The value of $$\sum_{r=1}^{20}\left(\left|\sqrt{\pi\left(\int_{0}^{r}x|\sin \pi x\right)}\right|\right)$$ is ______.

Let $$S=\sum_{r=1}^{20}\Bigl|\sqrt{\pi\,\bigl(\int_{0}^{r}x\,|\sin \pi x|\,dx\bigr)}\Bigr|$$.

Because $$r$$ is a positive integer, the outer absolute value and the square root will always give a non-negative result, so we first evaluate the integral
$$I(r)=\int_{0}^{r}x\,|\sin \pi x|\,dx.$$

The function $$|\sin \pi x|$$ has period $$1$$. Split the interval $$[0,r]$$ into $$r$$ sub-intervals $$[n,n+1]$$ where $$n=0,1,\dots ,r-1$$:

$$I(r)=\sum_{n=0}^{r-1}\int_{n}^{\,n+1}x\,|\sin \pi x|\,dx.$$

Inside each sub-interval set $$x=n+t$$ with $$0\le t\le 1$$. Then $$dx=dt$$ and

$$\int_{n}^{\,n+1}x\,|\sin \pi x|\,dx =\int_{0}^{1}(n+t)\,|\sin \pi(n+t)|\,dt.$$

Since $$\sin \pi(n+t)=(-1)^n\sin \pi t$$, its absolute value becomes $$|\sin \pi t|$$, independent of $$n$$. Hence

$$\int_{n}^{\,n+1}x\,|\sin \pi x|\,dx =n\!\!\int_{0}^{1}|\sin \pi t|\,dt +\!\int_{0}^{1}t\,|\sin \pi t|\,dt.$$

Define
$$A=\int_{0}^{1}|\sin \pi t|\,dt, \qquad B=\int_{0}^{1}t\,|\sin \pi t|\,dt.$$

Because $$\sin \pi t \ge 0$$ on $$[0,1]$$, the absolute signs can be dropped:

$$A=\int_{0}^{1}\sin \pi t\,dt =\left[-\frac{\cos \pi t}{\pi}\right]_{0}^{1} =\frac{2}{\pi}.$$

For $$B$$, integrate by parts:

$$B=\int_{0}^{1}t\sin \pi t\,dt =\left[-\frac{t\cos \pi t}{\pi}\right]_{0}^{1} +\frac{1}{\pi}\!\int_{0}^{1}\cos \pi t\,dt =\frac{1}{\pi}.$$

Therefore
$$\int_{n}^{\,n+1}x\,|\sin \pi x|\,dx =nA+B =n\Bigl(\frac{2}{\pi}\Bigr)+\frac{1}{\pi}.$$

Summing over $$n$$ from $$0$$ to $$r-1$$:

$$I(r)=\sum_{n=0}^{r-1}\Bigl(n\frac{2}{\pi}+\frac{1}{\pi}\Bigr) =\frac{2}{\pi}\sum_{n=0}^{r-1}n +\frac{1}{\pi}\sum_{n=0}^{r-1}1.$$

Using $$\sum_{n=0}^{r-1}n=\frac{r(r-1)}{2}$$ and $$\sum_{n=0}^{r-1}1=r$$, we get

$$I(r)=\frac{2}{\pi}\cdot\frac{r(r-1)}{2} +\frac{1}{\pi}\,r =\frac{r(r-1)+r}{\pi} =\frac{r^{2}}{\pi}.$$

Hence
$$\pi\,I(r)=\pi\cdot\frac{r^{2}}{\pi}=r^{2},\qquad \sqrt{\pi\,I(r)}=\sqrt{r^{2}}=r.$$

The summation becomes
$$S=\sum_{r=1}^{20}r =\frac{20\cdot21}{2}=210.$$

Final answer: $$210$$.