# Is this right? Finding x-bar

• Sep 6th 2007, 02:39 PM
circuscircus
Is this right? Finding x-bar
I don't have an answer to check against but could someone let me know if I did this right?

Formula for xbar:
$\displaystyle x bar = \frac{My}{m}$

Finding m:
$\displaystyle m = p \int_{\sqrt{3}}^{0} \frac{1}{x^2+1}$
$\displaystyle m = p ( arctan(\sqrt{3}))$

Finding My:
$\displaystyle My = p \int_{0}^{\sqrt{3}} x * \frac{1}{x^2+1}$
$\displaystyle My = p \int_{0}^{\sqrt{3}} \frac{x}{x^2+1}$

$\displaystyle u=x^2+1$
$\displaystyle du=2x$

$\displaystyle = p * \frac{1}{2} \int_{1}^{4} 1/u$
$\displaystyle = p * \frac{1}{2} ( ln|4| - ln|1|)$
$\displaystyle = p * \frac{1}{2} ln|4/1|$
$\displaystyle = p * \frac{1}{2} ln|4|$

x bar
$\displaystyle x bar = \frac{ln|4|}{2arctan\sqrt{3}}$
• Sep 6th 2007, 03:24 PM
topsquark
Quote:

Originally Posted by circuscircus
I don't have an answer to check against but could someone let me know if I did this right?

Formula for xbar:
$\displaystyle x bar = \frac{My}{m}$

Finding m:
$\displaystyle m = p \int_{\sqrt{3}}^{0} \frac{1}{x^2+1}$
$\displaystyle m = p ( arctan(\sqrt{3}))$

Finding My:
$\displaystyle My = p \int_{0}^{\sqrt{3}} x * \frac{1}{x^2+1}$
$\displaystyle My = p \int_{0}^{\sqrt{3}} \frac{x}{x^2+1}$

$\displaystyle u=x^2+1$
$\displaystyle du=2x$

$\displaystyle = p * \frac{1}{2} \int_{1}^{4} 1/u$
$\displaystyle = p * \frac{1}{2} ( ln|4| - ln|1|)$
$\displaystyle = p * \frac{1}{2} ln|4/1|$
$\displaystyle = p * \frac{1}{2} ln|4|$

x bar
$\displaystyle x bar = \frac{ln|4|}{2arctan\sqrt{3}}$

You mostly have it, but there are a few details.

First m is negative (unless you reversed the integration limits.)

Second, you can do better than $\displaystyle atn(\sqrt{3})$. What is this value?

Third, you know that |4| = 4 so you can drop the absolute value. And how does $\displaystyle \frac{1}{2}ln(4)$ compare to the value of $\displaystyle ln(2)$?

-Dan