Here is another approach which is somewhat simpler than the one given in another answer here.
I establish that $$\int_{0}^{1}f(x)\{nx\}\,dx\to\frac{1}{2}\int_{0}^{1}f(x)\,dx$$ as $n\to\infty $. The integral on left of above equation can be split as sum of $n$ integrals $$\sum_{k=0}^{n-1}\int_{k/n}^{(k+1)/n}f(x)\{nx\}\,dx=\frac{1}{n}\sum_{k=0}^{n-1}\int_{k}^{k+1}f(t/n)\{t\}\,dt$$ Using mean value theorem for integrals the right hand side of the above equation can be written as $$\frac{1}{n}\sum_{k=0}^{n-1}f(t_k/n)\int_{k}^{k+1}\{t\}\,dt$$ where $t_k\in[k,k+1]$ and since $\{t\} $ is periodic with period $1$ the above reduces to $$\left(\int_{0}^{1}\{t\}\,dt\right)\cdot\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{t_k}{n}\right)$$ The integral above is $1/2$ as $\{t\} =t$ if $t\in[0,1)$ and the next factor is Riemann sum for $f$ or $[0,1]$. Thus the above tends to $$\frac{1}{2}\int_{0}^{1}f(x)\,dx$$ Above derivation assume that $f$ is continuous on $[0,1]$. Putting $f(x) =x^{2019}$ we get the desired limit as $1/4040$.
More generally we can use same method to prove that $$\lim_{n\to\infty} \int_{0}^{1}f(x)g(\{nx\})\,dx=\left(\int_{0}^{1}f(x)\,dx\right)\left(\int_{0}^{1}g(x)\,dx\right)$$ where $f$ is continuous on $[0,1]$ and $g$ is of constant sign and Riemann integrable on $[0,1]$.
Going further we can also note that if $g$ is periodic with period $T$ and of constant sign and Riemann integrable on $[0,T]$ and $f$ is continuous on $[0,T]$ then $$\lim_{n\to\infty} \int_{0}^{T}f(x)g(nx)\,dx=\frac{1}{T}\left(\int_{0}^{T}f(x)\,dx\right)\left(\int_{0}^{T}g(x)\,dx\right)$$
Based on suggestion in comments, one can prove that the above result holds for Riemann integrable $f, g$ and $g$ also being periodic with period $T$.
The idea is to express the integral on left as a sum $$\frac{1}{n}\sum_{k=0}^{n-1}\int_{kT}^{(k+1)T}f(x/n)g(x)\,dx$$ which can be further rewritten as $$\frac{1}{n}\sum_{k=0}^{n-1}\int_{0}^{T}f((x+kT)/n)g(x+kT)\,dx$$ And since $g$ is periodic it follows that the above can be written as $$\frac{1}{T}\int_{0}^{T}\left(\frac{T}{n}\sum_{k=0}^{n-1}f\left(\frac{x+kT}{n}\right)g(x)\right)\,dx\tag{1}$$ Since $f$ is Riemann integrable on $[0,T]$ with integral $I=\int_{0}^{T}f(x)\,dx$ we can see that if $$P_n=\{0,T/n,2T/n,\dots,(n-1)T/n,T\} $$ is a partition of $[0,T]$ and $U(f, P_n), L(f, P_n) $ be corresponding upper and lower Darboux sums then we have $$L(f, P_n) \leq S(f, P_n) \leq U(f, P_n)$$ where $S(f, P_n) $ is any Riemann sum for $f$ over $P_n$. Since the integral $I$ is also sandwiched between both upper and lower sums we have $$|S(f, P_n) - I|\leq U(f, P_n) - L(f, P_n) $$ We can now observe that the integrand in equation $(1)$ is of the form $S(f, P_n) g(x) $ and hence $$\left|\int_{0}^{T}S(f,P_n)g(x)\,dx-I\int_{0}^{T}g(x)\,dx\right|\leq (U(f, P_n) - L(f, P_n)) \int_{0}^{T}|g(x)|\,dx$$ and clearly the right hand side above tends to $0$ so that the left hand side also does the same. It follows that the desired limit is $$\frac{1}{T}\int_{0}^{T}f(x)\,dx\int_{0}^{T}g(x)\,dx$$ Credit for the idea of above proof must go to the user WE Tutorial School.
If the integral $\int_{0}^{T}g(x)\,dx=0$ then the above can be used as a proof of Riemann-Lebesgue Lemma for Riemann integrable functions and therefore the above is a generalization of it.
As this is a Riemann sum, you can convert it into an integral.
This becomes:
$$\int_0^1 \left \lfloor \frac2{\sqrt x} \right \rfloor -2\left \lfloor\frac1{\sqrt x} \right \rfloor\,dx$$
Put $\sqrt x \rightarrow 1/t$ to get:
$$ = 2\int_1^\infty \frac{\left \lfloor 2t \right \rfloor}{t^3} -2\frac{\left \lfloor t \right \rfloor}{t^3}\,dt$$
$$ = 2\left(\sum_{r=1}^\infty\int_{(r+1)/2}^{r/2 + 1}\frac{r+1}{t^3}\,dt - 2\sum_{r=1}^\infty\int_{r}^{r + 1}\frac{r}{t^3}\,dt\right)$$
$$ = 2\sum_{r=1}^\infty\left(\frac{2(2r+3)}{(1+r)(2+r)^2} - \frac{2r+1}{r(1+r)^2}\right)$$
$$ = 2\sum_{r=1}^\infty\left(\frac{4}{(r+1)(r+2)} - \frac{2}{(r+1)(r+2)^2} - \frac{2}{r(r+1)} + \frac{1}{r(1+r)^2}\right)$$
$$ = 2\sum_{r=1}^\infty\left(\frac{1}{r(1+r)^2}-\frac{2}{(r+1)(r+2)^2}\right)$$
$$ = 1 - 2\sum_{r=1}^\infty\left(\frac{1}{r(1+r)^2}\right)$$
$$= \boxed{\frac{\pi^2}3 - 3}$$
Best Answer
You can do the integral explicitly by breaking the interval into subintervals: $$[0,\pi] = [0,1) \cup [1,2) \cup [2,3) \cup [3, \pi].$$ Note that for $0 < \epsilon \le 1$ and $k \in \mathbb N$, if $x \in [k,k+\epsilon)$, then $-x \in (-k-\epsilon, -k]$ and so $\{-x\} = -x + (k+1).$ Thus \begin{align*} \int^{k+\epsilon}_k \{-x\}^n dx &= \int^{k+\epsilon}_k (-x+k+1)^ndx \\ &= -\int^{1-\epsilon}_1 y^n dy = \frac{1- (1-\epsilon)^{n+1}}{n+1}.\end{align*} Then \begin{align*}\int^\pi_0 \{-x\}^ndx &= \int^1_0 (1-x)^n dx + \int^2_1 (2-x)^ndx +\int^3_2 (3-x)^ndx + \int^\pi_3 (4-x)^ndx\\&= \frac{1}{n+1} + \frac 1 {n+1} + \frac 1 {n+1} + \frac{1-(4-\pi)^{n+1}}{n+1}.\end{align*} The last term exhibits exponential decay, so in the limit we see $$n\int^\pi_0 \{-x\}^ndx = \frac{4n}{n+1} - \frac{n}{n+1}(4-\pi)^{n+1} \to 4.$$