| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Complete table then apply trapezium rule |
| Difficulty | Moderate -0.3 This is a standard C4 integration question combining trapezium rule approximation with integration by parts. Part (a) is calculator work, part (b) is routine trapezium rule application, and part (c) is a textbook integration by parts example with straightforward algebraic manipulation. While it requires multiple techniques, each step follows standard procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.08i Integration by parts1.09f Trapezium rule: numerical integration |
| \(x\) | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4 |
| \(y\) | 0 | 0.608 | 3.296 | 4.385 | 5.545 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(1.386, 2.291\) | awrt 1.386, 2.291 |
| (b) | \(A = \frac{1}{2} \times 0.5(\ldots) = \ldots(0 + 2(0.608 + 1.386 + 2.291 + 3.296 + 4.385) + 5.545) = 0.25(0 + 2(0.608 + 1.386 + 2.291 + 3.296 + 4.385) + 5.545) = 0.25 \times 29.477\ldots \approx 7.37\) | B1; M1; A1fft; cao A1 |
| (c)(i) | \(\int x \ln x \, dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2}\ln x - \int \frac{x}{2} \, dx = \frac{x^2}{2}\ln x - \frac{x^2}{4}(+C)\) | M1 A1; M1 A1 |
| (c)(ii) | \(\left[\frac{x^2}{2}\ln x - \frac{x^2}{4}\right]_1^4 = (8\ln 4 - 4) - \left(-\frac{1}{4}\right) = 8\ln 4 - \frac{15}{4} = 8(2\ln 2) - \frac{15}{4} = \frac{1}{4}(64\ln 2 - 15)\) | M1; M1; \(\ln 4 = 2\ln 2\) seen or implied M1; \(a = 64, b = -15\) A1 |
(a) | $1.386, 2.291$ | awrt 1.386, 2.291 | B1 B1 | (2 marks)
(b) | $A = \frac{1}{2} \times 0.5(\ldots) = \ldots(0 + 2(0.608 + 1.386 + 2.291 + 3.296 + 4.385) + 5.545) = 0.25(0 + 2(0.608 + 1.386 + 2.291 + 3.296 + 4.385) + 5.545) = 0.25 \times 29.477\ldots \approx 7.37$ | B1; M1; A1fft; cao A1 | (4 marks)
(c)(i) | $\int x \ln x \, dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2}\ln x - \int \frac{x}{2} \, dx = \frac{x^2}{2}\ln x - \frac{x^2}{4}(+C)$ | M1 A1; M1 A1 |
(c)(ii) | $\left[\frac{x^2}{2}\ln x - \frac{x^2}{4}\right]_1^4 = (8\ln 4 - 4) - \left(-\frac{1}{4}\right) = 8\ln 4 - \frac{15}{4} = 8(2\ln 2) - \frac{15}{4} = \frac{1}{4}(64\ln 2 - 15)$ | M1; M1; $\ln 4 = 2\ln 2$ seen or implied M1; $a = 64, b = -15$ A1 | (7 marks)
**Total: [13]**
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5ef3ae4a-a06d-48c1-8b79-7d7c3f95d120-03_623_1176_196_374}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curve with equation $y = x \ln x , x \geqslant 1$. The finite region $R$, shown shaded in Figure 1, is bounded by the curve, the $x$-axis and the line $x = 4$.
The table shows corresponding values of $x$ and $y$ for $y = x \ln x$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
$x$ & 1 & 1.5 & 2 & 2.5 & 3 & 3.5 & 4 \\
\hline
$y$ & 0 & 0.608 & & & 3.296 & 4.385 & 5.545 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Complete the table with the values of $y$ corresponding to $x = 2$ and $x = 2.5$, giving your answers to 3 decimal places.
\item Use the trapezium rule, with all the values of $y$ in the completed table, to obtain an estimate for the area of $R$, giving your answer to 2 decimal places.
\item \begin{enumerate}[label=(\roman*)]
\item Use integration by parts to find $\int x \ln x \mathrm {~d} x$.
\item Hence find the exact area of $R$, giving your answer in the form $\frac { 1 } { 4 } ( a \ln 2 + b )$, where $a$ and $b$ are integers.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2010 Q2 [13]}}