2 Fig. 8 shows the line \(y = x\) and parts of the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\), where
$$\mathrm { f } ( x ) = \mathrm { e } ^ { x - 1 } , \quad \mathrm {~g} ( x ) = 1 + \ln x$$
The curves intersect the axes at the points A and B , as shown. The curves and the line \(y = x\) meet at the point C .
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a55b82e6-3fcb-4283-bd36-06a17a9a7536-1_804_888_1061_662}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{figure}
- Find the exact coordinates of A and B . Verify that the coordinates of C are \(( 1,1 )\).
- Prove algebraically that \(\mathrm { g } ( x )\) is the inverse of \(\mathrm { f } ( x )\).
- Evaluate \(\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x\), giving your answer in terms of e .
- Use integration by parts to find \(\int \ln x \mathrm {~d} x\).
Hence show that \(\int _ { \mathrm { e } ^ { - 1 } } ^ { 1 } \mathrm {~g} ( x ) \mathrm { d } x = \frac { 1 } { \mathrm { e } }\).
- Find the area of the region enclosed by the lines OA and OB , and the arcs AC and BC .