6.
\begin{figure}[h]
\captionsetup{labelformat=empty}
\caption{Figure 3}
\includegraphics[alt={},max width=\textwidth]{c0c6303b-f527-4e68-91bc-5c9c6ffa8de8-09_442_805_283_555}
\end{figure}
Figure 3 shows a sketch of the curve with equation \(y = ( x - 1 ) \ln x , \quad x > 0\).
- Complete the table with the values of \(y\) corresponding to \(x = 1.5\) and \(x = 2.5\).
| \(x\) | 1 | 1.5 | 2 | 2.5 | 3 |
| \(y\) | 0 | | \(\ln 2\) | | \(2 \ln 3\) |
- use the trapezium rule
- with values of \(y\) at \(x = 1,2\) and 3 to find an approximate value for \(I\) to 4 significant figures,
- 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.
- Explain, with reference to Figure 3, why an increase in the number of values improves the accuracy of the approximation.
- Show, by integration, that the exact value of \(\int _ { 1 } ^ { 3 } ( x - 1 ) \ln x \mathrm {~d} x\) is \(\frac { 3 } { 2 } \ln 3\).