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]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{666dc19e-f293-4738-8530-fce90df23d17-5_707_876_440_593}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}

(i) 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 $x \geqslant 0$.\\
(ii) 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 $( 1 , \ln 2 )$.\\
(iii) 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.\\
(iv) 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 $x = 1$.

\hfill \mbox{\textit{OCR MEI C3 2007 Q8 [18]}}