Signal
$ y(t)=exp(t) $
Power
$ P = \int_{t_1}^{t_2} \! |y(t)|^2\ dt $
$ P = \int_0^2 \! |e^(2t)\ dt $
$ P = \int_0^2 \! exp(2t) dt $ (since exp(t) cannot be negative,take off the absolute value sign)
$ P = {1\over 2} ( exp(2t) )\mid_0^{2} $
$ P = {1\over 2}(exp(4) - 1)} $
