Answer
$2$
Work Step by Step
\[\begin{align}
& \int_{-\infty }^{\infty }{{{e}^{-\left| x \right|}}}dx \\
& \text{By the definition of absolute value} \\
& {{e}^{-\left| x \right|}}={{e}^{x}}\text{, }x<0\text{ } \\
& {{e}^{-\left| x \right|}}={{e}^{-x}}\text{, }x>0\text{ } \\
& \text{Therefore} \\
& \int_{-\infty }^{\infty }{{{e}^{-\left| x \right|}}}dx=\int_{-\infty }^{0}{{{e}^{x}}}dx+\int_{0}^{\infty }{{{e}^{-x}}}dx \\
& \text{Integrating} \\
& =\left[ {{e}^{x}} \right]_{-\infty }^{0}-\left[ {{e}^{-x}} \right]_{0}^{\infty } \\
& =\left( {{e}^{0}}-{{e}^{-\infty }} \right)-\left( {{e}^{-\infty }}-{{e}^{0}} \right) \\
& =1-0-1+1 \\
& =2 \\
\end{align}\]
