Question

What is the arc length of `f(x)= sqrt(5x+1)` on `x in [0,2]`?

Answer 1

`L=1/10(sqrt759-sqrt29)+5/4ln((2sqrt11+sqrt69)/(2+sqrt29))` units.

Explanation:

`f(x)=sqrt(5x+1)`

`f'(x)=5/(2sqrt(5x+1))`

Arc length is given by:

`L=int_0^2sqrt(1+25/(4(5x+1)))dx`

Apply the substitution `5x+1=u`:

`L=1/5int_1^11sqrt(1+25/(4u))du`

Rearrange:

`L=1/5int_1^11sqrt(4u+25)/(2sqrtu)du`

Apply the substitution `sqrtu=v`:

`L=1/5int_1^sqrt11sqrt(4v^2+25)dv`

Apply the substitution `2v=5tantheta`:

`L=5/2intsec^3thetad theta`

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

`L=5/4[secthetatantheta+ln|sectheta+tantheta|]`

Reverse the last substitution:

`L=[1/10vsqrt(4v^2+25)+5/4ln|2v+sqrt(4v^2+25)|]_1^sqrt11`

Insert the limits of integration:

`L=1/10(sqrt759-sqrt29)+5/4ln((2sqrt11+sqrt69)/(2+sqrt29))`