Find the area between the curve $y=\sin x$ and the $x$ -axis from $x=-\pi /3$ to $x=\pi /3$.

The function $y=\sin x$ is odd, so it is negative in $[-\pi /3,0]$ and positive in $[0,\pi /3]$.Total Area = ${\int }_{-\pi /3}^0|\sin x|dx+{\int }_0^{\pi /3}\sin xdx$ By symmetry, Area = $2{\int }_0^{\pi /3}\sin xdx$ $
$\text{Area}=2[-\cos x{]}_0^{\pi /3}=2[-\cos (\pi /3)-(-\cos 0)]$ $
$$\text{Area}=2[-\frac{1}{2}+1]=2[\frac{1}{2}]=1$$