Standard Brownian Conditional expectation (Given the process $(B(t))t≥0$ of Brownian motion, define the random variables
$$Y=\int_0^{1}B(s)\,ds $$
$$X=B(1) $$
Determine the quantities $E(Y|X)$, $Var(Y−E(Y|X))$ and the conditioned density 
$f_{Y|X}$
I already compute mean and variance using Fubini's theorem $E(Y) = \int_0^{1}E(B(s))\,ds =0$ and $Var(Y)= E(Y^2)=\dfrac{1}{3}$
I don't know if it's help me, I have this formula aswell   $E(Y|X) = E(Y)+Cov(Y,X) + Cov(X,X)^{-1} + (X-E(X))$ 
Thanks for your help,
As I can't edit my comment anymore, 
Thanks for the hint but I don't catch why it helps me. My problem is to deal with the quantities $\int_0^{1}E(B(s))\,ds$|X).  I tried to use the formula :   $Cov(Y,X) = E[(Y(t)-E(X)(X-E(X))=E[Y(t)B(1)]$                                                                 for $t<1$ $B(1)= Y(t)+(B1-Y(t))$ then we have $B(1)Y(t) = Y(t)^2+Y(t)(B(1)-Y(t))$  then $E[B(1)Y(t)]= E[Y^2] + E[Y(t)]E[B(1)-Y(t)]$ as increments are independent we find that $E[B(1)Y(t)]= \dfrac{1}{3}$             Then if I do the same for $t>1$ I find $E[B(1)Y(t)] = 1$.
 A: 2
$$Y=\int_0^{1}B(s)\,ds $$
$$X=B(1) $$
We use fubini theorem for continuous process $X(t)$, $t\in [0,T]$ ,under appropriate assumptions it holds that :
$E(\int_0^{T}X(t)\,dt) = \int_0^{T}E(X(t))\,dt$ and $E(\int_0^{T}X(t)\,dt)^2 = \int_0^{T} \int_0^{T}E(X(t)X(s))\,dtds=\int_0^{T} (\int_0^{T}E(X(t)X(s))\,dt)ds$
Thus we have $E(Y)= \int_0^{1}E(B(t))\,dt=0$ and we find $Var(Y) = E(Y^2)= E[\int_0^{1}B(t)\,dt\int_0^{1}B(s)\,ds]=\int_0^{1} \int_0^{1}E(B(t)B(s))\,dtds=$$\int_0^{1} \int_0^{1}min(t,s)\,dtds=\int_0^{1} (\int_0^{t}s\,ds+\int_t^{1}t\,ds)\,dt=\dfrac{1}{3}$
A stochastic process $X(t)\in D$ is called a Gaussian process if for each choice {t1,...,tn} $c$ D,$n \in N$ of indices , the vector $ (X(t1),...,X(tn)) $ follow a multivariate normal distribution.
By definition, a brownian motion is a gaussian process.
Theorem: Let random vectors $Y=(Y1,...,Ym) $ and $X=(X1,...,Xn)^{T}$ have a joint Gaussian distribution Then :
1) if $Cov(Y,X)=0$ then $X,Y$ are independent ( $Hint$ : show that characteristic function factorizes)
2) $G:= Y-E(Y)-Cov(Y,X)Cov(X,X)^{-1} (X-E(X))$ and X are independent.
$Cov(G,X)=E[G(X-E(X))^{T}]=Cov(Y,X)-Cov(Y,X)=0$ as G and X are normally distributed their uncorrelations implies independence.
3) Because of independence of G and X we have $E(G|X)=E(G)=0$
 by properties of conditionnal expectation we have : 
$0=E(G|X)=E(Y|X)-E(Y)-Cov(Y,X)Cov(X,X)^{-1}(X-E[X])$
$E(Y|X)=E(Y)+Cov(Y,X)Cov(X,X)^{-1}(X-E[X])$
4) $E[(Y-E(Y|X))^2]=Cov(Y,Y)-Cov(Y,X)[Cov(X,X)]^{-1}Cov(Y,X))$
So for our problem : 
$E(Y|X)=E(Y)+Cov(Y,X)Cov(X,X)^{-1}(X-E[X])$
$=0+Cov(Y,X)*1*X$
$Cov(Y,X)=E[(Y-E(Y))(X-E(X))]=E[YX]=E[B(1)\int_0^{1}B(s)]\,ds=E[\int_0^{1}B(1)B(s)\,ds]$
$1<s$  $B(1)=B(s)+(B(1)-B(s))$
$B(1)B(s)= B(s)^2+ B(s)(B(1)-B(s))$
$E[\int_0^{1}B(1)B(s)]=E[\int_0^{1}B(s)^2+ B(s)(B(1)-B(s))\,ds]=E[\int_0^{1}B(s)^2\,ds+\int_0^{1} B(s)(B(1)-B(s))\,ds]
=\int_0^{1}E[B(s)^2]\,ds+\int_0^{1} E[B(s)(B(1)-B(s))]\,ds]$
With the independence of the increments $E[B(s)(B(1)-B(s))]=E[B(s)]E[(B(1)-B(s))]=0$
Thus we have $Cov(Y,X)=E[XY]=E[\int_0^{1}B(1)B(s)]= \int_0^{1}s\,ds=\dfrac{1}{2}$ 
and then $ E(Y|X) = \dfrac{1}{2}X$
from $4)$ we have $E[(Y-E(Y|X))^2]=\dfrac{1}{3}-\dfrac{1}{2}X[1*\dfrac{1}{2}X]$
$=\dfrac{1}{3}-\dfrac{1}{4}X^2$
Finally, $Var[Y-E(Y|X)]$
$=E[(Y-E(Y|X))^2]-E[Y-E(Y|X)]^2$
$= \dfrac{1}{3}-\dfrac{1}{4}X^2-\dfrac{1}{4}X^2$
$=\dfrac{1}{3}-\dfrac{1}{2}X^2$
I think I did it correctly but I am not sure for the part with the independence of the increments as if I found the same than mookid ( merci d'ailleurs).  feel free to tell me if there is an error , I am going to compute density now.
A: Hint: you can write
$$
B(s) = sX + T(s)
$$
where $T$ is a Brownian bridge.

you get
$$
E\left[Y|X\right] = E\left[\int_0^1 sXds\right]  +
E\left[\int_0^1 T(s)ds\right] = \frac X2
$$because $T$ has a symetric law.
and then
$$
var(Y - E[Y|X]) = var\left(\int_0^1 T(s) ds\right)
$$
