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

Sep 22, 2017

Approximately $1.48$ units.

Explanation:

The arc length of a curve on $x \in \left[a , b\right]$ is given by

$A = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

Taking the derivative, we get

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

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

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

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

So now applying the formula, we get:

$A = {\int}_{-} {2}^{-} 1 \sqrt{1 + {\left(\frac{1}{\sqrt{{x}^{2} - 1} {x}^{2}}\right)}^{2}} \mathrm{dx}$

$A = {\int}_{-} {2}^{-} 1 \sqrt{1 + \frac{1}{{x}^{6} - {x}^{4}}} \mathrm{dx}$

$A = {\int}_{-} {2}^{-} 1 \sqrt{\frac{{x}^{6} - {x}^{4} + 1}{{x}^{6} - {x}^{4}}} \mathrm{dx}$

This integral doesn't have an elementary solution. According to Wolfram Alpha, the approximation for this integral would be $1.48$ units.

Hopefully this helps!