1. ## Lengths of Curves

Find the length of the curve

$y= \frac{1}{3}(x^2 + 2)^{\frac{3}{2}}$ from $x=0$ to $x=3$

$f'(x) = x(x^2 + 2)^{\frac{1}{2}}$

$L = \int \sqrt{1 + \left(x(x^2+2)^\frac{1}{2}\right)}dx$

$L = \int \sqrt{1 + x^4 + 2x^2}dx$

Let $u = 1 + x^4 + 2x^2$

After this I'm unsure of what to do. For example in the practice problems we had $1 + 8x$ under the square root and what we did was let $u = 1 + 8x$ and $dx = \frac{1}{8}du$ and continue on from there. But in this problem I'm a little bit confused. I guess because the concept hasn't fully sank in yet.

2. Originally Posted by VitaX
Find the length of the curve

$y= \frac{1}{3}(x^2 + 2)^{\frac{3}{2}}$ from $x=0$ to $x=3$

$f'(x) = x(x^2 + 2)^{\frac{1}{2}}$

$L = \int \sqrt{1 + \left(x(x^2+2)^\frac{1}{2}\right)}dx$

$L = \int \sqrt{1 + x^4 + 2x^2}dx$

Let $u = 1 + x^4 + 2x^2$

After this I'm unsure of what to do. For example in the practice problems we had $1 + 8x$ under the square root and what we did was let $u = 1 + 8x$ and $dx = \frac{1}{8}du$ and continue on from there. But in this problem I'm a little bit confused. I guess because the concept hasn't fully sank in yet.
$x^4+2x^2+1=(x^2+1)^2$

3. Originally Posted by VitaX
Find the length of the curve

$y= \frac{1}{3}(x^2 + 2)^{\frac{3}{2}}$ from $x=0$ to $x=3$

$f'(x) = x(x^2 + 2)^{\frac{1}{2}}$

$L = \int \sqrt{1 + \left(x(x^2+2)^\frac{1}{2}\right)}dx$

$L = \int \sqrt{1 + x^4 + 2x^2}dx$

Let $u = 1 + x^4 + 2x^2$

After this I'm unsure of what to do. For example in the practice problems we had $1 + 8x$ under the square root and what we did was let $u = 1 + 8x$ and $dx = \frac{1}{8}du$ and continue on from there. But in this problem I'm a little bit confused. I guess because the concept hasn't fully sank in yet.

$x^4+2x^2+1=(x^2+1)^2$ ...and your integral is way easier than it looks!

Tonio