$ 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)\! $
