Recent content by kekly

Thanks for all your hlep. I have been playing around with it and I understand it now! It was not remembering how to use the substitution method that was causing me problems! Thansk again for all your help! kek

I understand the substitution that is made and I can find du/dr = r/sigma^2 That can then be rearranged to rdr = sigma^2du How then do I get to this! 2\pi {A} \sigma\int_{o}^{\infinity}{e^{-u} du} Forgive me I think it is part of the substiution method that I don't...

Thanks for your help. I am still a little confused. As I said I haven't done integration like this for a long time! I can't see how you substituted rdr= \sigma^2du back into get 2\pi {A} \sigma\int_{o}^{\infinity}{e^{-u} du}. I can see that when r=0, u= 0, when r= infinity, u= infinity...

I see what he has done now! Thanks so much. You guys are lifesavers! Kek

It's true not all of the bible has been proven however nothing in it has been proven as false either.

Hi, Need help desperately! I am trying to figure out the area under a gaussian cone by finding the integral of 2PIArEXP[-(r^2)/(2sigma^2)] dr My supervisor thought it is 2PI [A sigma^2 EXP(-(r^2)/(2 sigma^2)] Between 0 and infinity and he came up with the...