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...
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...