Integration

**ixi** · September 23rd 2009, 10:28 AM

The question is:

The integration of e^(-abs (x)) from -infinitiy to + infinity

The absolute before the x is throwing me off...

I got

-e^(-abs(x)) from -infinitiy to + infinity

i graphed it but doesn't it go to 0?

so would the answer be 0?

thanks

**artvandalay11** · September 23rd 2009, 12:00 PM

$\int_{-\infty}^\infty e^{-|x|}dx=\int_{-\infty}^0 e^{-|x|}dx+\int_{0}^\infty e^{-|x|}dx$

Now when $x\in(-\infty,0], |x|= -x,$ so $-|x|= -(-x)=x$

And when $x\in[0,\infty), |x|=x$ and so $-|x|=-x$

So therefore $\int_{-\infty}^0 e^{-|x|}dx+\int_{0}^\infty e^{-|x|}dx=\int_{-\infty}^0 e^{x}dx+\int_{0}^\infty e^{-x}dx$

$=1+1=2$ if you need help with the integration let me know

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