What is the arc length of f(x)=sqrt(x-1) on x in [2,6] ?
1 Answer
Jun 27, 2018
Explanation:
Given
f(x)=sqrt(x-1) ,x in [2,6]
Let
x=y^2+1 ,y in [1,sqrt5]
Take the derivative with respect to
x'=2y
Arc length is given by:
L=int_1^sqrt5sqrt(1+4y^2)dy
Apply the substitution
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))