M 5 years ago given f(x) = 2e^-2x for x > 0 and 0 elsewhere. determine Mx(t) for exponential distribution (theta = 1/2)

M(t) is the expected value of e^tx. $2\int\limits_{0}^{\infty}e ^{tx}e ^{-2x}dx=2\int\limits_{0}^{\infty}e ^{(t-2)x}dx$ evaluate that integral looking at when t is less than 2. you should find that M(t) is : $M(t)=1/(1-(\theta)t)$