Line 6: Line 6:
 
   <math>E=\int_{1}^{3}\ |x ^2|^2\, dx , \,\!</math>
 
   <math>E=\int_{1}^{3}\ |x ^2|^2\, dx , \,\!</math>
  
E= <math>\int_{1}^{3}\ x ^4\, dx , \,\!</math>
+
  <math>E=\int_{1}^{3}\ x ^4\, dx , \,\!</math>
 
   
 
   
E= <math>\frac{1}{5}\!</math> *<math>((3^5)\!</math> - <math>(1^5))\!</math>
+
  <math>E=\frac{1}{5}\!</math> *<math>((3^5)\!</math> - <math>(1^5))\!</math>
 
   
 
   
E= <math>\frac{1}{5}\!</math>*<math>(243-1)\!</math>
+
  <math>E=\frac{1}{5}\!</math>*<math>(243-1)\!</math>
 
   
 
   
E= <math>48.4\!</math>
+
  <math>E=48.4\!</math>

Revision as of 18:07, 5 September 2008

Energy and Power

$ x(t)=x^2\! $
and the limits are from 1 to 3

Energy calculation

 $ E=\int_{1}^{3}\ |x ^2|^2\, dx , \,\! $
 $ E=\int_{1}^{3}\ x ^4\, dx , \,\! $

 $ E=\frac{1}{5}\! $ *$ ((3^5)\! $ - $ (1^5))\! $

 $ E=\frac{1}{5}\! $*$ (243-1)\! $

 $ E=48.4\! $

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett