Question

If $f\left(x\right)=\sin x\forall x\in \left[0,x/2\right]$, $f\left(x\right)+f\left(\pi -x\right)=2\forall x\in \left[x/2,\pi \right]$ and $f\left(x\right)=f\left(2\pi -x\right)\forall x\in \left[\pi ,2\pi \right]$ then find the area bounded by $y=f\left(x\right)$ and the $x-$ axis.

Answer 1

$\begin{array}{c}A={\int }_0^{2\pi }ydx={\int }_0^{\pi /2}\sin xdx+{\int }_{\pi /2}^{\pi }2-\sin xdx+{\int }_{\pi }^{2\pi }-\sin xdx\\ =-{\left[\cos x\right]}_0^{{}^{\pi /2}}+{\left[2x\right]}_{{}^{\pi /2}}^{\pi }+{\left[\cos x\right]}_{{}^{\pi /2}}^{\pi }+{\left[\cos x\right]}_{\pi }^{2\pi }\\ =-\left[-1\right]+2\left[\frac{\pi }{2}\right]+\cos \sqrt{x}+\cos 2\pi -\cos \pi \\ =1+\pi +\left(-1\right)+\left(1+1\right)\\ =2+\pi \end{array}$