Find the maximum and minimum values of $x+\sin 2x$ on $[0, 2\pi]$.

Let $f(x) = x + \sin 2x$.
$\therefore f'(x) = 1 + 2 \cos 2x$.
Now,$f'(x) = 0 \Rightarrow \cos 2x = -\frac{1}{2}$.
Since $\cos \frac{2\pi}{3} = -\frac{1}{2}$ and $\cos \frac{4\pi}{3} = -\frac{1}{2}$,we have $2x = \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}$ for $x \in [0, 2\pi]$.
Thus,the critical points are $x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$.
Evaluating $f(x)$ at critical points and endpoints:
$f(0) = 0 + \sin(0) = 0$.
$f(\frac{\pi}{3}) = \frac{\pi}{3} + \sin(\frac{2\pi}{3}) = \frac{\pi}{3} + \frac{\sqrt{3}}{2}$.
$f(\frac{2\pi}{3}) = \frac{2\pi}{3} + \sin(\frac{4\pi}{3}) = \frac{2\pi}{3} - \frac{\sqrt{3}}{2}$.
$f(\frac{4\pi}{3}) = \frac{4\pi}{3} + \sin(\frac{8\pi}{3}) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2}$.
$f(\frac{5\pi}{3}) = \frac{5\pi}{3} + \sin(\frac{10\pi}{3}) = \frac{5\pi}{3} - \frac{\sqrt{3}}{2}$.
$f(2\pi) = 2\pi + \sin(4\pi) = 2\pi$.
Comparing these values,the absolute maximum value is $2\pi$ at $x = 2\pi$ and the absolute minimum value is $0$ at $x = 0$.