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

Answer

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 1: Understand the Curve and Limits
The function \( y = \sin x \) oscillates between -1 and 1. We need to find the area between the curve and the x-axis from \( x = 0 \) to \( x = 2\pi \).

### Step 2: Identify the Intervals
The sine function is positive from \( x = 0 \) to \( x = \pi \) and negative from \( x = \pi \) to \( x = 2\pi \). Therefore, we will split the area into two parts:
1. From \( x = 0 \) to \( x = \pi \) (where \( \sin x \) is above the x-axis).
2. From \( x = \pi \) to \( x = 2\pi \) (where \( \sin x \) is below the x-axis).

### Step 3: Set Up the Integral
The area \( A \) can be expressed as:
\[ A = \int_0^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} -\sin x \, dx \]
Here, the negative sign in the second integral accounts for the area below the x-axis.

### Step 4: Calculate the First Integral
Calculate the integral from \( 0 \) to \( \pi \):
\[ \int_0^{\pi} \sin x \, dx \]
The antiderivative of \( \sin x \) is \( -\cos x \). Thus:
\[ \int_0^{\pi} \sin x \, dx = [-\cos x]_0^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \]

### Step 5: Calculate the Second Integral
Now calculate the integral from \( \pi \) to \( 2\pi \):
\[ \int_{\pi}^{2\pi} -\sin x \, dx = -\int_{\pi}^{2\pi} \sin x \, dx \]
Using the antiderivative:
\[ -\int_{\pi}^{2\pi} \sin x \, dx = -[-\cos x]_{\pi}^{2\pi} = [\cos x]_{\pi}^{2\pi} = \cos(2\pi) - \cos(\pi) = 1 - (-1) = 1 + 1 = 2 \]

### Step 6: Combine the Areas
Adding both areas together:
\[ A = 2 + 2 = 4 \]

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

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