If we consider the following wave equation for $a\in L^{\infty}((0,1)\times (0,T))$ and $(x,t) \in (0,1)\times (0,T)$: $$\begin{cases} f''-f_{xx}+a(t,x)f=0 \\ f(0,t)=f(1,t)=0 \\ f(x,0)=f^0(x) \\ f'(x,0)=f^1(x) \end{cases} $$ As we know that the wave equation is reversible in time, if $f(x,t)$ then also $f(x,-t)$ is also a solution.
And my question if we define the energy as the following: $$E(t)=\frac{1}{2}[||f(t)||^2_{L^2(0,1)}+||f'(t)||^2_{H^{-1}(0,1)}]$$ But why ;due to time reversibility; do we have:
$$\int_{t_1}^{t_2}E(t)dt=\int_{t_1}^{t_2}E(0)dt$$
I don't think that your question (as is stated) is correct. What I would do is to define the energy as: $$ E(t):=\dfrac{1}{2}\int_0^1\big(f_t^2+f_x^2+af^2\big)(t,x)dx. $$ Thus, formally, taking the time derivative of the energy functional, using the equation and then integrating in time we obtain $$ E(t_2)-E(t_1)=\int_{t_1}^{t_2}\int_0^1 a_t(t,x)f^2(t,x)dxdt. $$ I think that the time reversibility of the equation plays no role in your question (unless I haven't understood it correctly). Hope my answer helps you.