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

Dec 30, 2017

Arc length of $f \left(x\right)$ is $\sqrt{2}$ unit.

Explanation:

f(x) = (x^2-2x)/(2-x) ; x in [-2,-1]

$f ' \left(x\right) = \frac{\left(2 x - 2\right) \left(2 - x\right) - \left({x}^{2} - 2 x\right) \left(- 1\right)}{2 - x} ^ 2$ or

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

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

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

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

$\therefore {\left[f ' \left(x\right)\right]}^{2} = 1$ . Total length of the arc from x = a to x = b is

${\int}_{a}^{b} \sqrt{1 + {\left[f ' \left(x\right)\right]}^{2}} \mathrm{dx}$

$L = {\int}_{- 2}^{- 1} \sqrt{1 + 1} \mathrm{dx}$ or

$L = {\int}_{- 2}^{- 1} \sqrt{2} \mathrm{dx}$ or

$L = \sqrt{2} {\left[x\right]}_{-} {2}^{-} 1 \mathmr{and} L = \sqrt{2} \left[- 1 - \left(- 2\right)\right]$ or

$L = \sqrt{2} \left[- 1 + 2\right] = \sqrt{2}$

Arc length of $f \left(x\right)$ is $\sqrt{2}$ unit. [Ans]