| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Complete table then estimate |
| Difficulty | Standard +0.3 This is a standard C4 integration by parts question (7 marks for part a) requiring two applications of IBP with a product of polynomial and trig function. Parts (b) and (c) are routine trapezium rule applications requiring only substitution into a formula. While the integration requires careful algebraic manipulation, it follows a well-practiced technique with no novel insight needed, making it slightly easier than average. |
| Spec | 1.08i Integration by parts1.09f Trapezium rule: numerical integration |
| \(x\) | \(\pi\) | \(\frac{5\pi}{4}\) | \(\frac{3\pi}{2}\) | \(\frac{7\pi}{4}\) | \(2\pi\) |
| \(y\) | \(9.8696\) | \(14.247\) | \(15.702\) | \(G\) | \(0\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = \int_\pi^{2\pi} x^2 \sin\left(\frac{1}{2}x\right) dx = -2x^2\cos\left(\frac{1}{2}x\right) + \int 4x\cos\left(\frac{1}{2}x\right) dx\) | M1 A1 | |
| \(= -2x^2\cos\left(\frac{1}{2}x\right) + 8x\sin\left(\frac{1}{2}x\right) - \int 8\sin\left(\frac{1}{2}x\right) dx\) | M1 A1 | |
| \(= -2x^2\cos\left(\frac{1}{2}x\right) + 8x\sin\left(\frac{1}{2}x\right) + 16\cos\left(\frac{1}{2}x\right)\) | A1 | |
| Use limits to obtain \([8\pi^2 - 16] - [8\pi]\) | M1 A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Requires 11.567 | B1 | (1) |
| Answer | Marks |
|---|---|
| Area \(= \frac{\pi}{4}[9.8696 + 0 + 2 \times 15.702]\) (B1 for \(\frac{\pi}{4}\) in (i) or \(\frac{\pi}{8}\) in (iii)) | B1, M1 |
| \(= 32.42\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(= \frac{\pi}{8}[9.8696 + 0 + 2(14.247 + 15.702 + 11.567)]\) | M1 | |
| \(= 36.48\) | A1 | (5) |
## (a)
$R = \int_\pi^{2\pi} x^2 \sin\left(\frac{1}{2}x\right) dx = -2x^2\cos\left(\frac{1}{2}x\right) + \int 4x\cos\left(\frac{1}{2}x\right) dx$ | M1 A1 |
$= -2x^2\cos\left(\frac{1}{2}x\right) + 8x\sin\left(\frac{1}{2}x\right) - \int 8\sin\left(\frac{1}{2}x\right) dx$ | M1 A1 |
$= -2x^2\cos\left(\frac{1}{2}x\right) + 8x\sin\left(\frac{1}{2}x\right) + 16\cos\left(\frac{1}{2}x\right)$ | A1 |
Use limits to obtain $[8\pi^2 - 16] - [8\pi]$ | M1 A1 | (7)
## (b)
Requires 11.567 | B1 | (1)
## (c)(i)
Area $= \frac{\pi}{4}[9.8696 + 0 + 2 \times 15.702]$ (B1 for $\frac{\pi}{4}$ in (i) or $\frac{\pi}{8}$ in (iii)) | B1, M1 |
$= 32.42$ | A1 |
## (c)(ii)
Area $= \frac{\pi}{8}[9.8696 + 0 + 2(14.247 + 15.702 + 11.567)]$ | M1 |
$= 36.48$ | A1 | (5)
**Total: (13 marks)**
---\includegraphics{figure_2}
Figure 2 shows the curve with equation
$$y = x^2 \sin\left(\frac{1}{2}x\right), \quad 0 < x \leq 2\pi.$$
The finite region $R$ bounded by the line $x = \pi$, the $x$-axis, and the curve is shown shaded in Fig 2.
\begin{enumerate}[label=(\alph*)]
\item Find the exact value of the area of $R$, by integration. Give your answer in terms of $\pi$.
[7]
\end{enumerate}
The table shows corresponding values of $x$ and $y$.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$x$ & $\pi$ & $\frac{5\pi}{4}$ & $\frac{3\pi}{2}$ & $\frac{7\pi}{4}$ & $2\pi$ \\
\hline
$y$ & $9.8696$ & $14.247$ & $15.702$ & $G$ & $0$ \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $G$.
[1]
\item Use the trapezium rule with values of $x^2 \sin\left(\frac{1}{2}x\right)$
\begin{enumerate}[label=(\roman*)]
\item at $x = \pi$, $x = \frac{3\pi}{2}$ and $x = 2\pi$ to find an approximate value for the area $R$, giving your answer to 4 significant figures,
\item at $x = \pi$, $x = \frac{5\pi}{4}$, $x = \frac{3\pi}{2}$, $x = \frac{7\pi}{4}$ and $x = 2\pi$ to find an improved approximation for the area $R$, giving your answer to 4 significant figures.
\end{enumerate}
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q6 [13]}}