Question

Let $g _{ i }:\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow R , i =1,2$ and $f :\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] \rightarrow R$ be functions such that $g_{1}(x)=1, g_{2}(x)=|4 x-\pi|$ and $f(x)=\sin ^{2} x$, for all $x \in\left[\frac{\pi}{8}, \frac{3 \pi}{8}\right] .$ Define $S_{i}=\int\limits_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} f(x) \cdot g_{i}(x) d x, i=1,2$
The value of $\frac{48 \,S _{2}}{\pi^{2}}$ is_____

Answer 1

$S _{2}=\int\limits_{\pi / 8}^{3 \pi / 8} \sin ^{2} x |4 x -\pi| dx $
$=\int\limits_{\pi / 8}^{3 \pi / 8} \sin ^{2}\left(\frac{\pi}{2}- x \right)\left|4\left(\frac{\pi}{2}- x \right)-\pi\right| dx$
$=\int\limits_{\pi / 8}^{3 \pi / 8} \cos ^{2} x |4 x -\pi| dx$
$\Rightarrow 2 S _{2}=\int\limits_{\pi / 8}^{3 \pi / 8}|4 x -\pi| dx $
$=2 \int\limits_{\pi / 8}^{\pi / 4}(\pi-4 x ) dx$
$S _{2}=\left.\left(\pi x -2 x ^{2}\right)\right|_{\pi / 8} ^{\pi / 4}$
$=\pi\left(\frac{\pi}{4}-\frac{\pi}{8}\right)-2\left(\frac{\pi^{2}}{16}-\frac{\pi^{2}}{64}\right)$
$S _{2}=\frac{4 \pi^{2}}{8}-\frac{3 \pi^{2}}{32}=\frac{\pi^{2}}{32}$
$=\frac{48 S _{2}}{\pi^{2}}=\frac{48}{\pi^{2}} \times \frac{\pi^{2}}{32}=\frac{3}{2}$
$\frac{48 S _{2}}{\pi^{2}}=1.5$