Find the area of the region bounded by
`y=sinx` and the x - axis in the interval `[0,2pi]`

Answer

To find the area of the region bounded by the curve \( y = \sin x \) and the x-axis in the interval \([0, 2\pi]\), we can follow these steps:

### Step 1: Identify the intervals
The function \( y = \sin x \) is positive in the interval \([0, \pi]\) and negative in the interval \([\pi, 2\pi]\). Therefore, we need to calculate the area separately for these two intervals.

### Step 2: Set up the integral for area
The area \( A \) can be expressed as:
\[ A = \int_0^{2\pi} |\sin x| \, dx \]
This can be split into two parts:
\[ A = \int_0^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} -\sin x \, dx \]
Here, we take \(-\sin x\) in the second integral because \(\sin x\) is negative in that interval.

### Step 3: Calculate the first integral
Now, we calculate the first integral:
\[ \int_0^{\pi} \sin x \, dx \]
The integral of \(\sin x\) is \(-\cos x\). Thus,
\[ \int_0^{\pi} \sin x \, dx = \left[-\cos x\right]_0^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \]

### Step 4: Calculate the second integral
Next, we calculate the second integral:
\[ \int_{\pi}^{2\pi} -\sin x \, dx \]
Again, using the integral of \(-\sin x\):
\[ \int_{\pi}^{2\pi} -\sin x \, dx = \left[\cos x\right]_{\pi}^{2\pi} = \cos(2\pi) - \cos(\pi) = 1 - (-1) = 1 + 1 = 2 \]

### Step 5: Combine the areas
Now, we can combine the areas from both intervals:
\[ A = 2 + 2 = 4 \]

### Final Answer
The area of the region bounded by \( y = \sin x \) and the x-axis in the interval \([0, 2\pi]\) is \( \boxed{4} \) square units.

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