Find the area below the curve `y=[sqrt(2+2cos2x)]` but above the x-axis in `[-3pi,6pi]` is (where [.] denotes the greatest integer function)

Options

A. a.`2pi` square units
B. b.`pi`square units
C. c.`6pi` square units
D. d.`8pi` square units

Explanation

To find the area below the curve \( y = \sqrt{2 + 2 \cos(2x)} \) but above the x-axis in the interval \([-3\pi, 6\pi]\), we will follow these steps:

### Step 1: Simplify the function
We start by simplifying the function \( y = \sqrt{2 + 2 \cos(2x)} \).

Using the identity \( 1 + \cos(2x) = 2 \cos^2(x) \), we can rewrite the function as:
\[ y = \sqrt{2(1 + \cos(2x))} = \sqrt{2 \cdot 2 \cos^2(x)} = 2 |\cos(x)| \]

### Step 2: Determine the intervals where the function is positive
The function \( y = 2 |\cos(x)| \) is positive wherever \( \cos(x) \) is non-negative. The cosine function is non-negative in the intervals:
- \( [2k\pi, (2k + 1)\pi] \) for integer \( k \)

### Step 3: Find the number of complete cycles in the interval \([-3\pi, 6\pi]\)
The period of \( \cos(x) \) is \( 2\pi \). To find the number of complete cycles in the interval \([-3\pi, 6\pi]\):
- The length of the interval is \( 6\pi - (-3\pi) = 9\pi \).
- The number of complete cycles is \( \frac{9\pi}{2\pi} = 4.5 \), meaning there are 4 complete cycles and an additional half cycle.

### Step 4: Calculate the area for one complete cycle
The area under one complete cycle of \( y = 2 |\cos(x)| \) from \( 0 \) to \( \pi \) is given by:
\[ \text{Area} = \int_0^{\pi} 2 \cos(x) \, dx \]
Calculating this integral:
\[ \int_0^{\pi} 2 \cos(x) \, dx = 2 \left[ \sin(x) \right]_0^{\pi} = 2 (0 - 0) = 0 \]
However, we need to consider the area from \( 0 \) to \( \frac{\pi}{2} \) and from \( \frac{\pi}{2} \) to \( \pi \):
\[ \text{Area from } 0 \text{ to } \frac{\pi}{2} = \int_0^{\frac{\pi}{2}} 2 \cos(x) \, dx = 2 [\sin(x)]_0^{\frac{\pi}{2}} = 2(1 - 0) = 2 \]
\[ \text{Area from } \frac{\pi}{2} \text{ to } \pi = \int_{\frac{\pi}{2}}^{\pi} 2 (-\cos(x)) \, dx = 2 [-\sin(x)]_{\frac{\pi}{2}}^{\pi} = 2(0 - (-1)) = 2 \]
Thus, the total area for one complete cycle is:
\[ \text{Total Area} = 2 + 2 = 4 \]

### Step 5: Calculate the total area for the interval \([-3\pi, 6\pi]\)
Since there are 4 complete cycles and a half cycle, we need to calculate the area for the half cycle from \( 0 \) to \( \frac{\pi}{2} \):
\[ \text{Area for half cycle} = 2 \]
Thus, the total area is:
\[ \text{Total Area} = 4 \times 4 + 2 = 16 + 2 = 18 \]

### Step 6: Apply the greatest integer function
Finally, we apply the greatest integer function to the total area:
\[ \lfloor 18 \rfloor = 18 \]

### Conclusion
The area below the curve \( y = \sqrt{2 + 2 \cos(2x)} \) but above the x-axis in the interval \([-3\pi, 6\pi]\) is \( \lfloor 18 \rfloor = 18 \).