Find the area bounded by the curve `y=2 cosx` and the X-axis from x = 0 to `x=2pi`.

Explanation

To find the area bounded by the curve \( y = 2 \cos x \) and the x-axis from \( x = 0 \) to \( x = 2\pi \), we can follow these steps:

### Step 1: Understand the Function
The function \( y = 2 \cos x \) oscillates between -2 and 2. We need to find the area between this curve and the x-axis over the interval \( [0, 2\pi] \).

### Step 2: Identify the Points of Intersection
The curve intersects the x-axis where \( y = 0 \):
\[ 2 \cos x = 0 \implies \cos x = 0 \]
The solutions in the interval \( [0, 2\pi] \) are:
\[ x = \frac{\pi}{2}, \quad x = \frac{3\pi}{2} \]

### Step 3: Set Up the Integral
The area can be calculated by integrating the function from \( 0 \) to \( 2\pi \). However, since the function is negative between \( \frac{\pi}{2} \) and \( \frac{3\pi}{2} \), we will split the integral into three parts:
\[ \text{Area} = \int_0^{2\pi} 2 \cos x \, dx = \int_0^{\frac{\pi}{2}} 2 \cos x \, dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 2 \cos x \, dx + \int_{\frac{3\pi}{2}}^{2\pi} 2 \cos x \, dx \]
To account for the negative area, we take the absolute value of the integral from \( \frac{\pi}{2} \) to \( \frac{3\pi}{2} \):
\[ \text{Area} = \int_0^{\frac{\pi}{2}} 2 \cos x \, dx + \left| \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 2 \cos x \, dx \right| + \int_{\frac{3\pi}{2}}^{2\pi} 2 \cos x \, dx \]

### Step 4: Calculate Each Integral
1. **From \( 0 \) to \( \frac{\pi}{2} \)**:
\[ \int_0^{\frac{\pi}{2}} 2 \cos x \, dx = 2 \left[ \sin x \right]_0^{\frac{\pi}{2}} = 2 \left( \sin \frac{\pi}{2} - \sin 0 \right) = 2(1 - 0) = 2 \]

2. **From \( \frac{\pi}{2} \) to \( \frac{3\pi}{2} \)**:
\[ \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} 2 \cos x \, dx = 2 \left[ \sin x \right]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} = 2 \left( \sin \frac{3\pi}{2} - \sin \frac{\pi}{2} \right) = 2(-1 - 1) = -4 \]
Taking the absolute value gives \( 4 \).

3. **From \( \frac{3\pi}{2} \) to \( 2\pi \)**:
\[ \int_{\frac{3\pi}{2}}^{2\pi} 2 \cos x \, dx = 2 \left[ \sin x \right]_{\frac{3\pi}{2}}^{2\pi} = 2 \left( \sin 2\pi - \sin \frac{3\pi}{2} \right) = 2(0 - (-1)) = 2 \]

### Step 5: Sum the Areas
Now, we add the areas calculated:
\[ \text{Total Area} = 2 + 4 + 2 = 8 \]

### Final Answer
The area bounded by the curve \( y = 2 \cos x \) and the x-axis from \( x = 0 \) to \( x = 2\pi \) is \( 8 \) square units.

---