Solution:
Graph of $y=\sin x$ can be drawn as

$\therefore$ Required area $=$ Area OABO $+$ Area BCDB
$\begin{array}{l}
=\int_{0}^{\pi} \sin x d x+\left|\int_{k}^{2 \pi} \sin x d x\right| \\
=[-\cos x]_{0}^{\pi}+\left|[-\cos x]_{\pi}^{2 \pi}\right| \\
=[-\cos \pi+\cos 0]+|-\cos 2 \pi+\cos \pi| \\
=1+1+|(-1-1)| \\
=2+|-2| \\
=2+2=4 \text {sq. units }
\end{array}$