how to prove $\int_{0}^{a}B(t)dt\sim N(0,\frac{a^3}{3})$

Question

Let $B(t)$ is Brownian Motion. I want to prove the integral $\int_{0}^{a}B(t)dt$ has normal distribution , $N(0,\frac{a^3}{3})$.
means $\int_{0}^{a}B(t)dt\sim N(0,\frac{a^3}{3})$

Answer 1

This is a normal random variable as a barycenter of normal random variables. Its mean is $\int\limits_0^aE[B_t]\mathrm dt=$ $___$. Its variance is $\int\limits_0^a\int\limits_0^aE[B_tB_s]\mathrm dt\mathrm ds=\int\limits_0^a\int\limits_0^a\min(t,s)\mathrm dt\mathrm ds=$ $___$.

Answer 2

Ito's formula gives $aB(a) = \int_{0}^{a}B(t)dt + \int_{0}^{a}tdB(t)$

so $\int_{0}^{a}B(t)dt =aB(a) - \int_{0}^{a}tdB(t) = \int_{0}^{a}(a-t) dB(t)$.

and for a deterministic function $f$, $\int_{0}^{a}f(t)dB(t) \sim \mathcal{N}(0, \int_{0}^{a}f^2(t)dt)$