I learnt that if $u\in L^{\infty}$ is a weak solution of the IVP
$u_t +f(u)_x=0$ with $u(x,0)=g(x)$ $\forall x \in \mathbb{R}$
Then $\int_{\mathbb{R}}u(x,t)dx=\int_{\mathbb{R}}g(x)dx$ $\forall t>0$.
I have the following doubts
1) Can we say something about $\int\limits_a^b u(x,t)dx$ for given $a<b$
(Note: its clear $\int\limits_a^b u(x,t)dx \neq \int\limits_a^b g(x)dx$ but atleast for Riemann problem it looks like $\int\limits_a^b u(x,t)dx = \int\limits_a^b g(x)dx+t (f(u_l)-f(u_r))$ is this true for general $L^{\infty}$ data.
2) Is the function $t\rightarrow \int\limits_a^b u(x,t)dx $ differentiable
(Note :For Riemann data the above function is differeniable)
If so how to prove? Please suggest