6.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
  \includegraphics[alt={},max width=\textwidth]{c0c6303b-f527-4e68-91bc-5c9c6ffa8de8-09_442_805_283_555}
\end{center}
\end{figure}

Figure 3 shows a sketch of the curve with equation $y = ( x - 1 ) \ln x , \quad x > 0$.
\begin{enumerate}[label=(\alph*)]
\item Complete the table with the values of $y$ corresponding to $x = 1.5$ and $x = 2.5$.

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
$x$ & 1 & 1.5 & 2 & 2.5 & 3 \\
\hline
$y$ & 0 &  & $\ln 2$ &  & $2 \ln 3$ \\
\hline
\end{tabular}
\end{center}

Given that $I = \int _ { 1 } ^ { 3 } ( x - 1 ) \ln x \mathrm {~d} x$,
\item use the trapezium rule
\begin{enumerate}[label=(\roman*)]
\item with values of $y$ at $x = 1,2$ and 3 to find an approximate value for $I$ to 4 significant figures,
\item with values of $y$ at $x = 1,1.5,2,2.5$ and 3 to find another approximate value for $I$ to 4 significant figures.
\end{enumerate}\item Explain, with reference to Figure 3, why an increase in the number of values improves the accuracy of the approximation.
\item Show, by integration, that the exact value of $\int _ { 1 } ^ { 3 } ( x - 1 ) \ln x \mathrm {~d} x$ is $\frac { 3 } { 2 } \ln 3$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C4 2006 Q6 [13]}}