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

Oct 10, 2014

$L = {\int}_{0}^{1} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

$= {\int}_{0}^{1} \sqrt{1 + {e}^{2 x}} \mathrm{dx}$

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

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 $\sqrt{1 + {e}^{2}}$

$= {\int}_{\sqrt{2}}^{\sqrt{1 + {e}^{2}}} {u}^{2} / \left\{{u}^{2} - 1\right\} \mathrm{du}$

by partial fraction decomposition,

$= {\int}_{\sqrt{2}}^{\sqrt{1 + {e}^{2}}} \left[1 + \frac{1}{2} \left(\frac{1}{u - 1} - \frac{1}{u + 1}\right)\right] \mathrm{du}$

$= {\left[u + \frac{1}{2} \left(\ln | u - 1 | - \ln | u + 1 |\right)\right]}_{\sqrt{2}}^{\sqrt{1 + {e}^{2}}}$

$= {\left[u + \frac{1}{2} \ln | \frac{u - 1}{u + 1} |\right]}_{\sqrt{2}}^{\sqrt{1 + {e}^{2}}}$

$= \sqrt{1 + {e}^{2}} + \frac{1}{2} \ln \left(\frac{\sqrt{1 + {e}^{2}} - 1}{\sqrt{1 + {e}^{2}} - 1}\right) - \sqrt{2} - \frac{1}{2} \ln \left(\frac{\sqrt{2} - 1}{\sqrt{2} + 1}\right)$

I hope that this was helpful.