(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Find the arc length of the following curves on the given interval by integrating with respect to x

y=x^4/4+ 1/(8x^2); [1,2]

2. Relevant equations

Let f have a continuous first derivative on the interval [a,b]. The length of the curve from (a,f(a)) to (b,f(b)) is

L = integral[a,b] sqrt(1+f'(x)^2) dx

3. The attempt at a solution

y=x^4/4+ 1/(8x^2); [1,2]

I took the derivative and got

dy/dx = x^3 - 1/(4x^4)

I than square this and got

x^6 - 1/2 + 1/(16*x^6)

plunging into the formula I get

integral[1,2] sqrt(1/2 + x^6 + 1/(16*x^6)) dx

I have no idea were to go from here. I tried making several substitutions and than realized that there was nothing in my techniques of integration to evaluate this in any way or form. I plugged the indefinite integral into wolfram alpha to see what it got for the antiderivative and apparently it doesn't know how to integrate this neither.

http://www.wolframalpha.com/input/?i=integral+sqrt%281%2F2%2Bx^6%2B1%2F%2816x^6%29%29

well apparently it came up with a antiderivative it just has no steps to prove it to be correct when you click on show steps "(step-by-step results unavailable)". So I conclude that there is something really simple that I'm not seeing or something really complicated. This came from my textbook in -Applications of Integration-Lengths of Curves

Thanks for any help!

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Calculus II - Lengths of Curves - Hard

**Physics Forums | Science Articles, Homework Help, Discussion**