To find the area bounded by the curves $y = 2\sin x$ and $y = \sin 2x$ over the interval $0 \leq x \leq \pi$, we follow these steps:

1. Set up the area integral:

The area between two curves is given by the integral:

$\text{Area} = \int_a^b \left( f(x) - g(x) \right) dx$

Here, $f(x) = 2\sin x$ (the upper curve) and $g(x) = \sin 2x$ (the lower curve). The limits of integration are from $0$ to $\pi$.

Thus, the area is:

$\text{Area} = \int_0^\pi \left( 2\sin x - \sin 2x \right) dx$

2. Compute the integral:

We now compute the integral of each function separately:

$\int_0^\pi 2\sin x \, dx = 2 \left[ -\cos x \right]_0^\pi = 2 \left( -(-1) - (-1) \right) = 2 \times 2 = 4$

For the second term:

$\int_0^\pi \sin 2x \, dx = \frac{-1}{2} \left[ \cos 2x \right]_0^\pi = \frac{-1}{2} \left( \cos 2\pi - \cos 0 \right) = \frac{-1}{2} (1 - 1) = 0$

3. Final answer:

The area is:

$\text{Area} = 4 - 0 = 4$

Thus, the area bounded by the curves $y = 2\sin x$ and $y = \sin 2x$ between $0$ and $\pi$ is $\boxed{4}$.

Would you like more details on any step? Here are some related questions for further practice:

1. What happens to the area if we integrate over a different interval?
2. How does the area change if the upper curve is modified to $y = 3\sin x$?
3. What is the area between $y = \cos x$ and $y = \sin x$ over the interval $0 \leq x \leq \pi/2$?
4. How would you find the points of intersection between $y = 2\sin x$ and $y = \sin 2x$?
5. Can you generalize this method for any pair of sinusoidal functions?

Tip: When finding the area between curves, always make sure to carefully identify which curve is on top over the entire interval to avoid negative areas.