| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2018 |
| Session | October |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Apply trapezium rule to given table |
| Difficulty | Easy -1.3 This is a straightforward trapezium rule application with all values provided in a table, requiring only substitution into the standard formula. The sketch in part (a) is trivial (exponential decay crossing y-axis at (0,1)). No problem-solving or conceptual challenge—purely procedural calculation well below average A-level difficulty. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.06a Exponential function: a^x and e^x graphs and properties1.09f Trapezium rule: numerical integration |
| \(x\) | - 0.9 | - 0.8 | - 0.7 | - 0.6 | - 0.5 |
| \(y\) | 1.866 | 1.741 | 1.625 | 1.516 | 1.414 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape: curve in quadrants 1 and 2, moving smoothly from negative gradient (\(<-1\)) becoming less negative to approximately 0, no turning points, tends towards vertical on lhs, not asymptotic to \(x\)-axis on rhs | B1 | Allow curve to tend towards vertical on lhs as long as it does not go too far beyond vertical. Allow if it does not appear asymptotic to \(x\)-axis on rhs |
| Correct shape, position and intercept: asymptote at least below a horizontal line halfway between intercept and \(x\)-axis; intercept at \((0,1)\) marked | B1 | Shape and intercept as above. For position look for asymptote at least below a horizontal line halfway between intercept and \(x\)-axis |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h = 0.1\) | B1 | Correct \(h\) (allow \(h = -0.1\)). May be implied by trapezium rule. May be unsimplified e.g. \(\frac{(-0.5)-(-0.9)}{4}\) |
| \(A = \frac{1}{2}(0.1)[1.866 + 1.414 + 2(1.741 + 1.625 + 1.516)]\) | M1 | Correct application of trapezium rule using their \(h\). Bracketing must be correct but may be implied by final answer. Note \(1.866+1.414+2(1.741+1.625+1.516)=13.044\). Square brackets must contain first \(y\) value plus last \(y\) value, inner bracket multiplied by 2 containing remaining \(y\) values, no additional values. Copying error or omitting one value from inner bracket may be regarded as slip (M mark allowed). Extra repeated term forfeits M mark. M0 if \(x\) values used instead of \(y\) values |
| \(A = 0.6522\) or \(A = 0.652\) | A1 | Allow either answer (must be positive) and allow \(\frac{3261}{5000}\) if no decimal seen |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape: curve in quadrants 1 and 2, moving smoothly from negative gradient ($<-1$) becoming less negative to approximately 0, no turning points, tends towards vertical on lhs, not asymptotic to $x$-axis on rhs | B1 | Allow curve to tend towards vertical on lhs as long as it does not go too far beyond vertical. Allow if it does not appear asymptotic to $x$-axis on rhs |
| Correct shape, position and intercept: asymptote at least below a horizontal line halfway between intercept and $x$-axis; intercept at $(0,1)$ marked | B1 | Shape and intercept as above. For position look for asymptote at least below a horizontal line halfway between intercept and $x$-axis |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h = 0.1$ | B1 | Correct $h$ (allow $h = -0.1$). May be implied by trapezium rule. May be unsimplified e.g. $\frac{(-0.5)-(-0.9)}{4}$ |
| $A = \frac{1}{2}(0.1)[1.866 + 1.414 + 2(1.741 + 1.625 + 1.516)]$ | M1 | Correct application of trapezium rule using their $h$. Bracketing must be correct but may be implied by final answer. Note $1.866+1.414+2(1.741+1.625+1.516)=13.044$. Square brackets must contain first $y$ value plus last $y$ value, inner bracket multiplied by 2 containing remaining $y$ values, no additional values. Copying error or omitting one value from inner bracket may be regarded as slip (M mark allowed). Extra repeated term forfeits M mark. M0 if $x$ values used instead of $y$ values |
| $A = 0.6522$ or $A = 0.652$ | A1 | Allow either answer (must be positive) and allow $\frac{3261}{5000}$ if no decimal seen |
---6. (a) Sketch the graph of $y = \left( \frac { 1 } { 2 } \right) ^ { x } , x \in \mathbb { R }$, showing the coordinates of the point at which the graph crosses the $y$-axis.
The table below gives corresponding values of $x$ and $y$, for $y = \left( \frac { 1 } { 2 } \right) ^ { x }$ The values of $y$ are rounded to 3 decimal places.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & - 0.9 & - 0.8 & - 0.7 & - 0.6 & - 0.5 \\
\hline
$y$ & 1.866 & 1.741 & 1.625 & 1.516 & 1.414 \\
\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
$$\int _ { - 0.9 } ^ { - 0.5 } \left( \frac { 1 } { 2 } \right) ^ { x } d x$$
II\\
\hfill \mbox{\textit{Edexcel C12 2018 Q6 [5]}}