S StefanS meandering Velosoph Veteran Oct 9, 2005 #2 K.I.L.E.R said: dp / (p^0.5) = ln(p^0.5) Is this integral correct? Click to expand... Sorry but that's wrong the answer is §dp 1/(p^0.5)=2*p^(0.5) instead of d/dp ln(p^0.5)=1/2*d/dp ln(p)=1/2 *1/p
K.I.L.E.R said: dp / (p^0.5) = ln(p^0.5) Is this integral correct? Click to expand... Sorry but that's wrong the answer is §dp 1/(p^0.5)=2*p^(0.5) instead of d/dp ln(p^0.5)=1/2*d/dp ln(p)=1/2 *1/p
S StefanS meandering Velosoph Veteran Oct 9, 2005 #4 K.I.L.E.R said: Ta. Where does the 2 come from? Click to expand... Just look at the following differential d/dp (p^(1/2))=1/2*p^(-1/2) This is why the 2 is needed...
K.I.L.E.R said: Ta. Where does the 2 come from? Click to expand... Just look at the following differential d/dp (p^(1/2))=1/2*p^(-1/2) This is why the 2 is needed...
K K.I.L.E.R Retarded moron Veteran Oct 10, 2005 #5 Thanks. Unfortunately I have to spread some rep before giving you another point.
Simon F Tea maker Moderator Veteran Supporter Oct 10, 2005 #6 K.I.L.E.R said: Thanks. Unfortunately I have to spread some rep before giving you another point. Click to expand... Did it for you.
K.I.L.E.R said: Thanks. Unfortunately I have to spread some rep before giving you another point. Click to expand... Did it for you.
K K.I.L.E.R Retarded moron Veteran Oct 10, 2005 #7 Simon F said: Did it for you. Click to expand... Thanks. Now I give you another one unless B3D forum software decides I can't.
Simon F said: Did it for you. Click to expand... Thanks. Now I give you another one unless B3D forum software decides I can't.
