What is the arc length of f(x)=sqrt(x+2) on x in [0,2]?
I know that the integral is complex but I need it to be solved using trigonometric substitution
I know that the integral is complex but I need it to be solved using trigonometric substitution
1 Answer
Jun 13, 2018
Explanation:
f(x)=sqrt(x+2)
f'(x)=1/(2sqrt(x+2))
Arc length is given by:
L=int_0^2sqrt(1+1/(4x+8))dx
Apply the substitution
L=1/2int_(2sqrt2)^4sqrt(u^2+1)du
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[usqrt(u^2+1)+ln|u+sqrt(u^2+1)|]_(2sqrt2)^4
Hence
L=1/2(2sqrt17-3sqrt2)+1/4ln((4+sqrt17)/(3+2sqrt2))