Let `f(x)="max"{sin x,cos x,1/2}`, then determine the area of region bounded by the curves `y=f(x)`, X-axis, Y-axis and `x=2pi`.

Answer

To find the area of the region bounded by the curves \( y = f(x) = \max(\sin x, \cos x, \frac{1}{2}) \), the X-axis, the Y-axis, and the line \( x = 2\pi \), we will follow these steps:

### Step 1: Identify the points of intersection
We need to find the points where \( \sin x \), \( \cos x \), and \( \frac{1}{2} \) intersect.

1. **Intersection of \( \sin x \) and \( \cos x \)**:
\[ \sin x = \cos x \implies \tan x = 1 \implies x = \frac{\pi}{4} + n\pi, \, n \in \mathbb{Z} \]
In the interval \( [0, 2\pi] \), this gives us \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \).

2. **Intersection of \( \sin x \) and \( \frac{1}{2} \)**:
\[ \sin x = \frac{1}{2} \implies x = \frac{\pi}{6}, \frac{5\pi}{6} \]

3. **Intersection of \( \cos x \) and \( \frac{1}{2} \)**:
\[ \cos x = \frac{1}{2} \implies x = \frac{\pi}{3}, \frac{5\pi}{3} \]

### Step 2: Determine the intervals for \( f(x) \)
Now we can summarize the intervals where each function is the maximum:

- **Interval \( [0, \frac{\pi}{6}] \)**: Here, \( \cos x \) is the maximum.
- **Interval \( [\frac{\pi}{6}, \frac{\pi}{4}] \)**: Here, \( \sin x \) is the maximum.
- **Interval \( [\frac{\pi}{4}, \frac{5\pi}{6}] \)**: Here, \( \sin x \) is the maximum.
- **Interval \( [\frac{5\pi}{6}, \frac{5\pi}{3}] \)**: Here, \( \frac{1}{2} \) is the maximum.
- **Interval \( [\frac{5\pi}{3}, 2\pi] \)**: Here, \( \cos x \) is the maximum.

### Step 3: Set up the integral for the area
The area \( A \) can be computed as:
\[ A = \int_0^{\frac{\pi}{6}} \cos x \, dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \sin x \, dx + \int_{\frac{\pi}{4}}^{\frac{5\pi}{6}} \sin x \, dx + \int_{\frac{5\pi}{6}}^{\frac{5\pi}{3}} \frac{1}{2} \, dx + \int_{\frac{5\pi}{3}}^{2\pi} \cos x \, dx \]

### Step 4: Calculate each integral
1. **First integral**:
\[ \int_0^{\frac{\pi}{6}} \cos x \, dx = [\sin x]_0^{\frac{\pi}{6}} = \sin\left(\frac{\pi}{6}\right) - \sin(0) = \frac{1}{2} \]

2. **Second integral**:
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \sin x \, dx = [-\cos x]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = -\cos\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \]

3. **Third integral**:
\[ \int_{\frac{\pi}{4}}^{\frac{5\pi}{6}} \sin x \, dx = [-\cos x]_{\frac{\pi}{4}}^{\frac{5\pi}{6}} = -\cos\left(\frac{5\pi}{6}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \]

4. **Fourth integral**:
\[ \int_{\frac{5\pi}{6}}^{\frac{5\pi}{3}} \frac{1}{2} \, dx = \frac{1}{2} \left(\frac{5\pi}{3} - \frac{5\pi}{6}\right) = \frac{1}{2} \cdot \frac{5\pi}{2} = \frac{5\pi}{4} \]

5. **Fifth integral**:
\[ \int_{\frac{5\pi}{3}}^{2\pi} \cos x \, dx = [\sin x]_{\frac{5\pi}{3}}^{2\pi} = \sin(2\pi) - \sin\left(\frac{5\pi}{3}\right) = 0 + \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \]

### Step 5: Combine all areas
Now we can combine the areas:
\[ A = \frac{1}{2} + \left(-\frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\right) + \frac{5\pi}{4} + \frac{\sqrt{3}}{2} \]
This simplifies to:
\[ A = \frac{1}{2} + \frac{5\pi}{4} + 2\left(\frac{\sqrt{3}}{2}\right) = \frac{1}{2} + \frac{5\pi}{4} + \sqrt{3} \]

### Final Answer
Thus, the area of the region bounded by the curves is:
\[ A = \frac{1}{2} + \sqrt{3} + \frac{5\pi}{4} \]