\section*{Part C}
In this part, we consider the function $h$ defined on $\mathbb { R }$ by
$$h ( x ) = ( x - 1 ) \mathrm { e } ^ { - 2 x } + 1 .$$
We admit that the function $h$ is differentiable on $\mathbb { R }$.\\
We place ourselves in an orthonormal coordinate system ( O ; I, J).\\
We denote $\mathscr { C } _ { h }$ the representative curve of the function $h$ and $d$ the line with equation $y = x$.\\
We admit that the curve $\mathscr { C } _ { h }$ is above the line $d$ on the interval $[ 0 ; 1 ]$.\\
Let $\mathscr { D }$ be the region of the plane bounded by the curve $\mathscr { C } _ { h }$, the line $d$ and the vertical lines with equations $x = 0$ and $x = 1$. Let $\mathscr { A }$ be the area of $\mathscr { D }$ expressed in square units.
\begin{enumerate}
  \item On ANNEX 1, shade the region $\mathscr { D }$ and justify that
$$\mathscr { A } = \int _ { 0 } ^ { 1 } [ h ( x ) - x ] \mathrm { d } x$$
  \item a. Prove that, for all real $x$,
$$h ( x ) - x = ( 1 - x ) \left( 1 - \mathrm { e } ^ { - 2 x } \right) .$$
b. We admit that, for all real $x$, $\mathrm { e } ^ { - 2 x } \geqslant 1 - 2 x$.\\
Prove that, for all real $x$ in the interval $[ 0 ; 1 ]$,
$$h ( x ) - x \leqslant 2 x - 2 x ^ { 2 } .$$
c. Deduce that $\mathscr { A } \leqslant \frac { 1 } { 3 }$.
  \item Let $H$ be the function defined on $[ 0 ; 1 ]$ by
$$H ( x ) = \frac { 1 } { 4 } ( 1 - 2 x ) \mathrm { e } ^ { - 2 x } + x$$
We admit that the function $H$ is an antiderivative of the function $h$ on $[ 0 ; 1 ]$.\\
Determine the exact value of $\mathscr { A }$.
\end{enumerate}