What is the arclength of f(x)=(x-2)/(x^2-x-2) on x in [1,2]?
1 Answer
May 16, 2018
Explanation:
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
L=int_1^2sum_(n=0)^oo((1/2),(n))1/(x+1)^(4n)dx
Isolate the
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))