What is the arc length of the curve given by f(x)=xe^(-x) in the interval x in [0,ln7]?

1 Answer
Aug 24, 2016

$\approx 2.05$

Explanation:

$s = \int \dot{s} \setminus \mathrm{dt}$

$= {\int}_{a}^{b} \sqrt{\vec{v} \cdot \vec{v}} \setminus \mathrm{dt}$

In Cartesian:
$\vec{r} = \left(\begin{matrix}x \\ x {e}^{- x}\end{matrix}\right)$

$\vec{v} = \frac{d}{\mathrm{dt}} \left(\begin{matrix}x \\ x {e}^{- x}\end{matrix}\right) = \left(\begin{matrix}\dot{x} \\ \dot{x} {e}^{- x} - \dot{x} x {e}^{- x}\end{matrix}\right)$

$= \dot{x} \left(\begin{matrix}1 \\ {e}^{- x} \left(1 - x\right)\end{matrix}\right)$

$\implies s = {\int}_{a}^{b} \sqrt{1 + {e}^{- 2 x} {\left(1 - x\right)}^{2}} \setminus \frac{\mathrm{dx}}{\mathrm{dt}} \setminus \mathrm{dt}$

$\implies {\int}_{0}^{\ln 7} \sqrt{1 + {e}^{- 2 x} {\left(1 - x\right)}^{2}} \setminus \mathrm{dx}$

Horrendous integration.

Computer says $\approx 2.05$

graph{x e^(-x) [-1.25, 3.75, -0.73, 1.77]}