Normal Distribution CurveDate: 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 Check out our web site! http://mathforum.org/dr.math/ |
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