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

Question

Elijah Abram · Accepted Answer

#L=1+sum_(n=1)^oo((1/2),(n))1/(4n-1)(1/2^(4n-1)-1/3^(4n-1))#

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

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

Arclength is given by:

#L=int_1^2sqrt(1+1/(x+1)^4)dx#

For #x in [1,2]#, #1/(x+1)^4<1#. Take the series expansion of the square root:

#L=int_1^2sum_(n=0)^oo((1/2),(n))1/(x+1)^(4n)dx#

Isolate the #n=0# term and simplify:

#L=int_1^2dx+sum_(n=1)^oo((1/2),(n))int_1^2 1/(x+1)^(4n)dx#

Integrate directly:

#L=1+sum_(n=1)^oo((1/2),(n))1/(1-4n)[1/(x+1)^(4n-1)]_1^2#

Insert the limits of integration:

#L=1+sum_(n=1)^oo((1/2),(n))1/(4n-1)(1/2^(4n-1)-1/3^(4n-1))#

Emilia Aguilar · Answer

undefined To find the arc length of the function $ f(x) = \frac{x - 2}{x^2 - x - 2} $ on the interval $[1,2]$, we first need to compute its derivative and then apply the arc length formula. The derivative of $ f(x) $ is calculated as $ f'(x) = \frac{2x^2 - 6x - 1}{(x^2 - x - 2)^2} $. The arc length formula for a function $ f(x) $ on an interval $[a,b]$ is given by $ \int_a^b \sqrt{1 + (f'(x))^2} \, dx $. Substituting the derivative into this formula and integrating over the interval $[1,2]$, we obtain the arc length of $ f(x) $ on this interval.