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How do you integrate $\frac 1 2e^{-|x|}$?

Attempt: This equals the integral of $\frac 1 2e^{-\sqrt{(x^2)}}$

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    @lord12: You have symmetry across the $y$-axis, so the median is $0$, no computation needed.2012-02-28

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Note that $\frac{1}{2}e^{-|x|}=\frac{1}{2}e^{-|-x|}$, so the function is symmetric about $x=0$. Thus in the particular case of integration from $-\infty$ to $\infty$, we have $\int_{-\infty}^{\infty}\frac{1}{2}e^{-|x|}dx=2\int_0^\infty\frac{1}{2}e^{-|x|}dx=\int_0^\infty e^{-x}dx=\lim\limits_{y\to\infty}\int_0^ye^{-x}=\lim\limits_{y\to\infty}(-e^{-y}+e^0)=e^0=1$ since $e^{-y}$ becomes arbitrarily small for large $y$.