Question

Arc length of integration (1+8x)^1/2 from x=0 to x=1?

Answer 1

`L=1/8(15-sqrt17)+2ln((sqrt17-1)/2)` units.

Explanation:

Let

`f(x)=sqrt(1+8x)`

`f'(x)=4/sqrt(1+8x)`

Arc length is given by:

`L=int_0^1sqrt(1+16/(1+8x))dx`

Apply the substitution `1+8x=16u^2`:

`L=4int_(1/4)^(3/4)sqrt(u^2+1)du`

Apply the substitution `u=tantheta`:

`L=4intsec^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=2[secthetatantheta+ln|sectheta+tantheta|]`

Reverse the last substitution:

`L=2[usqrt(u^2+1)+ln|u+sqrt(u^2+1)|]_(1/4)^(3/4)`

Hence

`L=1/8(15-sqrt17)+2ln(8/(1+sqrt17))`

Simplify:

`L=1/8(15-sqrt17)+2ln((sqrt17-1)/2)`