# Thread: Length of a curve

1. ## Length of a curve

g(x)=x^3+(1/12x) for 1<x<3

I can get as far as integral of sqrt(1+9x^4-(1/12x^4)) before running into substitution problems... what should I set U to be? I tried using x^4 but I keep getting screwed up somewhere.. Thanks in advance!

2. Originally Posted by strawlion
g(x)=x^3+(1/12x) for 1<x<3

I can get as far as integral of sqrt(1+9x^4-(1/12x^4)) before running into substitution problems... what should I set U to be? I tried using x^4 but I keep getting screwed up somewhere.. Thanks in advance!
$\displaystyle \displaystyle g(x) = x^3 + \frac{1}{12x}$

$\displaystyle \displaystyle g'(x) = 3x^2 - \frac{1}{12x^2}$

$\displaystyle \displaystyle [g'(x)]^2 = 9x^4 - \frac{1}{2} + \frac{1}{144x^4}$

$\displaystyle \displaystyle 1 + [g(x)]^2 = 9x^4 + \frac{1}{2} + \frac{1}{144x^4} = \left(3x^2 + \frac{1}{12x^2}\right)^2$

take it from here?

3. Thanks a lot! I got the answer, but I'm just curious as to how the -(1/2) appears when you square, and disappears again when you simplify... sorry I'm probably missing something obvious here haha

4. Originally Posted by strawlion
Thanks a lot! I got the answer, but I'm just curious as to how the -(1/2) appears when you square, and disappears again when you simplify... sorry I'm probably missing something obvious here haha
basic algebra ... $\displaystyle (a-b)^2 = a^2 - 2ab + b^2$

$\displaystyle a^2 + 2ab + b^2 = (a+b)^2$