What is the arclength of f(x)=(x-2)/x^2 on x in [-2,-1]?

Nov 15, 2016

$f \left(x\right) = \frac{x - 2}{x} ^ 2$

$\therefore f ' \left(x\right) = \frac{\left({x}^{2}\right) \left(1\right) - \left(x - 2\right) \left(2 x\right)}{{x}^{2}} ^ 2$
$\therefore f ' \left(x\right) = \frac{{x}^{2} - 2 {x}^{2} + 4 x}{x} ^ 4$
$\therefore f ' \left(x\right) = \frac{4 x - {x}^{2}}{x} ^ 4$
$\therefore f ' \left(x\right) = \frac{4 - x}{x} ^ 3$

Arc Length is given by:
$L = {\int}_{- 2}^{- 1} \sqrt{1 + f ' {\left(x\right)}^{2}} \mathrm{dx}$
$\therefore L = {\int}_{- 2}^{- 1} \sqrt{1 + {\left(\frac{4 - x}{x} ^ 3\right)}^{2}} \mathrm{dx}$
$\therefore L = {\int}_{- 2}^{- 1} \sqrt{1 + {\left(4 - x\right)}^{2} / {x}^{6}} \mathrm{dx}$
$\therefore L = {\int}_{- 2}^{- 1} \sqrt{\frac{{x}^{6} + {\left(4 - x\right)}^{2}}{x} ^ 6} \mathrm{dx}$
$\therefore L = {\int}_{- 2}^{- 1} \frac{\sqrt{{x}^{6} + {\left(4 - x\right)}^{2}}}{x} ^ 3 \mathrm{dx}$

This definite integral does not have an elementary intrinsic solution and would need to be solved numerically, using either a computer or estimated using the Trapezium Rule or Simpson's Rule