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Math Help - Is this right? Finding x-bar

  1. #1
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    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:
    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}}
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  2. #2
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    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:
    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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