# What is the arclength of f(x)=x^3-xe^x on x in [-1,0]?

Arc length $s = 1.54116$ units

#### Explanation:

The formula to determine length of arc s:
$s = {\int}_{a}^{b} \sqrt{1 + f ' {\left(x\right)}^{2}} \mathrm{dx}$
$a = - 1$ and $b = 0$ the limits

$f \left(x\right) = {x}^{3} - x {e}^{x}$

$f ' \left(x\right) 3 {x}^{2} - \left[x \cdot {e}^{x} \cdot 1 + 1 \cdot {e}^{x}\right]$
$f ' \left(x\right) = 3 {x}^{2} - x {e}^{x} - {e}^{x}$

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

There is no simple formula to evaluate the integral, so try using Simpson's Rule:

$s = 1.54116$ units

Just observe the graph from $x = - 1$ to $x = 0$
graph{y=x^3-x e^x [-2.5, 2.5, -1.25, 1.25]}