What is the arc length of f(x) = x-xe^(x^2)  on x in [ 2,4] ?

Jan 1, 2018

${\int}_{2}^{4} \sqrt{1 + {\left(1 - {e}^{{x}^{2}} - 2 {x}^{2} {e}^{{x}^{2}}\right)}^{2}} \setminus \mathrm{dx} \approx 2.93$

Explanation:

The formula for arc length of $f \left(x\right)$ on the interval $\left[a , b\right]$ is:
${\int}_{a}^{b} \sqrt{1 - {\left(f ' \left(x\right)\right)}^{2}} \setminus \mathrm{dx}$

First we'll work out the derivative of our function:
$\frac{d}{\mathrm{dx}} \left(x - x {e}^{{x}^{2}}\right) = 1 - \frac{d}{\mathrm{dx}} \left(x {e}^{{x}^{2}}\right)$

To figure out the second part, we'll use the product rule:
$\frac{d}{\mathrm{dx}} \left(f \left(x\right) g \left(x\right)\right) = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

In our case, we get:
$1 - \left(1 \cdot {e}^{{x}^{2}} + x \cdot \frac{d}{\mathrm{dx}} \left({e}^{{x}^{2}}\right)\right)$

We can use the chain rule, which says:
$f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

This gives us:
$1 - \left({e}^{{x}^{2}} + x \cdot 2 x {e}^{{x}^{2}}\right)$

$1 - {e}^{{x}^{2}} - 2 {x}^{2} {e}^{{x}^{2}}$

If we then plug this into our formula, we get:
${\int}_{2}^{4} \sqrt{1 + {\left(1 - {e}^{{x}^{2}} - 2 {x}^{2} {e}^{{x}^{2}}\right)}^{2}} \setminus \mathrm{dx}$

This integral doesn't have an elementary answer that I've been able to find, but the value can be approximated to be roughly $2.93$