Interpretation of $\int_{0}^{x} 1-F(t) dt$, in particular when F=Gamma/Erlang dist I know that $\int_{0}^{\infty} 1-F(t) dt$ is the expectation of a random  variable. But what happens when the upper limit is some finite number like so? 
\begin{align*}
\int_{0}^{x} 1-F(t) dt
\end{align*}
where F is the CDF of a gamma distribution and $F(x)<1$. What's the interpretation? Are there any other, more intuitive, forms to represent this?
I tried to understand the meaning by calculating the above for Uniform[0,1] distribution over interval [0,1/2] which resulted in 1/8. But it still isn't intuitive.
For another form, I started doing similar derivation to the one here:
Firefeather's answer to Find expected value using cdf, hoping that it would somehow simplify my integral but that led me to no better place. I end up with the following
\begin{align*}
\int_{0}^{x} 1-F(t) dt &= \int_{0}^{x} Pr(T>t) dt \\
&= \int_{0}^{x}\int_{t}^{\infty}f(y) dy dt \\
&= \int_{0}^{\infty}\int_{0}^{Min(y,x)}f(y) dt dy \\
&= \int_{0}^{\infty}Min(y,x)f(y) dy \\
& =\int_{0}^{x}yf(y) dy + \int_{x}^{\infty}xf(y) dy \\
\end{align*}
There's still no intuition/interpretation.
Finally, in special case of Erlang distribution, since it results from adding a bunch of exponentially distributed random variables, I thought that maybe I need to look into renewal functions from stochastic processes to get a better understanding of the above integral but no success so far.
 A: You can still use the same argument as Firefeather Find expected value using cdf. In this answer I describe a more intuitive view of the integration while copying the proof for your case: 
$$\begin{array}\\
\int_0^a 1-F(t) dt &= \int_0^a \int_t^\infty f_X(x) dx dt \\& =  \int_0^a \int_0^x f_X(x) dt dx +\int_a^\infty \int_0^x f_X(a) dt dx \end{array}$$
The intuitive part is to look at the domain and see the integration as scanning 
The $\int_0^a \int_t^a$ is an integration on the domain of a triangle and a rectangular.

See in the image how the integration is on a triangle and you can make triangle by either vertical strips with x from t to a or with horizonttal strips with t from 0 to x. 
These horizontal strips $\int_0^x$ can be changed into $x f(x)\vert_0^x$.  We should emphasize that the integrand f_X(x) changes in the vertical direction and not the horizontal direction so $\int f_X(x) dt = t f_X(x)$.

So $\int_0^a 1-F(t) dt = \int_0^a t f(t) dt + \int_a^\infty a f(t) dt = \int_0^a t f(t) dt + a (1-F(a)) $ . 
For an interpretation I am still lost.
