- #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?

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