Normal Distribution Curve

Date: 06/12/97 at 14:23:07
From: Sridevi
Subject: Integration

What is the integral of e^(-x^2)?

I have tried using the UV rule, but I couldn't reach an answer.

Date: 06/12/97 at 16:41:00
From: Doctor Anthony
Subject: Re: Integration

This integral is the one giving the area under the normal distribution 
curve. It cannot be evaluated by elementary techniques, but by 
squaring and changing to polar coordinates. 

A quick runthrough of the method is as follows. We assume the limits 
of integration are 0 to infinity.

  I = INT[e^(-x^2)dx] then squaring

I^2 = INT[e^(-x^2)dx] INT[e^(-y^2)dy]

    = INT[INT[e^-(x^2+y^2)dxdy]]

Now we convert to polar coordinates. The unit of area dxdy in 
Cartesian coordinates becomes the unit of area rd(theta)dr in polar 
coordinates, since x^2+y^2 = r^2. The region of integration is the 
first quadrant with both x and y ranging from 0 to infinity. In polar 
coordinates this means that theta varies from 0 to pi/2, and r ranges 
from 0 to infinity.

I^2 = INT[d(theta)INT[e^(-r^2)rdr]]    

If we put r^2 = u then rdr = 1/2du:  

    = INT[d(theta)INT[1/2*e^(-u)du]]

    = INT[d(theta)[-1/2*e^(-u)]]

    = INT[d(theta)[-1/2(0 - 1)]]

    = 1/2INT[d(theta)]

    = 1/2(theta)   from 0 to pi/2

    = 1/2(pi/2 - 0)

    =  pi/4        and finally taking square roots

      sqrt(pi)
 I =  --------
         2

In the case of the standardized normal distribution, the range of 
integration is from negative to positive infinity and the expression 
is e^(-x^2/2). The area under this curve is then sqrt(2pi) and this 
explains why the the factor 1/sqrt(2pi) appears in front of the 
integral so as to make the area under the curve equal to 1.   

-Doctor Anthony,  The Math Forum
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