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Math Help - Integrate the semicircle probability distribution

  1. #1
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    Integrate the semicircle probability distribution

    Hi folks, I know its a long shot, but would anyone like to explain how to integrate the function

    f(x) = 2/(pi * r^2) sqrt(R^2 - x^2)

    I'm sorry about the poor notation. Please see Wigner semicircle distribution - Wikipedia, the free encyclopedia for a clearer idea.

    It's been a long time since I've done integration. I presume it involves a substitution like, say, u = R^2 - x^2. I just can't seem to figure it out.

    Got any ideas?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by sabatier View Post
    Hi folks, I know its a long shot, but would anyone like to explain how to integrate the function

    f(x) = 2/(pi * r^2) sqrt(R^2 - x^2)

    I'm sorry about the poor notation. Please see Wigner semicircle distribution - Wikipedia, the free encyclopedia for a clearer idea.

    It's been a long time since I've done integration. I presume it involves a substitution like, say, u = R^2 - x^2. I just can't seem to figure it out.

    Got any ideas?
    Make a Trig Substitution:

    x=R\sin\theta \implies dx=R\cos\theta\,d\theta

    Then the integral becomes

    <br />
\frac{2}{\pi R^2} \int \sqrt{R^2-R^2\sin^2\theta} R\cos\theta \,d\theta
    \implies \frac{2}{\pi}\int \cos^2\theta\,d\theta.

    To integrate this, recall that \cos^2\theta=\frac{1+\cos(2\theta)}{2}.

    After you integrate, get the answer back in terms of x.

    Hope this helped!!
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  3. #3
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    Hey thanks a million for replying so quickly! Cheers!
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by sabatier View Post
    Hi folks, I know its a long shot, but would anyone like to explain how to integrate the function

    f(x) = 2/(pi * r^2) sqrt(R^2 - x^2)

    I'm sorry about the poor notation. Please see Wigner semicircle distribution - Wikipedia, the free encyclopedia for a clearer idea.

    It's been a long time since I've done integration. I presume it involves a substitution like, say, u = R^2 - x^2. I just can't seem to figure it out.

    Got any ideas?
    Alternatively since \int_{-r}^{r}\frac{2}{\pi{(r_1)^2}}\sqrt{r^2-x^2}dx

    Just represents the integration of a semi-circle of radius r multiplied by a contant. And since integration is equivalent to finding area we can see that

    A_{semi-circle}=\frac{1}{2}\pi{r^2}

    So \frac{2}{\pi{(r_1)^2}}\int_{-r}^{r}\sqrt{r^2-x^2}dx=\frac{2}{\pi{(r_1)^2}}\cdot\bigg(\frac{1}{2  }\pi{r^2}\bigg)

    Now if r_1=r

    \int_{-r}^{r}\frac{2}{\pi(r_1)^2}\sqrt{r^2-x^2}dx=1

    if r_1\ne{r}

    then \int_{-r}^{r}\frac{2}{\pi{(r_1)^2}}\sqrt{r^2-x^2}dx=\bigg(\frac{r}{r_1}\bigg)^2


    NOTE this only applies to if the integration limits are -r and r...that is what I have gathered...if they are not DO NOT APPLY this....even if they are not I hope this has helped gap a bride between integration and area and how sometimes we dont eve need to integrate
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by sabatier View Post
    Hi folks, I know its a long shot, but would anyone like to explain how to integrate the function

    f(x) = 2/(pi * r^2) sqrt(R^2 - x^2)

    I'm sorry about the poor notation. Please see Wigner semicircle distribution - Wikipedia, the free encyclopedia for a clearer idea.

    It's been a long time since I've done integration. I presume it involves a substitution like, say, u = R^2 - x^2. I just can't seem to figure it out.

    Got any ideas?
    Sinve f(x) is the density of the Wigner distribution we know that:

    I=\int_{-r}^r \frac{2}{\pi R^2} \sqrt{R^2-x^2}~dx=1

    So we might be interested in the cumulative distribution:

    <br />
I(\rho)=\int_{-r}^{\rho} \frac{2}{\pi R^2} \sqrt{R^2-x^2}~dx<br />
=\frac{2}{\pi R} \int_{-r}^{\rho} \sqrt{1-(x/R)^2}~dx<br />

    Now put u=x/R :

    <br />
I(\rho) = \frac{2}{\pi} \int_{-1}^{\rho/R} \sqrt{1-u^2}~du<br />

    Now a trig substitution like u=\sin(\theta) will finish this by turning it into a known integral (or at least one that you can see how to do).

    RonL
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