Integral of an upsampled signal, without actually resampling it
If you upsample a causal discrete-time signal $x[n]$ by an integer factor $L$ and you use an interpolation filter with a causal impulse response $h[n]$, the upsampled and interpolated signal is given by

$$y[n]=\sum_{k=0}^{\infty}x[k]h[n-kL]\tag{1}$$

(Note that the requirement of causality can be easily removed; I just use it to have the lower summation indices start with the value $0$.)

If you want to compute the sum of $y[n]$, you would need to compute

$$s_y=\sum_{n=0}^{\infty}y[n]=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}x[k]h[n-kL]\tag{2}$$

However, you can interchange the sums in (2), resulting in

$$s_y=\sum_{k=0}^{\infty}x[k]\sum_{n=0}^{\infty}h[n-kL]\tag{3}$$

The value of the inner sum in (3) is independent of the index $k$, so we can write

$$s_y=s_h\cdot \sum_{k=0}^{\infty}x[k]=s_h\cdot s_x\tag{4}$$

with

$$s_h=\sum_{n=0}^{\infty}h[n]$$

So the sum of the upsampled and interpolated signal is simply the sum of the original signal times a factor which is given by the sum over the impulse response values of the interpolation filter. Note that for the ideal sinc filter, this factor equals the upsampling factor $L$. This is also at least approximately true for all practical interpolation filters, because the sum over the impulse response equals the DC gain of the interpolation filter, which should ideally equal $L$.

So if you want to keep things simple (and why wouldn't you?), you could use

$$s_y\approx L\cdot s_x$$

which is a very good approximation (or even exactly true) for any reasonable interpolation filter.

In the above derivations I've assumed that all the (infinite) sums exist, which in practice is always the case if we deal with finite length signals and with stable interpolation filters.