| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Standard +0.3 This is a straightforward C2 integration question requiring substitution of values into a formula, application of the trapezium rule (a standard numerical method), and integration using power rules. All techniques are routine for this level, though the algebraic manipulation and rewriting terms for integration requires care. Slightly easier than average due to being a standard textbook-style question with clear steps. |
| Spec | 1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals1.09f Trapezium rule: numerical integration |
| \(x\) | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4 |
| \(y\) | 0 | 5.866 | 5.210 | 1.856 | 0 |
| Answer | Marks |
|---|---|
| \(6.272, 3.634\) | B1, B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}\times\frac{1}{2}\) or \(\frac{1}{4}\) | B1 | |
| \(...\{(0+0)+2(5.866+"6.272"+5.210+"3.634"+1.856)\}\) | M1A1ft | Need \(\{\}\) or implied later for A1ft |
| \(\frac{1}{2}\times0.5\{(0+0)+2(5.866+"6.272"+5.210+"3.634"+1.856)\} = \frac{1}{4}\times45.676\) | ||
| \(= 11.42\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int y\,dx = 27x - x^2 - 6x^{\frac{3}{2}} + 16x^{-1}(+c)\) | M1A1A1A1A1 | M1: \(x^n\to x^{n+1}\) on any term; A1: \(27x-x^2\); A1: \(-6x^{\frac{3}{2}}\); A1: \(+16x^{-1}\) |
| \(\left(27(4)-(4)^2-6(4)^{\frac{3}{2}}+16(4)^{-1}\right) - \left(27(1)-(1)^2-6(1)^{\frac{3}{2}}+16(1)^{-1}\right)\) | dM1 | Attempt to subtract either way round using limits 4 and 1; dependent on previous M1 |
| \(= (48-36) = 12\) | A1 | cao |
## Question 9:
$y = 27-2x-9\sqrt{x}-\frac{16}{x^2}$
### Part (a):
| $6.272, 3.634$ | B1, B1 | |
|---|---|---|
### Part (b):
| $\frac{1}{2}\times\frac{1}{2}$ or $\frac{1}{4}$ | B1 | |
|---|---|---|
| $...\{(0+0)+2(5.866+"6.272"+5.210+"3.634"+1.856)\}$ | M1A1ft | Need $\{\}$ or implied later for A1ft |
| $\frac{1}{2}\times0.5\{(0+0)+2(5.866+"6.272"+5.210+"3.634"+1.856)\} = \frac{1}{4}\times45.676$ | | |
| $= 11.42$ | A1 | cao |
### Part (c):
| $\int y\,dx = 27x - x^2 - 6x^{\frac{3}{2}} + 16x^{-1}(+c)$ | M1A1A1A1A1 | M1: $x^n\to x^{n+1}$ on any term; A1: $27x-x^2$; A1: $-6x^{\frac{3}{2}}$; A1: $+16x^{-1}$ |
|---|---|---|
| $\left(27(4)-(4)^2-6(4)^{\frac{3}{2}}+16(4)^{-1}\right) - \left(27(1)-(1)^2-6(1)^{\frac{3}{2}}+16(1)^{-1}\right)$ | dM1 | Attempt to subtract either way round using limits 4 and 1; dependent on previous M1 |
| $= (48-36) = 12$ | A1 | cao |9. $y$
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6f31b6f1-33b5-4bca-9030-cf93760b454d-13_895_1308_207_294}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The finite region $R$, as shown in Figure 2, is bounded by the $x$-axis and the curve with equation
$$y = 27 - 2 x - 9 \sqrt { } x - \frac { 16 } { x ^ { 2 } } , \quad x > 0$$
The curve crosses the $x$-axis at the points $( 1,0 )$ and $( 4,0 )$.
\begin{enumerate}[label=(\alph*)]
\item Complete the table below, by giving your values of $y$ to 3 decimal places.
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | l | l | }
\hline
$x$ & 1 & 1.5 & 2 & 2.5 & 3 & 3.5 & 4 \\
\hline
$y$ & 0 & 5.866 & & 5.210 & & 1.856 & 0 \\
\hline
\end{tabular}
\end{center}
\item Use the trapezium rule with all the values in the completed table to find an approximate value for the area of $R$, giving your answer to 2 decimal places.
\item Use integration to find the exact value for the area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2013 Q9 [12]}}