Asked by Anonymous
what is the integration of e^-|x| from negative infinity to x ?
Answers
Answered by
Steve
since |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
∫[-∞,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
Answered by
Anonymous
What if we have .. integration of xe^(|x|) dx from negative infinity to x.
Answered by
Steve
No 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
You should wind up with
-(x+1)e^-x for x<0
(x-1)e^x for x>=0
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