Question

If $f\left(x\right)=\sin x,\forall x\in [0,\frac{\pi }{2}],f\left(x\right)+f(\pi -x)=2$, $\forall x(\frac{\pi }{2},\pi ]$ and $f\left(x\right)=f(2\pi -x),\forall x\in (\pi ,2\pi )$, then the area enclosed by $y=f\left(x\right)$ and x-axis is

• A. $\pi$ sq. units
• B. $2\pi$ sq. units
• C. $2$ sq. units
• D. $4$ sq. units

Answer 1

The correct option is B $2\pi$ sq. units

$f\left(x\right)=\sin x$

$f\left(x\right)+f(\pi -x)=2$

$f\left(x\right)=2-f(\pi -x)=2-\sin (\pi -x)=2-\sin x$, where $x\in (\frac{\pi }{2},\pi ]$

$f\left(x\right)=f(2\pi -x)=2-\sin (2\pi -x)$, where $x\in (\pi ,\frac{3\pi }{2}]$

$f\left(x\right)=f(2\pi -x)=-\sin x$, where $x\in (\frac{3\pi }{2},2\pi ]$

$f\left(x\right)=\{\begin{array}{c}\sin x,x\in [0,\frac{\pi }{2}]\\ 2-\sin x,x\in (\frac{\pi }{2},\pi ]\\ 2+\sin x,x\in (\pi ,\frac{3\pi }{2}]\\ -\sin x,x\in (\frac{3\pi }{2},2\pi ]\end{array}$

Area $={\int }_0^{\pi /2}\sin x\sin dx+{\int }_{\pi /2}^{\pi }(2-\sin x)dx+{\int }_{\pi }^{3\pi /2}(2+\sin x)dx+{\int }_{3\pi /2}^{2\pi }(-\sin x)dx$

$=1+2\times \frac{\pi }{2}-1+2.\frac{\pi }{2}-1=2\pi$ sq. units.