# Determining if a system involving integrals is time varying or time-invariant

I've been studying for a signals and systems class coming this fall and am trying to understand the following problem.  Show whether the following system is time invariant or time varying: $$S_1:y = T_1[x],\;\; y(t) = \int_0^t{[x(\tau)]^3\,d\tau},\;\;t\geq 0$$ solution: $$z(x)=t[x(t-\sigma)] = \int_0^t{[x(\tau-\sigma)]^3\,d\tau}$$ $$= \int_{-\sigma}^{t-\sigma}{[x(\tau')]^3\,d\tau'}$$ Where the change of variable $\tau-\sigma \rightarrow \tau'$ is applied. and $$y(t-\sigma) = \int_0^{t-\sigma}{[x(\tau)]^3\,d\tau}$$ So I know that the answer is that $z(x) \neq y(t-\sigma)$ and thus the system is time varying. So my question is why doesn't 0 get subtracted by $\sigma$ in integral expressed in $y(t-\sigma)$?

Furthermore if I switch the limits of integration like so: $$S_2:y = T_2[x],\;\; y(t) = \int_{t}^{t+1}{[x(\tau)]^3\,d\tau},\;\; t\geq 0$$ Would $S_2$ be time invariant?

The fixed bound at 0 in $\int_0^t$ indeed suggests a time-variance.You only need a counter-example for $S_1$, like the constant $x(t) = 1$. $y(t+1) = t+1$, while $S_1(x(t+1)) = S_1(x(t)) = y(t)$.
For $S_2$, the two bounds moving at the same time are promising, and a standard variable change shows the time invariance.