# How do you find the arc length of the curve y=e^(3x) over the interval [0,1]?

Sep 24, 2016

$19.14$

#### Explanation:

simple arc length formulation:

$s = {\int}_{C} \mathrm{ds} = {\int}_{{x}_{1}}^{{x}_{2}} \sqrt{1 + {\left(y '\right)}^{2}} \mathrm{dx}$

here $y ' = 3 {e}^{3 x}$

$s = {\int}_{0}^{1} \sqrt{1 + {\left(3 {e}^{3 x}\right)}^{2}} \mathrm{dx}$

the integration is not nice, here is computer solution:

$\int \sqrt{1 + {\left(3 {e}^{3 x}\right)}^{2}} \mathrm{dx} = \frac{1}{3} \left(\sqrt{9 {e}^{6 x} + 1} - {\tanh}^{- 1} \left(\sqrt{9 {e}^{6 x} + 1}\right)\right) + C$

The numerical solution is: $19.14$