1. 

   msmat999 Member

   I was just wonder how I would integrate 2cxexp[-cx^2] where c is a constant. Would I want to use integration by parts? Along the lines of;
   
   u = 2cx => du/dx = x
   
   dv = exp[-cx^2]

    

   msmat999, Aug 25, 2012

   #1
2. 

   tiger Member

   Yes to integration by parts, but I would say
   u=cx^2
   du= 2cxdx
   2cxexp[-cx^2]
   becomes
   exp[-u]du

    

   tiger, Aug 26, 2012

   #2
3. 

   John Lee ActEd Tutor Staff Member

   We'd use integration by substitution with u = -cx².
   
   In which case du/dx = -2cx. Hence "du = -2cx dx"
   
   So the integral would become -exp(u) which would integrate to -exp(u) = -exp(-cx²).

    

   John Lee, Aug 28, 2012

   #3

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