L1 and L2 norm of a signal

Question

I have a question concerning a signal norm. I have the following derivative, which fulfil following inequality: $$\dot{V}_i(t)\le-kS_i^2(t)-\bar{k}|S_i(t)|$$ We also know that $V_i(0)\ge 0$ is lower bounded and $k$, $\bar{k}$ are positive constants, hence $\dot{V}\le0$. Then it is said that $S_i\in L_1 \cap L_2$. I am not sure how to understand it, I would appriciate some help. $$\int_{0}^{T} -\dot{V}_i(t) dt \ge \int_{0}^{T} (kS_i^2(t)+\bar{k}|S_i(t)|)dt$$ $$V_i(0)-V_i(T) \ge \int_{0}^{T}(kS_i^2(t)+\bar{k}|S_i(t)|)dt$$ And finally, $$V_i(0) \ge V_i(0)-V_i(T) \ge \int_{0}^{T}(kS_i^2(t)+\bar{k}|S_i(t)|)dt$$ Here I stuck and I do not know how to explain to myself that indeed $S_i \in L_1 \cap L_2$. Does it come from the fact that $$\int_{0}^{T}(kS_i^2(t)+\bar{k}|S_i(t)|)dt = \int_{0}^{T}(kS_i^2(t))dt+\int_{0}^{T}(\bar{k}|S_i(t)|)dt$$?