13.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{7d07e1ad-d87a-4eb5-a15e-05b927892915-40_495_634_207_657}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{figure}
Figure 4 shows a sketch of part of the curve \(C\) with equation
$$y = \frac { 1 } { 2 x } \ln 2 x , \quad x > \frac { 1 } { 2 }$$
The finite region \(R\), shown shaded in Figure 4, is bounded by the curve \(C\), the \(x\)-axis and the lines with equations \(x = \mathrm { e }\) and \(x = 5 \mathrm { e }\).
The table below shows corresponding values of \(x\) and \(y\) for \(y = \frac { 1 } { 2 x } \ln 2 x\). The values for \(y\) are given to 4 significant figures.
| \(x\) | e | 2 e | 3 e | 4 e | 5 e |
| \(y\) | 0.3114 | 0.2195 | 0.1712 | 0.1416 | 0.1215 |
- Use the trapezium rule with all the \(y\) values in the table to find an approximate value for the area of \(R\), giving your answer to 3 significant figures.
- Using the substitution \(u = \ln 2 x\), or otherwise, find \(\int \frac { 1 } { 2 x } \ln 2 x \mathrm {~d} x\)
- Use your answer to part (b) to find the true area of \(R\), giving your answer to 3 significant figures.
- Using calculus, find an equation for the tangent to the curve at the point where \(x = \frac { \mathrm { e } ^ { 2 } } { 2 }\), giving your answer in the form \(y = m x + c\) where \(m\) and \(c\) are exact multiples of powers of e.