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Normal Distribution Curve

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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.
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```
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
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
College Calculus

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