| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Moderate -0.3 This is a straightforward C2 question requiring basic calculator work to complete a table, standard application of the trapezium rule formula, and simple subtraction to find an area between a curve and line. All techniques are routine with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.09f Trapezium rule: numerical integration |
| \(x\) | 0 | 0.5 | 1 | 1.5 | 2 |
| \(y\) | 0 | 0.530 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) \(1.414\) (or exact \(\sqrt{2}\)), \(\quad 3.137\) | B1, B1 | Allow awrt |
| (b) \(\dfrac{1}{2}(0.5)\ldots\) | B1 | |
| \(\ldots\{0 + 6 + 2(0.530 + 1.414 + 3.137)\}\) | M1 A1ft | |
| \(= 4.04\) | A1 | Must be 3 s.f. |
| (c) Area of triangle \(= \dfrac{1}{2}(2\times6) = 6\) | B1 | Can be found by integration or trapezium rule on \(y=3x\) |
| Area required \(=\) Area of triangle \(-\) Answer to (b) | M1 | Subtract either way round |
| \(6 - 4.04 = 1.96\) | A1ft | ft from (b); answer to (b) must be less than 6 |
# Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** $1.414$ (or exact $\sqrt{2}$), $\quad 3.137$ | B1, B1 | Allow awrt |
| **(b)** $\dfrac{1}{2}(0.5)\ldots$ | B1 | |
| $\ldots\{0 + 6 + 2(0.530 + 1.414 + 3.137)\}$ | M1 A1ft | |
| $= 4.04$ | A1 | Must be 3 s.f. |
| **(c)** Area of triangle $= \dfrac{1}{2}(2\times6) = 6$ | B1 | Can be found by integration or trapezium rule on $y=3x$ |
| Area required $=$ Area of triangle $-$ Answer to (b) | M1 | Subtract either way round |
| $6 - 4.04 = 1.96$ | A1ft | ft from (b); answer to (b) must be less than 6 |
**Total: 9 marks**5. The curve $C$ has equation
$$y = x \sqrt { } \left( x ^ { 3 } + 1 \right) , \quad 0 \leqslant x \leqslant 2$$
\begin{enumerate}[label=(\alph*)]
\item Complete the table below, giving the values of $y$ to 3 decimal places at $x = 1$ and $x = 1.5$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 0 & 0.5 & 1 & 1.5 & 2 \\
\hline
$y$ & 0 & 0.530 & & & 6 \\
\hline
\end{tabular}
\end{center}
\item Use the trapezium rule, with all the $y$ values from your table, to find an approximation for the value of $\int _ { 0 } ^ { 2 } x \sqrt { } \left( x ^ { 3 } + 1 \right) \mathrm { d } x$, giving your answer to 3 significant figures.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{22ebc302-765c-4734-b312-b286ccb20be9-06_1110_644_1119_648}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows the curve $C$ with equation $y = x \sqrt { } \left( x ^ { 3 } + 1 \right) , 0 \leqslant x \leqslant 2$, and the straight line segment $l$, which joins the origin and the point $( 2,6 )$. The finite region $R$ is bounded by $C$ and $l$.
\item Use your answer to part (b) to find an approximation for the area of $R$, giving your answer to 3 significant figures.\\
(3)
\section*{LU}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2007 Q5 [9]}}