Question

Let $g_i:\left[\frac{\pi }{8},\frac{3\pi }{8}\right]\to R,i=1,2$ and $f:\left[\frac{\pi }{8},\frac{3\pi }{8}\right]\to R$ be functions such that $g_1(x)=1,g_2(x)=|4x-\pi |$ and $f(x)={\sin }^{2}x$, for all $x\in \left[\frac{\pi }{8},\frac{3\pi }{8}\right].$ Define $S_i=\underset{\frac{\pi }{8}}{\overset{\frac{3\pi }{8}}{\int }}f(x)\cdot g_i(x)dx,i=1,2$
The value of $\frac{48S_2}{{\pi }^2}$ is_____

Answer 1

Answer: 1.5

$S_2=\underset{\pi /8}{\overset{3\pi /8}{\int }}{\sin }^{2}x|4x-\pi |dx$
$=\underset{\pi /8}{\overset{3\pi /8}{\int }}{\sin }^2\left(\frac{\pi }{2}-x\right)\left|4\left(\frac{\pi }{2}-x\right)-\pi \right|dx$
$=\underset{\pi /8}{\overset{3\pi /8}{\int }}{\cos }^{2}x|4x-\pi |dx$
$\Rightarrow 2S_2=\underset{\pi /8}{\overset{3\pi /8}{\int }}|4x-\pi |dx$
$=2\underset{\pi /8}{\overset{\pi /4}{\int }}(\pi -4x)dx$
$S_2={\left(\pi x-2x^2\right)|}_{\pi /8}^{\pi /4}$
$=\pi \left(\frac{\pi }{4}-\frac{\pi }{8}\right)-2\left(\frac{{\pi }^2}{16}-\frac{{\pi }^2}{64}\right)$
$S_2=\frac{4{\pi }^2}{8}-\frac{3{\pi }^2}{32}=\frac{{\pi }^2}{32}$
$=\frac{48S_2}{{\pi }^2}=\frac{48}{{\pi }^2}\times \frac{{\pi }^2}{32}=\frac{3}{2}$
$\frac{48S_2}{{\pi }^2}=1.5$