What is the arc length of f(x)=sqrt(x-1) on x in [2,6] ?

1 Answer
Jun 27, 2018

L=sqrt5/2(sqrt21-1)+1/4ln((2sqrt5+sqrt21)/(2+sqrt5))

Explanation:

Given

f(x)=sqrt(x-1), x in [2,6]

Let y=f(x), we thus have:

x=y^2+1, y in [1,sqrt5]

Take the derivative with respect to y:

x'=2y

Arc length is given by:

L=int_1^sqrt5sqrt(1+4y^2)dy

Apply the substitution 2y=tantheta:

L=1/2intsec^3thetad theta

This is a known integral. If you do not have it memorized apply integration by parts or look it up in a table of integrals:

L=1/4[secthetatantheta+ln|sectheta+tantheta|]

Reverse the substitution:

L=1/4[2ysqrt(1+4y^2)+ln|2y+sqrt(1+4y^2)|]_1^sqrt5

Insert the limits of integration:

L=sqrt5/2(sqrt21-1)+1/4ln((2sqrt5+sqrt21)/(2+sqrt5))