Determining time variance of integrator Say we have an integrator $y(t)= \int_{- \infty}^{t} x(t) dt$

It is incorrect to use the following method to determine if it is time variant/invariant?

First, when we have input signal $x(t),$ 

$y_1(t)= \int_{- \infty}^{t} x(t) dt$

Then, we shift $y(t)$ by $t_0$

$y_1(t-t_0)=\int_{- \infty}^{t-t_0} x(\tau-t_0) d\tau$

If we have input signal $x(t-t_0),$ we get

$y_2(t)=\int_{- \infty}^{t} x(\tau-t_0) d\tau$

And since $y_1(t-t_0)\neq y_2(t)$

Then the system must be time variant.
This is not correct. First, you have to write it like that:

$$
\begin{align}
y(t) = \int_{-\infty}^t x(\tau)d\tau
\end{align}
$$

I.e. your variable in the integrand is different from the integral limits. Accordingly, your calculation will be different, as you will only replace $t$ with $t-t_0$ and not $\tau$.