Q:

Expected Value and Distribution Functions

Suppose you want to find $E(X)$ for a non-negative random variable $X$. So you can use $\int_{0}^{\infty} [1-F(x)] \ dx$. To get $1-F(x)$, all you have to do is integrate as follows: $\int_{x}^{\infty} f(t) \ dt$? In other words, instead of integrating from $t=0$ to $t= x$, we can just integrate from $t = x$ to $t = \infty$? 

A:

I guess your question is: Why is $E(X)=\int_0^\infty 1-F(x) dx$?
The answer is as follows: By definition $$E(X)=\int_0^\infty tf(t) dt.$$
Now \begin{eqnarray*} & & \int_0^\infty tf(t) dt\
&=&\int_0^\infty f(t)\int_0^t 1 dx dt\
&=&\int_0^\infty\int_0^t f(t) dx dt\
&=&\int_0^\infty\int_x^\infty f(t) dt dx \
&=&\int_0^\infty 1-F(x) dx.
\end{eqnarray*}

