$ x(t)=e^t $ [1,0]
Signal Energy
$ E=\int_{t_1}^{t_2}x(t)dt $
find the signal energy of $ x(t)=e^{4t}\! $ on $ [0,1]\! $
$ E = \int_{0}^{1} |e^{4t}|^2\ dt \! $
$ = \int_{0}^{2} e^{8t}\ dt \! $
$ = \frac{1}{8}[e^{8t}]_{t=0}^{t=1} \! $ $ = \frac{1}{8}(e^8 -1)\! $
Signal Power
Average signal power between $ [t_1,t_2]\! $ is $ P_{avg}=\frac{1}{t_2-t_1}\int_{t_1}^{t_2} |x(t)|^2\ dt \! $.
$ P_{avg}=\frac{1}{1-0} \int_{0}^{1} |e^{4t}|^2\ dt \! $
$ = \int_{0}^{1} e^{8t}\ dt \! $
$ = \frac{1}{8}[e^{8t}]_{t=0}^{t=1} \! $
$ = \frac{1}{8}(e^8 -1)\! $
