What is the arc length of #f(x)=xsqrt(x^2-1) # on #x in [3,4] #?

1 Answer
May 20, 2018

#L=7+1/8ln(15/8)+sum_(n=2)^oo((1/2),(n))1/4^n(1/8^(n-1)-1/15^(n-1))#

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 #x^2-1=u#:

#L=int_8^15sqrt(1+1/(4u))du#

For #u in [8,15]#, #1/(4u)<1#. Take the series expansion of the square root:

#L=int_8^15sum_(n=0)^oo((1/2),(n))1/(4u)^ndu#

Isolate the #n=0# and #n=1# terms:

#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))#