$ 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 $
$ E=\int_{0}^{1}|e^(8t)dt $
$ = \frac{1}{8}[e^{8t}]_{t=0}^{t=1} \! $ $ = \frac{1}{8}(e^8 -1)\! $
