Question

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

• A. $2$
• B. $\frac{3}{2}$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\frac{3}{2}$

$S_2=\underset{\pi /8}{\overset{3\pi /8}{\int }}f\left(x\right)\cdot g_2\left(x\right)dx$
Given : $g_2\left(x\right)=|4x-\pi |$
$S_2=\underset{\pi /8}{\overset{3\pi /8}{\int }}{\sin }^{2}x\cdot |4x-\pi |dx\cdots \left(i\right)$
Using property
$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$
$S_2=\underset{\pi /8}{\overset{3\pi /8}{\int }}{\cos }^{2}x\cdot |2\pi -4x-\pi |dx$
$S_2=\underset{\pi /8}{\overset{3\pi /8}{\int }}{\cos }^{2}x\cdot |\pi -4x|dx\cdots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right)$, we get
$2S_2=\underset{\pi /8}{\overset{3\pi /8}{\int }}|4x-\pi |dx$
Substitute $4x-\pi =t\Rightarrow 4dx=dt$
$\Rightarrow 2S_2=\frac{1}{4}\underset{-\pi /2}{\overset{\pi /2}{\int }}|t|dt$
$\Rightarrow 2S_2=\frac{1}{4}\cdot 2\underset{0}{\overset{\pi /2}{\int }}tdt$
$\Rightarrow S_2=\frac{{\pi }^2}{32}$
$\therefore \frac{48S_2}{{\pi }^2}=\frac{3}{2}$