Monday

September 1, 2014

September 1, 2014

Posted by **Anonymous** on Monday, June 16, 2014 at 5:18pm.

- MATH -
**Steve**, Monday, June 16, 2014 at 5:45pmsince |x| = -x for x < 0.

∫[-∞,x] e^-|t| dt

= ∫[-∞,x] e^t dt if x < 0

= e^t

So, for x>=0,

∫[-∞,x] e^-|t| dt

= ∫[-∞,0] e^t dt + ∫[0,x] e^-t dt

= 1 + (1-e^-x)

= 2 - e^-x

- MATH -
**Anonymous**, Monday, June 16, 2014 at 5:47pmWhat if we have .. integration of xe^(|x|) dx from negative infinity to x.

- MATH -
**Steve**, Monday, June 16, 2014 at 6:20pmNo idea. Do it the way I did, but you have to use integration by parts. If you get stuck, show how far you got.

You should wind up with

-(x+1)e^-x for x<0

(x-1)e^x for x>=0

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