Originally Posted by

**Kimberly** Calculate the length of the curve:

$\displaystyle y=\frac{x^4}{4} + \frac{1}{8x^2}$ from $\displaystyle 1\leq{x}\leq3$

I'll base myself on the formula: $\displaystyle \int_{a}^{b}\sqrt{1-(f'(x))^2}dx$

Oops! It is not minus but __plus__ into the square root!

Step 1: Derive the function.

$\displaystyle y=\frac{x^4}{4} + \frac{1}{8x^2}$

$\displaystyle y' = x^3 - \frac{1}{4x^3}$

Step 2: Square the derivative.

$\displaystyle (x^3 - \frac{1}{4x^3})^2$

$\displaystyle x^6 + \frac{1}{16x^6} - \frac{1}{2}$

Step 3: Subtract the derivative from $\displaystyle 1$

$\displaystyle 1 - (x^6 + \frac{1}{16x^6} - \frac{1}{2})$

$\displaystyle \frac{3}{2} - \frac{1}{16x^6} - x^6$

This yields a monster of a integral...

$\displaystyle \int_{1}^{3}\sqrt{\frac{3}{2}-\frac{1}{16x^6}-x^6}dx$

And THAT I need help with...

EDIT: Sorry for the double post, Admin... my internet acted up. Please delete the other post. Thank you and sorry!