Ouput SNR of an integrator
I hope you do not mind, but I am going to change terminology a little bit. Since you mentioned simulations, suppose you are sampling the Gaussian white noise, at the input to the integrator, at a constant rate of $f_s$ samples per second. The point spacing, $Δt$, is $1/f_s$ seconds. Suppose $N$ independent consecutive samples are collected. Then the $N$ samples span a period of $τ_a$ seconds and this equals $N Δt$. The sample mean of the $N$ noisy inputs to the integrator obviously provides an estimate of the known $V_{in}$ input voltage. The sample variance of the $N$ noisy inputs to the integrator provides an estimate of the known $σ^2$ variance of the Gaussian white noise. The power SNR at the input is $(V_{in}/σ)^2$. The empirical estimate would be the square of the sample mean divided by the sample variance and bigger $N$ values are almost surely better.

The Gaussian white noise has zero mean and $σ^2$ variance, so the bilateral noise power spectral density, $η$, is $σ^2 Δt$. The ideal integrator has $τ_i$ integration time constant. (For an ideal op amp integrator, with just an input resistor, $R$, and feedback capacitor, $C$, $τ_i = RC$. The sign inversion is trivial and can be ignored.) Now suppose that the integration period, i.e., integration aperture or gate duration, is $τ_a$ seconds, exactly as above. Assume the integrator has initial condition zero, integration starts at $t = 0$ and ends at $t = τ_a$, so $τ_a$ is simply the reciprocal of the OP’s $F_s$. N.B. My $f_s$ is not the OP’s $F_s$.

Then the DC power gain of the integrator is $(τ_a/τ_i)^2$ and the noise equivalent bandwidth is $1/2τ_a$. Now consider two cases.

Case 1: Let $τ_i = τ_a$.  Then the signal power at the output of the integrator, at $t = τ_a$, is simply $(V_{in})^2$, because the DC power gain is unity. The noise power is $2η$ times unity DC power gain times $1/2τ_a$, which equals $η/τ_a$. But $η = σ^2 Δt$ and $τ_a = N Δt$. So the output power SNR is $(V_{in})^2 N/σ^2$. This is $N$ times larger than the input power SNR.

Case 2: Let $τ_i ≠ τ_a$.  Then the signal power at the output of the integrator, at $t = τ_a$, is $(V_{in} τ_a/τ_i)^2$, because the DC power gain is $(τ_a/τ_i)^2$. The noise power is $2η$ times $(τ_a/τ_i)^2$ times $1/2τ_a$, which equals $η τ_a /(τ_i)^2$. But $η = σ^2 Δt$ and $τ_a = N Δt$. So the output power SNR is $(V_{in} τ_a/τ_i)^2/[σ^2 Δt τ_a /(τ_i)^2]$, which again equals $(V_{in})^2 N/σ^2$. This is again $N$ times larger than the input power SNR.

In Case 1, the signal power is independent of $N$, while the noise power is inversely proportional to $N$, via $τ_a = N Δt$. So the power SNR is proportional to $N$. This is simply averaging independently sampled white noise. In Case 2, signal power is proportional to $N^2$, while noise power is proportional to $N$. So the power SNR is again proportional to $N$.

Relationship to matched filtering

The figure below shows that, if nothing was known about the input waveform for $t &gt; τ_a$, then the input could be treated as though it was a rectangular pulse starting at $t = 0$ and ending at $t = τ_a$. The integration would effectively be gated integration, which would be a simple case of matched filtering: the impulse response of the (aperture synchronized) gated integrator is a rectangular pulse of duration $τ_a$ and reversing it changes nothing because the gated integrator's impulse response is symmetric. The figure below shows the calculation of the matched filter power SNR. The integration time constant, $τ_i$, has been set equal to unity because it affects both the pulse energy, and the noise, the same way, so it divides out.



I hope this helps!