Apply the <Feynman-Kac formula> with $V(x)=\sigma x$ and terminal function $1$. The ansatz $u(t,x)=e^{A(t)x+C(t)}$ gives
$$
A'=-\sigma,\qquad C'=\frac12A^2,qquad A(0)=C(0)=0.
$$
Thus $A(t)=-\sigma t$ and $C(t)=\sigma^2t^3/6$, so
$$
\boxed{\mathbb E_x\exp\left(-\sigma\int_0^tB_sds\right)
=\exp\left(-\sigma tx+\frac{\sigma^2t^3}{6}\right).}
$$
Equivalently, the <Integral of Brownian motion> is Gaussian with mean $xt$ and variance $t^3/3$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 202 / 3 / 3 / 4 / Solution

{c}
{title2=$\int_0^tB_s\,ds$}

For Brownian motion started at $x$, the time integral is a <Gaussian random variable> with
$$
\mathbb E_x\int_0^tB_sds=xt,
\qquad
\operatorname{Var}\left(\int_0^tB_sds\right)=\frac{t^3}{3}.
$$
The variance follows by integrating the covariance kernel $\operatorname{cov}(B_r,B_s)=\min(r,s)$ twice.

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 Incoming links: Integral of Brownian motion