# What is the arc length of f(x) = (x^2-x)^(3/2)  on x in [2,3] ?

Mar 1, 2016

$L = {\int}_{2}^{3} \sqrt{1 + \frac{9}{4} \left(x \left(x - 1\right) {\left(2 x - 1\right)}^{2}\right)} \mathrm{dx}$
Integrating we find (use Wolfram Alpha there is no closed integral for this integral):
L = 11.9145

#### Explanation:

We are going to use the Arc Length formula for L:
$L = {\int}_{a}^{b} \mathrm{ds}$
where
$\mathrm{ds} = \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$
$L = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$
so let's differentiate $y = f \left(x\right) = {\left({x}^{2} - x\right)}^{\frac{3}{2}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) = \frac{3}{2} \sqrt{x \left(x - 1\right)} \cdot \left(2 x - 1\right)$
Now find:
${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = {\left(f ' \left(x\right)\right)}^{2} = \frac{9}{4} \left(x \left(x - 1\right) {\left(2 x - 1\right)}^{2}\right)$
$L = {\int}_{2}^{3} \sqrt{1 + \frac{9}{4} \left(x \left(x - 1\right) {\left(2 x - 1\right)}^{2}\right)} \mathrm{dx}$
Integrating we find (use Wolfram Alpha there is no closed integral for this integral):
L = 11.9145