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.