I'm trying to find

$$E \left[\int_3^4 B_t \, dt \mid \int_1^2 B_t \, dt= c\right]$$

where $B_t$ is standard Brownian motion and integral is just a Riemann integral and constant $c$ is a known real number and bounderies of integrals represent times.

If we write integrals in a Riemann sum approximation, then I think the above expectation can be found by solving a similar expectation: $ E [\sum_1^N B_{t_j} \Delta t_j \mid \sum_1^M B_{t_k} \Delta t_k=c]. $

So for a simple example to find $E[a_4 B_4 + a_3 B_3 \mid a_2 B_2+ a_1 B_1 =c]$ for some real coefficients $a_1, a_2, a_3, a_4$, we can write \begin{align*} E[a_4 B_4 + a_3 B_3 \mid a_2 B_2+ a_1 B_1 =c]&= \\E[ a_4 (B_4-B_3) + (a_3 +a_4)B_3 \mid a_2 (B_2-B_1) + (a_2+a_1)B_1=c] & = \\a_4 E[B_4 -B_3 \mid a_2 (B_2 -B_1) + (a_2+a_1) B_1=c]+ (a_3+a_4) E[B_3 \mid a_2 (B_2-B_1) + (a_2+a_1)B_1=c] \end{align*}

Since $B_4- B_3$ is independent of $B_2-B_1$ and $B_1$ so is independent of linear combination of both and in particular $a_2 (B_2-B_1) + (a_2+a_1)B_1$. So we have \begin{align*} E[a_4 B_4 + a_3 B_3 \mid a_2 B_2+ a_1 B_1 =c] &= \\a_4 E[B_4-B_3] + (a_3+a_4) E[B_3 \mid a_2 (B_2-B_1) + (a_2+a_1)B_1=c]=(a_3+a_4) E[B_3 \mid a_2 (B_2-B_1) + (a_2+a_1)B_1=c] =(a_3+a_4) E[B_3-B_2 +B_2 \mid a_2 (B_2-B_1) + (a_2+a_1)B_1=c]=(a_3+a_4) E[B_2 |a_2 (B_2-B_1) + (a_2+a_1)B_1=c] \end{align*} So according to these calculations I think we have

$$E \left[\int_3^4 B_t \,dt \mid \int_1^2 B_t \, dt= c\right]=E\left[B_2\mid \int_1^2 B_t \, dt= c \right]$$

Is it correct? if so, how can I finish this?

Thank you in advance!