what is the integration of e^-|x| from negative infinity to x ?
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
What if we have .. integration of xe^(|x|) dx from negative infinity to x.
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
To find the integral of the function e^(-|x|) from negative infinity to x, you can break it into two separate integrals based on the intervals of x.
For x < 0:
The absolute value of a negative number is its positive counterpart. Therefore, when x < 0, the integral becomes:
∫[negative infinity to x] e^(-|x|) dx = ∫[negative infinity to x] e^(-(-x)) dx = ∫[negative infinity to x] e^x dx
To integrate e^x, we can use the fact that the derivative of e^x is itself:
∫[negative infinity to x] e^x dx = e^x + C1
For x > 0:
When x > 0, the absolute value of x is just x. So, the integral becomes:
∫[negative infinity to x] e^(-|x|) dx = ∫[negative infinity to x] e^(-x) dx
To integrate e^(-x), we again use the fact that the derivative of e^(-x) is itself:
∫[negative infinity to x] e^(-x) dx = -e^(-x) + C2
Combining the two cases, we have:
∫[negative infinity to x] e^(-|x|) dx = e^x + C1 (for x < 0) and -e^(-x) + C2 (for x > 0)
Please note that since the integral bounds are from negative infinity to x, the result will also depend on the value of x. So, the final result will be different for different values of x.