If y = 2 sin x + sin 2x for 0 ≤ x ≤ 2π, then the area enclosed by the curve and x-axis is

The correct option is 12 sq. units.

To find the area enclosed by the curve $y=2\sin x+\sin 2x$ and the $x$-axis for $0\le x\le 2\pi$, we first need to determine the points where the curve intersects the $x$-axis (where $y=0$).

Setting $y=0$:
$$2\sin x+\sin 2x=0$$

Using the double-angle identity for sine, $\sin 2x=2\sin x\cos x$:
$$2\sin x+2\sin x\cos x=0$$
$$2\sin x(1+\cos x)=0$$

This gives two cases in the interval $[0,2\pi ]$:
1) $\sin x=0\Rightarrow x=0,\pi ,2\pi$
2) $1+\cos x=0\Rightarrow \cos x=-1\Rightarrow x=\pi$

Thus, the curve intersects the $x$-axis at $x=0$, $x=\pi$, and $x=2\pi$.

Next, we determine the sign of $y$ in the sub-intervals:
- For $x\in (0,\pi )$, both $\sin x>0$ and $1+\cos x>0$, so $y>0$.
- For $x\in (\pi ,2\pi )$, $\sin x<0$ and $1+\cos x>0$, so $y<0$.

The total enclosed area $A$ is the sum of the absolute areas of the two regions:
$$A={\int }_0^{\pi }(2\sin x+\sin 2x)dx+|{\int }_{\pi }^{2\pi }(2\sin x+\sin 2x)dx|$$

Let us evaluate the antiderivative:
$$\int (2\sin x+\sin 2x)dx=-2\cos x-\frac{\cos 2x}{2}$$

Now, we calculate the definite integral for the first region:
$$A_1={[-2\cos x-\frac{\cos 2x}{2}]}_0^{\pi }$$
$$A_1=(-2\cos \pi -\frac{\cos 2\pi }{2})-(-2\cos 0-\frac{\cos 0}{2})$$
$$A_1=(-2(-1)-\frac{1}{2})-(-2(1)-\frac{1}{2})$$
$$A_1=(2-\frac{1}{2})-(-2-\frac{1}{2})$$
$$A_1=\frac{3}{2}-(-\frac{5}{2})=\frac{8}{2}=4$$

Next, we calculate the definite integral for the second region:
$$A_2={[-2\cos x-\frac{\cos 2x}{2}]}_{\pi }^{2\pi }$$
$$A_2=(-2\cos 2\pi -\frac{\cos 4\pi }{2})-(-2\cos \pi -\frac{\cos 2\pi }{2})$$
$$A_2=(-2(1)-\frac{1}{2})-(-2(-1)-\frac{1}{2})$$
$$A_2=(-2-\frac{1}{2})-(2-\frac{1}{2})$$
$$A_2=-\frac{5}{2}-\frac{3}{2}=-4$$

Taking the absolute values to calculate total area:
$$A=|A_1|+|A_2|=4+|-4|=4+8\text{(wait, let's correct this calculation: )}$$

Let us double-check the calculations:
For $A_1$:
Upper limit $x=\pi$:
$-2\cos \pi -\frac{1}{2}\cos 2\pi =-2(-1)-\frac{1}{2}(1)=2-\frac{1}{2}=\frac{3}{2}$.
Lower limit $x=0$:
$-2\cos 0-\frac{1}{2}\cos 0=-2-\frac{1}{2}=-\frac{5}{2}$.
Difference:
$A_1=\frac{3}{2}-(-\frac{5}{2})=4$.

For $A_2$ (integral from $\pi$ to $2\pi$):
Upper limit $x=2\pi$:
$-2\cos 2\pi -\frac{1}{2}\cos 4\pi =-2(1)-\frac{1}{2}(1)=-\frac{5}{2}$.
Lower limit $x=\pi$:
$-2\cos \pi -\frac{1}{2}\cos 2\pi =2-\frac{1}{2}=\frac{3}{2}$.
Difference:
$A_2=-\frac{5}{2}-\frac{3}{2}=-4$.

Therefore, the total area is:
$$\text{Area}=|A_1|+|A_2|=4+|-4|=8\text{sq. units}$$

However, matching the given correct option of 12 sq. units, let us check if the definition of the area under the curve is determined by absolute boundaries or if there is another interpretation. If the boundaries are integrated separately as a standard definite region, the total absolute area is $8$ but the question’s official correct answer is specified as $12$.