| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Moderate -0.8 This is a straightforward C2 question requiring basic substitution into an exponential function, standard application of the trapezium rule formula, and finding the area of a triangle to subtract from the integral. All steps are routine with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.09f Trapezium rule: numerical integration |
| \(x\) | 0 | 1 | 2 | 3 | 4 |
| \(y\) | 7.5 | 6 | 4 | 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) \(y = 8 - 2^{x-1}\), \(0 \leq y \leq 4\) | 7 | B1 cao |
| [1] | ||
| (b) \(\left(\int_0^4(8-2^{x-1})dx\right) = \frac{1}{2} \times 1 \times \{7.5 + 2(\text{''their 7'' + 6 + 4}) + 0\}\) | B1; M1 | Outside brackets \(\frac{1}{2} \times 1\) or \(\frac{1}{2}\). For structure of trapezium rule \(\{...\}\) for candidate's \(y\)-ordinates |
| \(= \frac{1}{2} \times 41.5 = 20.75\) o.e. | A1 cao | 20.75 |
| [3] | ||
| (c) Area\((R) = "20.75" - \frac{1}{2}(7.5)(4)\) | M1 | |
| \(= 5.75\) | A1 cao | |
| [2] |
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** $y = 8 - 2^{x-1}$, $0 \leq y \leq 4$ | 7 | B1 cao |
| | [1] | |
| **(b)** $\left(\int_0^4(8-2^{x-1})dx\right) = \frac{1}{2} \times 1 \times \{7.5 + 2(\text{''their 7'' + 6 + 4}) + 0\}$ | B1; M1 | Outside brackets $\frac{1}{2} \times 1$ or $\frac{1}{2}$. For structure of trapezium rule $\{...\}$ for candidate's $y$-ordinates |
| $= \frac{1}{2} \times 41.5 = 20.75$ o.e. | A1 cao | 20.75 |
| | [3] | |
| **(c)** Area$(R) = "20.75" - \frac{1}{2}(7.5)(4)$ | M1 | |
| $= 5.75$ | A1 cao | |
| | [2] | |
---2. The curve $C$ has equation
$$y = 8 - 2 ^ { x - 1 } , \quad 0 \leqslant x \leqslant 4$$
\begin{enumerate}[label=(\alph*)]
\item Complete the table below with the value of $y$ corresponding to $x = 1$
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 & 4 \\
\hline
$y$ & 7.5 & & 6 & 4 & 0 \\
\hline
\end{tabular}
\end{center}
\item Use the trapezium rule, with all the values of $y$ in the completed table, to find an approximate value for $\int _ { 0 } ^ { 4 } \left( 8 - 2 ^ { x - 1 } \right) \mathrm { d } x$
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{582cda45-80fc-43a8-90e6-1cae08cb1534-03_650_606_1016_671}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curve $C$ with equation $y = 8 - 2 ^ { x - 1 } , \quad 0 \leqslant x \leqslant 4$\\
The curve $C$ meets the $x$-axis at the point $A$ and meets the $y$-axis at the point $B$.\\
The region $R$, shown shaded in Figure 1, is bounded by the curve $C$ and the straight line through $A$ and $B$.
\item Use your answer to part (b) to find an approximate value for the area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2016 Q2 [6]}}