Answer
$$\frac{1}{8}$$
Work Step by Step
Given $$y=\sin (2 x), \quad y=\sin (4 x), \quad x=0, \quad x=\frac{\pi}{6}$$
Since
\begin{align*}
\sin (4 x)&=\sin (2 x)\\
2\sin ( 2x)\cos(2x)-\sin (2x)&=0\\
\sin(2x)(2\cos (2x)-1)&=0
\end{align*}
then $x=0,\ \ x=\pi/6$ and $\sin4x\geq \sin2x $ for $ 0\leq x\leq \pi/6$
Hence,
\begin{aligned}
A &=\int_{0}^{\pi / 6}(\sin (4 x)-\sin (2 x)) d x \\
&=\left.\left(\frac{-\cos (4 x)}{4}+\frac{\cos (2 x)}{2}\right)\right|_{0} ^{\pi / 6} \\
&=\frac{1}{8}
\end{aligned}
