Let $g : [ 0,1 ] \rightarrow \mathbb { R }$ be the function such that $g ( x ) = 1 / x$ if $x \geqslant \mathrm { e } ^ { - 1 }$ and $g ( x ) = 0$ otherwise. We fix a real $\varepsilon \in ] 0 , \mathrm { e } ^ { - 1 } [$. We define two continuous applications $g ^ { + } , g ^ { - } : [ 0,1 ] \rightarrow \mathbb { R }$ as follows:
$g ^ { + }$ is affine on $\left[ \mathrm { e } ^ { - 1 } - \varepsilon , \mathrm { e } ^ { - 1 } \right]$ and coincides with $g$ on $\left[ 0 , \mathrm { e } ^ { - 1 } - \varepsilon \right] \cup \left[ \mathrm { e } ^ { - 1 } , 1 \right]$;
$g ^ { - }$ is affine on $\left[ \mathrm { e } ^ { - 1 } , \mathrm { e } ^ { - 1 } + \varepsilon \right]$ and coincides with $g$ on $\left[ 0 , \mathrm { e } ^ { - 1 } \left[ \cup \left[ \mathrm { e } ^ { - 1 } + \varepsilon , 1 \right] \right. \right.$.
Calculate $\int _ { 0 } ^ { 1 } g ^ { + } ( t ) \mathrm { d } t$ and $\int _ { 0 } ^ { 1 } g ^ { - } ( t ) \mathrm { d } t$.
Let $g : [ 0,1 ] \rightarrow \mathbb { R }$ be the function such that $g ( x ) = 1 / x$ if $x \geqslant \mathrm { e } ^ { - 1 }$ and $g ( x ) = 0$ otherwise. We fix a real $\varepsilon \in ] 0 , \mathrm { e } ^ { - 1 } [$. We define two continuous applications $g ^ { + } , g ^ { - } : [ 0,1 ] \rightarrow \mathbb { R }$ as follows:
\begin{itemize}
\item $g ^ { + }$ is affine on $\left[ \mathrm { e } ^ { - 1 } - \varepsilon , \mathrm { e } ^ { - 1 } \right]$ and coincides with $g$ on $\left[ 0 , \mathrm { e } ^ { - 1 } - \varepsilon \right] \cup \left[ \mathrm { e } ^ { - 1 } , 1 \right]$;
\item $g ^ { - }$ is affine on $\left[ \mathrm { e } ^ { - 1 } , \mathrm { e } ^ { - 1 } + \varepsilon \right]$ and coincides with $g$ on $\left[ 0 , \mathrm { e } ^ { - 1 } \left[ \cup \left[ \mathrm { e } ^ { - 1 } + \varepsilon , 1 \right] \right. \right.$.
\end{itemize}
Calculate $\int _ { 0 } ^ { 1 } g ^ { + } ( t ) \mathrm { d } t$ and $\int _ { 0 } ^ { 1 } g ^ { - } ( t ) \mathrm { d } t$.