Line 24: Line 24:
 
   <math>P=\frac{1}{2}\int_{1}^{3}\ x ^4\, dx , \,\!</math>
 
   <math>P=\frac{1}{2}\int_{1}^{3}\ x ^4\, dx , \,\!</math>
  
   <math>P=\frac{1}{2}\!</math> <math>\*</math> <math>\frac{1}{5}</math> *<math>((3^5)\!</math> - <math>(1^5))\!</math>
+
   <math>P=\frac{1}{2}\!</math> * <math>\frac{1}{5}</math> *<math>((3^5)\!</math> - <math>(1^5))\!</math>
  
  

Revision as of 17:25, 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\! $

Power calculation


 $ P=\frac{1}{(3-1)}\int_{1}^{3}\ |x^2|^2\,dx,\,\! $
 
 $ P=\frac{1}{2}\int_{1}^{3}\ x ^4\, dx , \,\! $
 $ P=\frac{1}{2}\! $ * $ \frac{1}{5} $ *$ ((3^5)\! $ - $ (1^5))\! $


 $ P=\frac{1}{10}\! $*$ (243-1)\! $


 $ P=24.2\! $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood