811. Given the mathematical expectation $m_{x}(t)=3 t^{2}+1$ of the random function $X(t)$, find the mathematical expectation of the integral $Y(t)=\int_{0} X(s) \mathrm{d} s$.
Solution. The desired expected value

$$
m_{u}(t)=\int_{0}^{t} m_{x}(s) d s=\int_{0}^{t}\left(3 s^{2}+1\right) \cdot d s=t^{2}+t
$$