Do Nyquist samples have the same sum as the integral of the signal? Assume I sample a signal according to the Nyquist criterion. Then I perform a simple summation / integral over a linearly interpolated signal. Is this equivalent to the integral over the continuous signal, i.e.?

$$\sum_{n=-\infty}^{\infty}x(nT) T = \int_{-\infty}^{\infty}x(t)\text{d}t$$

I hope my representation is correct for what I mean. My question relates to whether a reconstruction, e.g. Whittaker–Shannon interpolation, is necessary   in practical application. I would also appreciate any literature pointers for discussions on how this might affect practical cases with finite time and imperfect filters.
The equation in your question is correct. If we assume that $x(t)$ is an ideally band-limited signal with bandwidth $B$, then, according to the sampling theorem, we have

$$x(t)=\sum_{n=-\infty}^{\infty}x(nT)\frac{\sin\left[\pi (t-nT)/T\right]}{\pi (t-nT)/T}\tag{1}$$

where the sampling rate $1/T$ satisfies $1/T&gt;2B$.

If we assume that the integral over $x(t)$ exists, using $(1)$ we obtain

$$\begin{align}\int_{-\infty}^{\infty}x(t)dt&amp;=\int_{-\infty}^{\infty}\sum_{n=-\infty}^{\infty}x(nT)\frac{\sin\left[\pi (t-nT)/T\right]}{\pi (t-nT)/T}dt\\&amp;=\sum_{n=-\infty}^{\infty}x(nT)\int_{-\infty}^{\infty}\frac{\sin\left[\pi (t-nT)/T\right]}{\pi (t-nT)/T}dt\\&amp;=\int_{-\infty}^{\infty}\frac{\sin(\pi t/T)}{\pi t/T}dt\cdot \sum_{n=-\infty}^{\infty}x(nT)\\&amp;=T\sum_{n=-\infty}^{\infty}x(nT)\tag{2}\end{align}$$

where the last equality follows from the fact that the integral in the one but last line just equals the DC value of an ideal low pass filter with gain $T$.