# What is the arclength of (sqrt(e^(t^2)/(e^t)+1),t^3) on t in [-1,1]?

$s = 2.86658$ units

#### Explanation:

Given $x \left(t\right) = \sqrt{{e}^{{t}^{2} - t} + 1}$ and $y \left(t\right) = {t}^{3}$

dx/dt=((e^(t^2-t))(2t-1))/(2 sqrt(e^(t^2-t) +1)

$\frac{\mathrm{dy}}{\mathrm{dt}} = 3 {t}^{2}$

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

$s = {\int}_{-} {1}^{1} \sqrt{{\left(\frac{\left({e}^{{t}^{2} - t}\right) \left(2 t - 1\right)}{2 \sqrt{{e}^{{t}^{2} - t} + 1}}\right)}^{2} + {\left(3 {t}^{2}\right)}^{2}} \mathrm{dt}$

the integral is complicated so Simpson's Rule is used:

so that $s = 2.86658$