Arc Length problem: y= (x^5)/6 + 1/(10x^3)

[Nineball]

I'm solving an arc length problem, I got this far on differentiation and algebra alone...

Original problem:

y= (x^5)/6 + 1/(10x^3), 1 <= x >= 2

What I got to:

1/30 integral from 1 to 2 of the square root of...

this mess: [(675x^8 + 625x^16 + 81) / (x^8)] dx of course

I'm having trouble figuring out what to do with this square root and that nasty function that came from 1+f'(x)

 

[royhaas]

I'm not sure where your mistake is. Go back and be very careful. You will find that \(\displaystyle 1+(y')^2\) is a perfect square.

 

[stapel]

y= (x^5)/6 + 1/(10x^3), 1 <= x >= 2

I'm guessing the highlighted portion should be a "less than or equal to"...?

Eliz.

 

[Deleted member 4993]

Just to add to royhaas's hint:

1 + (f'(x))^2 = (5/6 x^4 + 3/10 x^-4)^2