Expected value of random variable $X$ truncated at $t$. I have a random variable $x$ distributed between $0$ and $1$ with continuous and differentiable density $f(x)$. 
I want to express the expected value of $x$ conditional on $x < t$ where $0 < t < 1$ using only $t$, $F(t)$, and $E(X)$, i.e.,
$$E(X|x < t) = \int_0^t\frac{ f(x) x}{F(t)} dx = g(t, F(t), E(X)).$$ 
I tried using integration by parts 
$$
E(X|x < t)
 = \int_0^t\frac{ f(x) x}{F(t)} dx
 = \frac{F(t)\cdot t}{F(t)} - \int_0^t \frac{ F(x)}{F(t)} dx
$$
but I don't know what to do with $\int_0^t \frac{ F(x)}{F(t)} dx$.
I know $E(X) = \int_0^1 1 - F(x) dx$, but I'm not sure if/how I can use that.
I feel like I'm blanking on something simple here.
Edit: Corrected the integration by parts.
 A: EDIT: You can have a look at a family of my CDFs here.
This is unfortunately not possible. Consider for example $X$ uniformly distributed on $[0,1]$ and let $Y$ have the following CDF:
$$F_Y(x)=\frac{g\left(\frac14+\frac x2\right)}{g\left(\frac34\right)}+(1-x),$$
where $$g(x)=\ln(x)-\ln(1-x).$$
We have $\int_0^1 F_X = \frac12$ and
\begin{split}
\int_0^1 F_Y&=\frac12+\frac1{g\left(\frac34\right)}\int_0^1 g\left(\frac14+\frac x2\right)\,\mathrm dx
\\&=\frac12+\frac2{g\left(\frac34\right)}\int_{\frac14}^{\frac34} g(y)\,\mathrm dy
\\&=\frac12
\end{split}
since $g(x+\frac12)=-g(-x+\frac12)$.
Hence, $\mathbb EX=\mathbb EY$. Also, it is easy to check that $F_Y\left(\frac12\right)=\frac12=F_X\left(\frac12\right)$.
However, since $F_Y(x)>F_X(x)$ (see footnote) for $0<x<\frac12$, we have $\mathbb E(X\mid X<\frac12)>\mathbb E(Y\mid Y<\frac12)$ so there is no way to express the conditional expectation based solely on $t, \mathbb EX$ and $F(t)$.
Plot of $\color{red}{F_X}$ and $\color{green}{F_Y}$:

Footnote: We have
$$F_Y(x)-F_X(x)=1-2x+\frac{\ln\left(\frac{1+2x}{3-2x}\right)}{\ln(3)}$$
which is clearly positive on $[0,0.5]$.
A: This isn't enough information. 
Consider the uniform distribution, and the triangular distribution whose peak is at $1/2$, both on $[0,1]$ They both have mean equal to $1/2$, and have $F(t) = 1/2$ at $t=1/2$.
Thus, you cannot write $E(X\mid X<t)$ solely as a function of the numbers $E(X)$, $F(t)$ and $t$.
