# Thread: Integrate the semicircle probability distribution

1. ## 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?

2. Originally Posted by sabatier
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

$
\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!!

3. Hey thanks a million for replying so quickly! Cheers!

4. Originally Posted by sabatier
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

5. Originally Posted by sabatier
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:

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

Now put $u=x/R$ :

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

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