If `f(x) = max {sin x, cos x,1/2},` then the area of the region bounded by the curves `y =f(x),` x-axis, Y-axis and `x=(5pi)/3` is

Options

A. `(sqrt(2)-sqrt(3)/2+(5pi)/(12))`sq units
B. `(sqrt(2)+sqrt(3)+(5pi)/(2))`sq units
C. `(sqrt(2)+sqrt(3)+(5pi)/(2))`sq units
D. None of these

Answer

To find the area of the region bounded by the curves \( y = f(x) \), the x-axis, the y-axis, and the line \( x = \frac{5\pi}{3} \), we will follow these steps:

### Step 1: Define the function \( f(x) \)
Given \( f(x) = \max \{ \sin x, \cos x, \frac{1}{2} \} \), we need to determine the intervals where each function is the maximum.

### Step 2: Identify the intervals
1. **From \( 0 \) to \( \frac{\pi}{4} \)**:
- Here, \( \cos x \) is greater than both \( \sin x \) and \( \frac{1}{2} \).
- Thus, \( f(x) = \cos x \).

2. **From \( \frac{\pi}{4} \) to \( \frac{5\pi}{6} \)**:
- In this interval, \( \sin x \) is greater than \( \cos x \) and \( \frac{1}{2} \).
- Thus, \( f(x) = \sin x \).

3. **From \( \frac{5\pi}{6} \) to \( \frac{5\pi}{3} \)**:
- Here, \( \frac{1}{2} \) is greater than both \( \sin x \) and \( \cos x \).
- Thus, \( f(x) = \frac{1}{2} \).

### Step 3: Set up the area integral
The area \( A \) can be computed as the sum of the areas under each segment of the curve:
\[ A = \int_{0}^{\frac{\pi}{4}} \cos x \, dx + \int_{\frac{\pi}{4}}^{\frac{5\pi}{6}} \sin x \, dx + \int_{\frac{5\pi}{6}}^{\frac{5\pi}{3}} \frac{1}{2} \, dx \]

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

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

3. **Third integral**:
\[ \int_{\frac{5\pi}{6}}^{\frac{5\pi}{3}} \frac{1}{2} \, dx = \frac{1}{2} \left( \frac{5\pi}{3} - \frac{5\pi}{6} \right) = \frac{1}{2} \left( \frac{10\pi}{6} - \frac{5\pi}{6} \right) = \frac{1}{2} \left( \frac{5\pi}{6} \right) = \frac{5\pi}{12} \]

### Step 5: Sum the areas
Now, we sum all the areas calculated:
\[ A = \frac{\sqrt{2}}{2} + \left( \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \right) + \frac{5\pi}{12} \]
\[ = \sqrt{2} + \frac{\sqrt{3}}{2} + \frac{5\pi}{12} \]

### Final Answer
Thus, the area of the region bounded by the curves is:
\[ A = \sqrt{2} + \frac{\sqrt{3}}{2} + \frac{5\pi}{12} \]