The value of $$\frac{16S_{1}}{\pi}$$ is _______.

The given integral for $$S_1$$ is

$$S_1=\int_{\frac{\pi}{8}}^{\frac{3\pi}{8}}\sin^{2}x\,dx.$$

First, rewrite $$\sin^{2}x$$ using the double-angle identity:

$$\sin^{2}x=\frac{1-\cos2x}{2}.$$

Hence

$$S_1=\int_{\frac{\pi}{8}}^{\frac{3\pi}{8}}\frac{1-\cos2x}{2}\,dx =\frac12\int_{\frac{\pi}{8}}^{\frac{3\pi}{8}}1\,dx -\frac12\int_{\frac{\pi}{8}}^{\frac{3\pi}{8}}\cos2x\,dx.$$

Integrate term by term:

$$\int1\,dx=x,$$
$$\int\cos2x\,dx=\frac{\sin2x}{2}.$$

Therefore

$$S_1=\frac12\Bigl[x\Bigr]_{\frac{\pi}{8}}^{\frac{3\pi}{8}} -\frac12\Bigl[\frac{\sin2x}{2}\Bigr]_{\frac{\pi}{8}}^{\frac{3\pi}{8}} =\frac12\left(\frac{3\pi}{8}-\frac{\pi}{8}\right) -\frac14\left(\sin\frac{3\pi}{4}-\sin\frac{\pi}{4}\right).$$

Compute the sine values:

$$\sin\frac{3\pi}{4}=\frac{\sqrt2}{2},\quad \sin\frac{\pi}{4}=\frac{\sqrt2}{2}.$$

Thus the second bracket is zero, and we get

$$S_1=\frac12\left(\frac{2\pi}{8}\right)=\frac{\pi}{8}.$$

Now evaluate the required expression:

$$\frac{16S_1}{\pi}=\frac{16\left(\frac{\pi}{8}\right)}{\pi}=\frac{16}{8}=2.$$

Hence the value lies in the interval 1.99 - 2.01.