Is this right? Finding x-bar

**circuscircus** · September 6th 2007, 02:39 PM

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

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

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

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

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

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

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

**topsquark** · September 6th 2007, 03:24 PM

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:
$x bar = \frac{My}{m}$

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

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

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

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

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

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 $atn(\sqrt{3})$ . What is this value?

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

-Dan

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