent |
| $\left\{1 + 0.333 + 2(0.667 + 0.5 + 0.4)\right\}$ | M1A1 | Correct $\{\ldots\}$ bracket structure; first $y$ value + last $y$ value + 2 times sum of remaining $y$ values; accept rounding to 2dp |
| $\frac{1}{2} \times 0.5 \times \{4.467\} = \text{awrt } 1.12$ | A1cao | — |
---4. (a) Sketch the graph of $y = \frac { 1 } { x } , x > 0$
The table below shows corresponding values of $x$ and $y$ for $y = \frac { 1 } { x }$, with the values for $y$ rounded to 3 decimal places where necessary.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 1 & 1.5 & 2 & 2.5 & 3 \\
\hline
$y$ & 1 & 0.667 & 0.5 & 0.4 & 0.333 \\
\hline
\end{tabular}
\end{center}
(b) Use the trapezium rule with all the values of $y$ from the table to find an approximate value, to 2 decimal places, for $\int _ { 1 } ^ { 3 } \frac { 1 } { x } \mathrm {~d} x$
\hfill \mbox{\textit{Edexcel C12 2015 Q4 [6]}}