Find the area enclosed the curve y=sin x and the X-axis between `x=0 and x=pi`.

Explanation

To find the area enclosed by the curve \( y = \sin x \) and the X-axis between \( x = 0 \) and \( x = \pi \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Function and Limits**:
We are given the function \( y = \sin x \) and we need to find the area between this curve and the X-axis from \( x = 0 \) to \( x = \pi \).

2. **Set Up the Integral**:
The area under the curve can be found using the definite integral of the function from the lower limit \( x_1 = 0 \) to the upper limit \( x_2 = \pi \):
\[ \text{Area} = \int_{0}^{\pi} \sin x \, dx \]

3. **Calculate the Integral**:
To calculate the integral, we find the antiderivative of \( \sin x \):
\[ \int \sin x \, dx = -\cos x + C \]
Now, we evaluate this from \( 0 \) to \( \pi \):
\[ \text{Area} = \left[-\cos x\right]_{0}^{\pi} \]

4. **Evaluate at the Limits**:
Now we substitute the limits into the antiderivative:
\[ = -\cos(\pi) - (-\cos(0)) \]
We know that \( \cos(\pi) = -1 \) and \( \cos(0) = 1 \):
\[ = -(-1) - (-1) = 1 + 1 = 2 \]

5. **Final Result**:
Therefore, the area enclosed by the curve \( y = \sin x \) and the X-axis between \( x = 0 \) and \( x = \pi \) is:
\[ \text{Area} = 2 \text{ square units} \]