Doing an old exam question for one of the courses I'm reading, I was faced with the following problem:
Let $\left\{B(t); t \geq 0\right\}$ be a Brownian motion, with variance parameter $\sigma^2=1$. Define the process $\left\{X(t); t \geq 0\right\}$ by: $$X(t)=\int_0^t B(u)du, \quad t\geq 0.$$ Calculate $P(X(2) > X(1)+1)$.
Obviously, $X(2)-X(1)=\int_1^2 B(u)du$. I do understand why $X(2)-X(1)\sim N(0,\tilde{\sigma})$, for some $\tilde{\sigma}$. What I fail to understand is why one could calculate $\tilde{\sigma}$ in this way:
$$\tilde{\sigma}^2= E((X(2)-X(1))^2)=E((\int_1^2 B(u)du)^2){\color{red}=}\int_1^2\int_1^2E(B(t)B(u))dudt=...$$
When calculating $\tilde{\sigma}$ myself, I simply used limits of sums, which worked, though it didn't really feel satisfactory.
What am I failing to see? How did they go from the left-hand side to the right-hand side of the red equality sign above? Could someone do the manipulations in a more step-by-step manner?