What is the arc length of #f(x)=xsqrt(x^2-1) # on #x in [3,4] #?
1 Answer
May 20, 2018
Explanation:
#f(x)=xsqrt(x^2-1)#
#f'(x)=(2x^2-1)/sqrt(x^2-1)#
Arc length is given by:
#L=int_3^4sqrt(1+(2x^2-1)^2/(x^2-1))dx#
Simplify:
#L=int_3^4xsqrt((4x^2-3)/(x^2-1))dx#
Rearrange:
#L=int_3^4(2x)sqrt(1+1/(4(x^2-1)))dx#
Apply the substitution
#L=int_8^15sqrt(1+1/(4u))du#
For
#L=int_8^15sum_(n=0)^oo((1/2),(n))1/(4u)^ndu#
Isolate the
#L=int_8^15(1+1/(8u))du+sum_(n=2)^oo((1/2),(n))1/4^nint_8^15 1/u^ndu#
Integrate term by term:
#L=[u+1/8lnu]_ 8^15+sum_(n=2)^oo((1/2),(n))1/4^n[-1/u^(n-1)]_8^15#
Insert the limits of integration:
#L=7+1/8ln(15/8)+sum_(n=2)^oo((1/2),(n))1/4^n(1/8^(n-1)-1/15^(n-1))#