Passage
Question Stem for Question Nos. 11 and 12
Let gi:[π8,3π8]→Rg_i: \left[\frac\pi8, \frac{3\pi}8\right] \to \mathbb Rgi:[8π,83π]→R, i=1,2i = 1, 2i=1,2, and f:[π8,3π8]→Rf: \left[\frac\pi8, \frac{3\pi}8\right] \to \mathbb Rf:[8π,83π]→R be functions such thatg1(x)=1,g2(x)=∣4x−π∣andf(x)=sin2x,for all x∈[π8,3π8]g_1(x) = 1,\quad g_2(x) = |4x - \pi| \quad\text{and}\quad f(x) = \sin^2 x,\quad\text{for all } x \in \left[\frac\pi8, \frac{3\pi}8\right]g1(x)=1,g2(x)=∣4x−π∣andf(x)=sin2x,for all x∈[8π,83π]DefineSi=∫π/83π/8f(x)⋅gi(x) dx,i=1,2S_i = \int_{\pi/8}^{3\pi/8} f(x)\cdot g_i(x)\,dx,\qquad i = 1, 2Si=∫π/83π/8f(x)⋅gi(x)dx,i=1,2
The value of 16S1π\dfrac{16S_1}{\pi}π16S1 is ____.
Official answer: 1.99
Compute S1S_1S1. Using sin2x=1−cos2x2\sin^2x=\frac{1-\cos2x}2sin2x=21−cos2x,S1=∫π/83π/8sin2x dx=12[x−sin2x2]π/83π/8=12⋅π4=π8.S_1=\int_{\pi/8}^{3\pi/8}\sin^2x\,dx=\frac12\left[x-\frac{\sin2x}2\right]_{\pi/8}^{3\pi/8}=\frac12\cdot\frac\pi4=\frac\pi8.S1=∫π/83π/8sin2xdx=21[x−2sin2x]π/83π/8=21⋅4π=8π.(The sine terms cancel, since sin3π4=sinπ4\sin\frac{3\pi}4=\sin\frac\pi4sin43π=sin4π.)
Value. 16S1π=2\dfrac{16S_1}\pi=2π16S1=2.