Let g_i: left[frac{pi}{8}, frac{3pi}{8}right] | vedclass

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(C) For $S_1$,we have $S_1 = \int_{\pi/8}^{3\pi/8} \sin^2 x dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we get $S_1 = \int_{\pi/

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$2, 1.50$

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(C) For $S_1$,we have $S_1 = \int_{\pi/8}^{3\pi/8} \sin^2 x dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we get $S_1 = \int_{\pi/8}^{3\pi/8} \sin^2(\frac{\pi}{8} + \frac{3\pi}{8} - x) dx = \int_{\pi/8}^{3\pi/8} \cos^2 x dx$.Adding these,$2S_1 = \int_{\pi/8}^{3\pi/8} (\sin^2 x + \cos^2 x) dx = \int_{\pi/8}^{3\pi/8} 1 dx = \frac{3\pi}{8} - \frac{\pi}{8} = \frac{\pi}{4}$.Thus,$S_1 = \frac{\pi}{8}$,so $\frac{16S_1}{\pi} = \frac{16}{\pi} \cdot \frac{\pi}{8} = 2$.For $S_2$,$S_2 = \int_{\pi/8}^{3\pi/8} \sin^2 x |4x - \pi| dx$. Using the same property,$S_2 = \int_{\pi/8}^{3\pi/8} \sin^2(\frac{\pi}{2}-x) |4(\frac{\pi}{2}-x) - \pi| dx = \int_{\pi/8}^{3\pi/8} \cos^2 x |\pi - 4x| dx = \int_{\pi/8}^{3\pi/8} \cos^2 x |4x - \pi| dx$.Adding these,$2S_2 = \int_{\pi/8}^{3\pi/8} |4x - \pi| (\sin^2 x + \cos^2 x) dx = \int_{\pi/8}^{3\pi/8} |4x - \pi| dx$.The integral represents the area of two triangles with base $\frac{\pi}{8}$ and height $\frac{\pi}{2}$,so $2S_2 = 2 \cdot (\frac{1}{2} \cdot \frac{\pi}{8} \cdot \frac{\pi}{2}) = \frac{\pi^2}{16}$.Thus,$S_2 = \frac{\pi^2}{32}$,so $\frac{48S_2}{\pi^2} = \frac{48}{\pi^2} \cdot \frac{\pi^2}{32} = \frac{48}{32} = 1.5$.

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