Solution
= Solution
The <probability density function> is normalized because $\int_0^\infty\alpha e^{-\alpha t}dt=1$. <Integration by parts> gives $\mathbb E[T]=\int_0^\infty t\alpha e^{-\alpha t}dt=1/\alpha$ and $\mathbb E[T^2]=2/\alpha^2$. Therefore
$$
\boxed{\mathbb E[T]=\frac1\alpha,\qquad\operatorname{Var}(T)=\mathbb E[T^2]-\mathbb E[T]^2=\frac1{\alpha^2}.}
$$
These are the mean and <variance> of the <exponential distribution> with rate $\alpha$.