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

##### 1 Answer

Sep 22, 2017

Approximately

#### Explanation:

The arc length of a curve on

#A = int_a^b sqrt(1 + (dy/dx)^2)dx#

Taking the derivative, we get

#f'(x) = (x/sqrt(x^2 - 1)(x) - sqrt(x^2 - 1)(1))/x^2#

#f'(x) = (x^2/sqrt(x^2 - 1) - sqrt(x^2 - 1))/x^2#

#f'(x) = ((x^2 - x^2 + 1)/sqrt(x^2 - 1))/x^2#

#f'(x) = 1/(sqrt(x^2 - 1)x^2)#

So now applying the formula, we get:

#A = int_-2^-1 sqrt(1 + (1/(sqrt(x^2 - 1)x^2))^2) dx#

#A = int_-2^-1 sqrt(1 + 1/(x^6 - x^4)) dx#

#A = int_-2^-1 sqrt((x^6 - x^4 + 1)/(x^6 - x^4)) dx#

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

Hopefully this helps!