- #1
Bazman
- 21
- 0
When solving for the variance of the normal distribution one needs to evaluate the following integral:
INT(-infinfity to infinity)[x^2*e^(-x^2/2).dx]
I proceed using integration by parts:
[-x.e^(-x^2/2)|(infin to -infin) + INT(-infin to infin) 2*e(-x^2/2)dx]
However apparently correct answer is:
[-x.e^(-x^2/2)|(infin to -infin) + INT(-infin to infin) e(-x^2/2)dx]
but I don't see how the 2 cancels?
INT(-infinfity to infinity)[x^2*e^(-x^2/2).dx]
I proceed using integration by parts:
[-x.e^(-x^2/2)|(infin to -infin) + INT(-infin to infin) 2*e(-x^2/2)dx]
However apparently correct answer is:
[-x.e^(-x^2/2)|(infin to -infin) + INT(-infin to infin) e(-x^2/2)dx]
but I don't see how the 2 cancels?
Last edited: