| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | October |
| 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 two-part question requiring (a) sketching an exponential function and finding the y-intercept by substituting x=0, and (b) applying the trapezium rule formula with values provided in a table. Both parts are routine procedural tasks with no problem-solving or conceptual challenges—simpler than the average A-level question. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.06a Exponential function: a^x and e^x graphs and properties1.09f Trapezium rule: numerical integration |
| \(x\) | 0.5 | 1 | 1.5 | 2 | 2.5 | 3 |
| \(y\) | 0.192 | 0.333 | 0.577 | 1 | 1.732 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Curve in quadrant one and two, gradient approaching zero on lhs and increasing as \(x\) increases | B1 | Shape and position correct; curve just in quadrants one and two |
| Point \((0, \frac{1}{9})\) on curve | B1 | Accept \(\frac{1}{9}\) marked on \(y\)-axis; do not accept \(3^{-2}\) or 0.11; condone \(\left(0, 0.\dot{1}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| State \(h = 0.5\), or use of \(\frac{1}{2} \times 0.5\) | B1 aef | For using \(\frac{1}{2} \times 0.5\) or \(h=0.5\) or equivalent such as \((1-0.5)\) |
| \(\left\{0.192 + 3 + 2\left(0.333 + 0.577 + 1 + 1.732\right)\right\}\) | M1, A1 | For correct structure of \(\{\ldots\}\); first \(y\) value plus last \(y\) value plus 2 times bracket containing sum of remaining \(y\) values |
| \(\frac{1}{2} \times 0.5 \times \{10.476\} =\) awrt \(2.62\) | A1 | Correct answer implies all 4 marks |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Curve in quadrant one and two, gradient approaching zero on lhs and increasing as $x$ increases | B1 | Shape and position correct; curve just in quadrants one and two |
| Point $(0, \frac{1}{9})$ on curve | B1 | Accept $\frac{1}{9}$ marked on $y$-axis; do not accept $3^{-2}$ or 0.11; condone $\left(0, 0.\dot{1}\right)$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| State $h = 0.5$, or use of $\frac{1}{2} \times 0.5$ | B1 aef | For using $\frac{1}{2} \times 0.5$ or $h=0.5$ or equivalent such as $(1-0.5)$ |
| $\left\{0.192 + 3 + 2\left(0.333 + 0.577 + 1 + 1.732\right)\right\}$ | M1, A1 | For correct structure of $\{\ldots\}$; first $y$ value plus last $y$ value plus 2 times bracket containing sum of remaining $y$ values |
| $\frac{1}{2} \times 0.5 \times \{10.476\} =$ awrt $2.62$ | A1 | Correct answer implies all 4 marks |
---\begin{enumerate}
\item (a) Sketch the graph of $y = 3 ^ { x - 2 } , x \in \mathbb { R }$
\end{enumerate}
Give the exact values for the coordinates of the point where your graph crosses the $y$-axis.
The table below gives corresponding values of $x$ and $y$, for $y = 3 ^ { x - 2 }$\\
The values of $y$ are rounded to 3 decimal places where necessary.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$x$ & 0.5 & 1 & 1.5 & 2 & 2.5 & 3 \\
\hline
$y$ & 0.192 & 0.333 & 0.577 & 1 & 1.732 & 3 \\
\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
$$\int _ { 0.5 } ^ { 3 } 3 ^ { x - 2 } \mathrm {~d} x$$
Give your answer to 2 decimal places.
\hfill \mbox{\textit{Edexcel C12 2016 Q7 [6]}}