Recent content by Predz

Anyone?

I re-did it again (this time using infinity and 0). This is exactly what I did: \intke(-|x|) k\inte(-|x|) =k[-e(-|x|)] k[-e(-|infinity|) - (-e(-|0|)] = 1 k[0 - (-1)] = 1 k[0 +1] = 1 k = 1 Please tell me exactly WHERE I went wrong and how to fix it. Also, can you tell me how to start the...

Has to be equal to 1. So I integrated ke(-|x|) and got -ke(-|x|). Then I did -ke(-|x|) = 1 and replaced x by 0 and negative infinity. I solved and got k = -1. Is my integration correct? Was I right to replace x by 0 and negative infinity? Am I completely off?

Is it negative infinity? See the post above yours.

So I worked out the integration part by myself. When I integrated ke(-|x|), I got -ke(-|x|). To find k, I substituted x by 0 and -\infty. I worked it out and I found out that k=-1. I'm not sure I did this right though...

Homework Statement The random variable X has the probability density f(x) defined by: f(x) = ke(-|x|) Compute the constant k, then find the probability density of Y=X2 Homework Equations The Attempt at a Solution I'm completely stumped by this question. I know I need to...