I have the following:
Let $F(x)$ be the cumulative distribution function of the random variable $x$ for which it holds that $Pr(x>0)=1$, i.e. $F(a)=\int_{-\infty}^{a}f(x)dx$.
As a result, I have:
E$[x]=\int_{0}^{\infty}(1-F(x))dx$ where $E[]$ is the usual unconditional expectation operator.
Why should there be $1-F(x)$ as integrand instead of just $F(x)$, i.e. why not $E[x]=\int_{0}^{\infty}F(x)dx$?
You have a positive and continuous random variable (As it admits a pdf and $P(X>0)=1$). For positive random variables (discrete as well) we have that $$E[X]= \int_0^\infty P(X>x)dx $$ and being the cdf $F(x)=P(X\leq x)$ you have that $P(X>x) = 1-P(X\leq x) = 1-F(x)$.
For more details on why you have such an equality for positive r.vs. : Intuition behind using complementary CDF to compute expectation for nonnegative random variables.
