Find the area bounded by `y=| sin x -(1)/(2)| and y= 1" for "x in [0,pi]`

Explanation

To find the area bounded by the curves \( y = | \sin x - \frac{1}{2} | \) and \( y = 1 \) for \( x \) in the interval \([0, \pi]\), we can follow these steps:

### Step 1: Analyze the function \( y = | \sin x - \frac{1}{2} | \)

The function \( y = | \sin x - \frac{1}{2} | \) can be broken down into two cases based on the value of \( \sin x \):

1. When \( \sin x \geq \frac{1}{2} \), then \( | \sin x - \frac{1}{2} | = \sin x - \frac{1}{2} \).
2. When \( \sin x < \frac{1}{2} \), then \( | \sin x - \frac{1}{2} | = \frac{1}{2} - \sin x \).

### Step 2: Determine the points of intersection

We need to find the points where \( y = | \sin x - \frac{1}{2} | \) intersects with \( y = 1 \).

- Set \( \sin x - \frac{1}{2} = 1 \):
\[ \sin x = \frac{3}{2} \quad \text{(not possible)} \]

- Set \( \frac{1}{2} - \sin x = 1 \):
\[ \sin x = -\frac{1}{2} \quad \text{(not in the interval [0, π])} \]

### Step 3: Identify intervals for integration

We find the critical points where \( \sin x = \frac{1}{2} \):
- \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \).

Thus, we can break the integral into three parts:
1. From \( 0 \) to \( \frac{\pi}{6} \): \( y = \frac{1}{2} - \sin x \)
2. From \( \frac{\pi}{6} \) to \( \frac{5\pi}{6} \): \( y = \sin x - \frac{1}{2} \)
3. From \( \frac{5\pi}{6} \) to \( \pi \): \( y = \frac{1}{2} - \sin x \)

### Step 4: Set up the area calculation

The total area \( A \) can be expressed as:
\[ A = \int_0^{\frac{\pi}{6}} \left(1 - \left(\frac{1}{2} - \sin x\right)\right) dx + \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \left(1 - \left(\sin x - \frac{1}{2}\right)\right) dx + \int_{\frac{5\pi}{6}}^{\pi} \left(1 - \left(\frac{1}{2} - \sin x\right)\right) dx \]

### Step 5: Calculate each integral

1. **First Integral**:
\[ \int_0^{\frac{\pi}{6}} \left(1 - \left(\frac{1}{2} - \sin x\right)\right) dx = \int_0^{\frac{\pi}{6}} \left(\frac{1}{2} + \sin x\right) dx \]
\[ = \left[\frac{1}{2}x - \cos x\right]_0^{\frac{\pi}{6}} = \left(\frac{1}{2} \cdot \frac{\pi}{6} - \cos\left(\frac{\pi}{6}\right)\right) - \left(0 - 1\right) \]

2. **Second Integral**:
\[ \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \left(1 - \left(\sin x - \frac{1}{2}\right)\right) dx = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \left(\frac{3}{2} - \sin x\right) dx \]

3. **Third Integral**:
\[ \int_{\frac{5\pi}{6}}^{\pi} \left(1 - \left(\frac{1}{2} - \sin x\right)\right) dx = \int_{\frac{5\pi}{6}}^{\pi} \left(\frac{1}{2} + \sin x\right) dx \]

### Step 6: Combine the results

After evaluating the integrals, we can combine the results to find the total area.

### Final Area Calculation
The final area will be:
\[ A = \text{Area from first integral} + \text{Area from second integral} + \text{Area from third integral} \]

### Final Answer
The area bounded by the curves is \( \frac{\pi}{2} + 2 \) square units.

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