Question

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

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

Answer 1

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

$f\left(x\right)=\sin x$
$f\left(x\right)+f(\pi -x)=2$
$f\left(x\right)=2-f(\pi -x)=2-\sin (\pi -x)=2-\sin x,$ 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 $=\underset{0}{\overset{\pi /2}{\int }}\sin xdx+\underset{\pi /2}{\overset{\pi }{\int }}(2-\sin x)dx+$
$\underset{\pi }{\overset{3\pi /2}{\int }}(2+\sin x)dx+\underset{3\pi /2}{\overset{2\pi }{\int }}(-\sin x)dx$
$=1+2\times \frac{\pi }{2}-1+2\times \frac{\pi }{2}-1+1$
$=2\pi \text{sq. units}$