Changes

1 byte removed ,  06:47, September 18, 2011
Seperatint -> Separating, Substitutiing -> Substituting
Line 13: Line 13:  
<math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy.</math>
 
<math>\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2-y^2}dxdy.</math>
   −
Seperating it,
+
Separating it,
    
<math>\int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy = (\int_{-\infty}^{\infty} e^{-x^2}dx)^2 </math>
 
<math>\int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy = (\int_{-\infty}^{\infty} e^{-x^2}dx)^2 </math>
Line 27: Line 27:  
<math> = 2\pi \int_{0}^{\infty} e^{-r^2}rdr.</math>
 
<math> = 2\pi \int_{0}^{\infty} e^{-r^2}rdr.</math>
   −
Substitutiing <math>z=r^2</math> into the integral,
+
Substituting <math>z=r^2</math> into the integral,
    
<math> = 2\pi \int_{0}^{\infty} e^{-z}(z^{1/2})(\frac{dz}{2z^{1/2}})</math>
 
<math> = 2\pi \int_{0}^{\infty} e^{-z}(z^{1/2})(\frac{dz}{2z^{1/2}})</math>
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