Question

What is the arclength of `f(x)=x-sqrt(x+3)` on `x in [1,3]`?

Answer 1

`L=(4sqrt6-1)/(8sqrt2)sqrt((4sqrt6-1)^2+1)-35/8+1/(8sqrt2)ln((4sqrt6-1+sqrt((4sqrt6-1)^2+1))/(7+5sqrt2))` units.

Explanation:

`f(x)=x-sqrt(x+3)`

`f'(x)=1-1/(2sqrt(x+3))`

Arclength is given by:

`L=int_1^3sqrt(1+(1-1/(2sqrt(x+3)))^2)dx`

Apply the substitution `2sqrt(x+3)=u`:

`L=int_4^(2sqrt6)sqrt(1+(1-1/u)^2)(1/2udu)`

Simplify:

`L=1/2int_4^(2sqrt6)sqrt(2u^2-2u+1)du`

Complete the square in the square root:

`L=1/(2sqrt2)int_4^(2sqrt6)sqrt((2u-1)^2+1)du`

Apply the substitution `2u-1=tantheta`:

`L=1/(4sqrt2)intsec^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/(8sqrt2)[secthetatantheta+ln|sectheta+tantheta|]`

Reverse the substitution:

`L=1/(8sqrt2)[(2u-1)sqrt((2u-1)^2+1)+ln|(2u-1)+sqrt((2u-1)^2+1)|]_4^(2sqrt6)`

Insert the limits of integration:

`L=(4sqrt6-1)/(8sqrt2)sqrt((4sqrt6-1)^2+1)-35/8+1/(8sqrt2)ln((4sqrt6-1+sqrt((4sqrt6-1)^2+1))/(7+5sqrt2))`