Alternating summation and subtraction of square roots

For an approximation, we might start (using a not generally adopted, but not unfamiliar notation) with$$H^{(-1/2)}_n=1 + \sqrt{2} + \ldots + \sqrt{n}=\frac{2}{3}n^{3/2} + \frac{\sqrt{n}}{2} + \zeta\left(-\frac12\right)+o(1),$$ mentioned in this question ($\zeta$ is the famous Riemann zeta function).
Then, $$\sqrt{2n}-\sqrt{2n-1}\pm\ldots=-H^{(-1/2)}_{2n}+2\sqrt{2}\,H^{(-1/2)}_n=\frac12\sqrt{2n}+(2\sqrt{2}-1)\,\zeta\left(-\frac12\right)+o(1).$$ For $n=50$, the main term would be $\displaystyle5+(2\sqrt{2}-1)\,\zeta\left(-\frac12\right)=4.619895\ldots$, while the original sum is $4.6323951\ldots$, as @Henry pointed out, already, $\displaystyle(2\sqrt{2}-1)\,\zeta\left(-\frac12\right)=-0.380105\ldots$ being slightly more accurate. The error (of the order $O(1/n)$) is no surprise.

We want to find the integer part of $\sum_{x=1}^{50} f(x)$ where $f(x) = \sqrt{2x}-\sqrt{2x-1}$.

For decreasing $f(x)$, $$\int_a^{b+1} f(x) \; dx < \sum_{x=a}^b f(x) < \int_{a-1}^b f(x) \; dx$$ (It may help to make a sketch to see this.)

It's easier to work with the sum from $x=2$ to $50$ instead of $x=1$ to 50 because $f(x)$ is not defined when $x=0$. We can adjust for the missing $x=1$ term later.

By elementary calculus, $$\int f(x) \; dx = \frac{1}{3} (2x)^{3/2} - \frac{1}{3} (2x-1)^{3/2} + C$$ so $$\int_2^{51} f(x) \; dx < \sum_{x=2}^{50} f(x) < \int_{1}^{50} f(x) \; dx$$ yields $$4.10 < \sum_{x=2}^{50} f(x) < 4.38$$ Now to adjust for the missing $x=1$ term. Since $$\sum_{x=2}^{50} f(x) + \sqrt{2} - \sqrt{1} = \sum_{x=1}^{50} f(x)$$ we have $$4.10 + \sqrt{2} - \sqrt{1} < \sum_{x=1}^{50} f(x) < 4.38 + \sqrt{2} - \sqrt{1}$$ which yields $$4.51 < \sum_{x=1}^{50} f(x) < 4.79$$ so the integer part of the sum is $\boxed{4}$.