Evaluation of Scaled Definite Integral for Constant Weight Function

To find the value of $\frac{16S_1}{\pi}$, we first evaluate the definite integral $S_1$.

The function $g_1(x)$ is given as $g_1(x) = 1$ and $f(x) = \sin^2 x$ for all $x \in \left[\frac{\pi}{8}, \frac{3\pi}{8}\right]$.

Thus, $S_1$ is defined as: $S_1 = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x) \cdot g_1(x) \, dx = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sin^2 x \, dx$

Using the property of definite integrals $\int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx$, where $a = \frac{\pi}{8}$ and $b = \frac{3\pi}{8}$, we have: $a + b = \frac{\pi}{8} + \frac{3\pi}{8} = \frac{\pi}{2}$

Substituting $x \to \frac{\pi}{2} - x$ into the integral: $S_1 = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sin^2\left(\frac{\pi}{2} - x\right) \, dx = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \cos^2 x \, dx$

Adding the two expressions for $S_1$: $2S_1 = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sin^2 x \, dx + \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \cos^2 x \, dx$ $2S_1 = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} (\sin^2 x + \cos^2 x) \, dx$ $2S_1 = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} 1 \, dx$

Evaluating the simple integral: $2S_1 = \left[ x \right]_{\frac{\pi}{8}}^{\frac{3\pi}{8}} = \frac{3\pi}{8} - \frac{\pi}{8} = \frac{\pi}{4}$

Dividing by 2 gives: $S_1 = \frac{\pi}{8}$

Now, we substitute $S_1$ into the required expression: $\frac{16S_1}{\pi} = \frac{16}{\pi} \cdot \left(\frac{\pi}{8}\right) = 2$

Thus, the value of $\frac{16S_1}{\pi}$ is $2$.