How do you find the length of the curve #y=e^x# between #0<=x<=1# ?

1 Answer
Oct 10, 2014

#L=int_0^1sqrt{1+({dy}/{dx})^2} dx#

#=int_0^1sqrt{1+e^{2x}}dx#

by the substitution #u=sqrt{1+e^{2x}}#.

#Rightarrow {du}/{dx}=e^{2x}/sqrt{1+e^{2x}}={u^2-1}/u Rightarrow dx={u}/{u^2-1}du#

As #x# goes from #0# to #1#, #u# goes from #sqrt{2}# to #sqrt{1+e^2}#

#=int_{sqrt{2}}^{sqrt{1+e^2}}u^2/{u^2-1} du#

by partial fraction decomposition,

#=int_{sqrt{2}}^{sqrt{1+e^2}}[1+1/2(1/{u-1}-1/{u+1})] du#

#=[u+1/2(ln|u-1|-ln|u+1|)]_{sqrt{2}}^{sqrt{1+e^2}}#

#=[u+1/2ln|{u-1}/{u+1}|]_{sqrt{2}}^{sqrt{1+e^2}}#

#=sqrt{1+e^2}+1/2 ln({sqrt{1+e^2}-1}/{sqrt{1+e^2}-1})-sqrt{2}-1/2ln({sqrt{2}-1}/{sqrt{2}+1})#

I hope that this was helpful.