Find the area of region by the curve `y=sinx" between "x=0" and "x=2pi`.

Explanation

To find the area of the region bounded by the curve \( y = \sin x \) between \( x = 0 \) and \( x = 2\pi \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Curve**:
The function \( y = \sin x \) oscillates between 1 and -1. It crosses the x-axis at \( x = 0 \), \( x = \pi \), and \( x = 2\pi \).

2. **Identify the Area to be Calculated**:
The area between \( x = 0 \) and \( x = 2\pi \) consists of two parts:
- From \( x = 0 \) to \( x = \pi \), where \( \sin x \) is positive.
- From \( x = \pi \) to \( x = 2\pi \), where \( \sin x \) is negative.

3. **Set Up the Integral**:
The area \( A \) can be expressed as:
\[ A = \int_{0}^{2\pi} |\sin x| \, dx \]
Since \( \sin x \) is positive from \( 0 \) to \( \pi \) and negative from \( \pi \) to \( 2\pi \), we can split the integral:
\[ A = \int_{0}^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} -\sin x \, dx \]

4. **Calculate the First Integral**:
\[ \int_{0}^{\pi} \sin x \, dx \]
The integral of \( \sin x \) is \( -\cos x \):
\[ = [-\cos x]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \]

5. **Calculate the Second Integral**:
\[ \int_{\pi}^{2\pi} -\sin x \, dx \]
Again, using the integral of \( -\sin x \):
\[ = [-\cos x]_{\pi}^{2\pi} = -\cos(2\pi) - (-\cos(\pi)) = -1 - 1 = -2 \]
Since we are taking the absolute value, this area is \( 2 \).

6. **Combine the Areas**:
\[ A = 2 + 2 = 4 \]

### Final Answer:
The area of the region bounded by the curve \( y = \sin x \) between \( x = 0 \) and \( x = 2\pi \) is \( 4 \) square units.