Arc Length

  • #1
1,235
1
If [tex] y = \frac{x^{3}}{6} + \frac{1}{2x}\ [/tex] and [tex] \frac{1}{2}\leq x\leq 1 [/tex]. Find the arc length.

So [tex] \frac{dy}{dx} = \frac{x^{2}}{2} - \frac{1}{2x^{2}} [/tex]. So I got [tex] \frac{1}{2} \int^{1}_{\frac{1}{2}} \sqrt{2+x^{4} + x^{-4}} dx [/tex]. How would you evaulate this?

Thanks
 
Last edited:

Answers and Replies

  • #2
1,235
1
nvm got it. just a perfect square
 

Related Threads on Arc Length

  • Last Post
Replies
1
Views
975
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
4
Views
875
  • Last Post
Replies
2
Views
696
  • Last Post
Replies
8
Views
951
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
1K
Top