Proof of differentialbility in mean square calculus?

Question

let $x_t$ be a mean squared Riemann integrable over $[a, t]$ for every $t\in[a,b]$. Then $y_t=\int\limits_a^t x_\tau d\tau\ $ is mean squared continuous on $[a, b]$. Furthermore, if $x_t $ is mean squared continuous on [a, b],then $y_t$ is mean squared differentiable on $(a, b) $ and $\dot{y_t}= x_t$..