Results 1 to 2 of 2

Thread: Is this right? Finding x-bar

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    64

    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
    $\displaystyle x bar = \frac{ln|4|}{2arctan\sqrt{3}}$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1
    Quote Originally Posted by circuscircus View Post
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: Dec 8th 2011, 10:27 AM
  2. Replies: 1
    Last Post: Jul 3rd 2010, 10:40 PM
  3. Finding a limit. Finding Maclaurin series.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 18th 2010, 10:04 PM
  4. Finding the radius, solving, and finding theta?
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Jun 13th 2009, 02:37 PM
  5. Replies: 1
    Last Post: Apr 9th 2009, 09:02 AM

Search Tags


/mathhelpforum @mathhelpforum