8 Fig. 8 shows part of the curve \(y = \mathrm { f } ( x )\), where
$$\mathrm { f } ( x ) = \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \text { for } x \geqslant 0 .$$
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{666dc19e-f293-4738-8530-fce90df23d17-5_707_876_440_593}
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\caption{Fig. 8}
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- Find \(\mathrm { f } ^ { \prime } ( x )\), and hence calculate the gradient of the curve \(y = \mathrm { f } ( x )\) at the origin and at the point \(( \ln 2,1 )\).
The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \ln ( 1 + \sqrt { x } )\) for \(x \geqslant 0\).
- Show that \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are inverse functions. Hence sketch the graph of \(y = \mathrm { g } ( x )\).
Write down the gradient of the curve \(y = \mathrm { g } ( x )\) at the point \(( 1 , \ln 2 )\).
- Show that \(\int \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } - 2 \mathrm { e } ^ { x } + x + c\).
Hence evaluate \(\int _ { 0 } ^ { \ln 2 } \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \mathrm {~d} x\), giving your answer in an exact form.
- Using your answer to part (iii), calculate the area of the region enclosed by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis and the line \(x = 1\).