# Thread: root integral

1. ## root integral

Hi. In one of my homework problems, I have to find the arclength of

$\displaystyle y = \frac{3}{2}x^\frac{2}{3} + 2, [1, 8]$

If y' = x^(-1/3), then I believe that translates into

$\displaystyle \int{\sqrt{1+x^\frac{-2}{3}}dx}$

(Not sure how to add the bounds with TEX formatting.)

But I'm not sure what to do with that integral, as u-substitution doesn't seem to apply. Could someone point me in the right direction?

2. ## Re: root integral

Hello, infraRed!

$\displaystyle \text{Find the arclength of: }\:y \:=\:\tfrac{3}{2}x^{\frac{2}{3}} + 2,\;\; [1, 8]$

$\displaystyle \displaystyle \text{If }y' \,=\, x^{\text{-}\frac{1}{3}},\,\text{ then that translates into: }\: \int^8_1\!\! \sqrt{1+x^{\text{-}\frac{2}{3}}}\,dx$

$\displaystyle \text{But I'm not sure what to do with that integral.}$

We have: .$\displaystyle \displaystyle \sqrt{1+x^{\text{-}\frac{2}{3}}} \;=\;\sqrt{1 + \frac{1}{x^{\frac{2}{3}}}} \;=\;\sqrt{\frac{x^{\frac{2}{3}} + 1}{x^{\frac{2}{3}}}} \;=\;\frac{\sqrt{x^{\frac{2}{3}}+1}} {\sqrt{x^{\frac{2}{3}}}}} \;=\;\frac{\sqrt{x^{\frac{2}{3}}+1}}{x^{\frac{1}{3 }}}$

Hence: .$\displaystyle \displaystyle L \;=\;\int^8_1\left(x^{\frac{2}{3}}+1\right)^{\frac {1}{2}} \left(x^{\text{-}\frac{1}{3}}dx\right)$

Let $\displaystyle u \,=\,x^{\frac{2}{3}} + 1 \quad\Rightarrow\quad du \,=\,\tfrac{2}{3}x^{\text{-}\frac{1}{3}}dx \quad\Rightarrow\quad \begin{Bmatrix}x=8 & \to & u=5 \\ x=1 & \to & u=2 \end{Bmatrix}$

Substitute: .$\displaystyle L \;=\;\int^5_2 u^{\frac{1}{2}}\left(\tfrac{3}{2}\,du\right) \;=\;\tfrac{3}{2}\int^5_2 u^{\frac{1}{2}}\,du \;=\;u^{\frac{3}{2}}\,\bigg]^5_2$

. . . . . . . . . $\displaystyle L \;=\;5^{\frac{3}{2}} - 2^{\frac{3}{2}} \;=\; 5\sqrt{5} - 2\sqrt{2}$

3. ## Re: root integral

Black magic if I ever saw it! Thanks!

BTW, what is the TEX syntax for adding bounds to your integrals? The tutorial says integrals are \int{...}, but that is all I know.

4. ## Re: root integral

You do \int_0^1f(x)\,dx to get $\displaystyle \int_0^1f(x)\,dx$. The "\," just adds a little space - without it, it looks like $\displaystyle \int_0^1f(x)dx$.

If you mouse over the graphical formula, it gives you the source text. It also usually works to copy the graphical formula and paste it into an input box - it gives you the source text.

- Hollywood