Expected value of exponential distribution with integration by parts (non-textbook way)

Question

I try to calculate expected value of exponential distribution with integration by parts without success. I know that the text book way to do it so that first take lambda out of the integral and then do the integration by parts. What if I would like to calculate it directly: \begin{align} E[X]=\int_{0}^\infty x\lambda e^{-\lambda x} dx \end{align} And now \begin{align} f(x)=x, \quad f'(x)=dx \end{align} and \begin{align} g'(x)=-\lambda e^{-\lambda x}dx, \quad g(x)= e^{-\lambda x} \end{align} So I would like to start with \begin{align} E[X]=-\int_{0}^\infty x (-\lambda e^{-\lambda x}) dx. \end{align} Is this possible and if not why not?

Answer 1

In this case it is much easier to use the fact (which follows from Tonelli's theorem) that for a non-negative random variable $X$ with finite mean, $\mathbb E[X] = \int_0^\infty (1-F_X(x))\ \mathsf dx$, with $F_X$ being the distribution function of $X$. Hence $$ \mathbb E[X] = \int_0^ \infty e^{-\lambda t}\ \mathsf dt = \frac1\lambda. $$