Let us find the energy and average power of a signal $ x(t) = 5e^{5t} $ for the time interval [0,5]
Energy
Energy expanded from [0,5]
$ E = \int_0^{5}{(5e^{5t})^2dt} $
$ E =\int_0^{5}{25e^{10t}dt} $
$ E =25\int_0^{5}{e^{10t}dt} $
$ E =25|\frac{e^{10t}}{10}|_{t=0}^{t=5} $
$ E =\frac{25}{10}|e^{50}-e^0| $
$ E =2.5(e^{50}-1) $
Average Power
$ P =\frac{1}{5-0}\int_0^{5}{(5e^{5t})^2dt} $
$ P =\frac{1}{5}\int_0^{5}{25e^{10t}dt} $
$ P =5\int_0^{5}{e^{10t}dt} $
$ P =5\int_0^{5}{e^{10t}dt} $
$ P =5|\frac{e^{10t}}{10}|_{t=0}^{t=5} $
$ P =0.5[e^{50} - e^0] $
$ P = 0.5(e^{50} - 1) $
