# What is the arclength of f(t) = (t^2-5,e^t) on t in [1,3]?

Jun 2, 2018

$\approx 19.2253$

#### Explanation:

We get
$x ' \left(t\right) = 2 t$ so $x ' {\left(t\right)}^{2} = 4 {t}^{2}$

$y ' \left(t\right) = {e}^{t}$ so $y ' {\left(t\right)}^{2} = {e}^{2 t}$
and the integral is given by
${\int}_{1}^{3} \sqrt{4 {t}^{2} + {e}^{2 t}} \mathrm{dt}$
With a numerical method we get
$\approx 19.2253$