Time-averaging as a low-pass filter?
first of all, if i were the OP, i would fix the notation a bit.  $x(t)$, $y(t)$ are continuous-time signals and $t\in\mathbb{R}$ is a continuous argument.  $x[n] \triangleq x(nT)$ and $y[n] \triangleq y(nT)$ are discrete=time signals and and $n\in\mathbb{Z}$ is a discrete argument, an integer.  square brackets mean that an integer goes into them.  the normal convention is that $T=\frac{1}{f_s}$ is your sampling period.

a simple moving-average filter with equal-weighting in the average is

$$ \begin{align} 
   y[n] &amp; = \frac{1}{N} \sum\limits_{i=0}^{N-1} x[n-i] \\
        &amp; = \frac{1}{N} \left(x[0] + \sum\limits_{i=1}^{N} x[n-i] - x[n-N]\right) \\
        &amp; =  \frac{1}{N} \sum\limits_{i=1}^{N} x[n-i] + \frac{1}{N} \left(x[0]  - x[n-N]\right) \\
        &amp; =  \frac{1}{N} \sum\limits_{i=0}^{N-1} x[n-1-i] + \frac{1}{N} \left(x[0]  - x[n-N]\right) \\
        &amp; =  y[n-1] + \frac{1}{N} \left(x[0]  - x[n-N]\right) \\
\end{align} $$

that is representing both the non-recursive implementation and the recursive.  the recursive implementation works as long as you subtract exactly what you earlier added, something that might not happen in a floating-point context.  both are FIR, even though the recursive implementation is something we would call a Truncated IIR,  still an FIR.

to get the frequency response, apply the Z-Transform to either form, and solve for $Y(z)$ in terms of $X(z)$.  then compute the transfer function:

$$ H(z) \triangleq \frac{Y(z)}{X(z)} $$

then substitute $z = e^{j \omega}$  and solve for the $\omega$ such that $\left|H(e^{j \omega})\right|^2 = \frac{1}{2}$.  that $\omega= 2\pi f T$ is your -3.01 dB cut-off frequency.

try doing that math yourself.  we'll be here to help guide you.