The area of the region bounded by the curve y=cosx, X-axis and the lines x=0, x=`2pi` is

Options

A. 2
B. 4
C. 0
D. 3

Answer

To find the area of the region bounded by the curve \( y = \cos x \), the x-axis, and the lines \( x = 0 \) and \( x = 2\pi \), we can break the problem down into manageable steps.

### Step-by-Step Solution:

1. **Identify the Area to be Calculated**:
The area we want to find is bounded by the curve \( y = \cos x \), the x-axis, and the vertical lines \( x = 0 \) and \( x = 2\pi \).

2. **Analyze the Behavior of the Function**:
The function \( y = \cos x \) oscillates between 1 and -1. Specifically:
- From \( x = 0 \) to \( x = \pi/2 \), \( \cos x \) is positive.
- From \( x = \pi/2 \) to \( x = 3\pi/2 \), \( \cos x \) is negative.
- From \( x = 3\pi/2 \) to \( x = 2\pi \), \( \cos x \) is positive again.

3. **Set Up the Integral**:
Since the area above the x-axis contributes positively and the area below the x-axis contributes negatively, we need to split the integral into parts:
- From \( x = 0 \) to \( x = \pi/2 \): Area is \( \int_0^{\pi/2} \cos x \, dx \)
- From \( x = \pi/2 \) to \( x = 3\pi/2 \): Area is \( \int_{\pi/2}^{3\pi/2} -\cos x \, dx \) (negative because it is below the x-axis)
- From \( x = 3\pi/2 \) to \( x = 2\pi \): Area is \( \int_{3\pi/2}^{2\pi} \cos x \, dx \)

4. **Calculate Each Integral**:
- **First Integral**:
\[ A_1 = \int_0^{\pi/2} \cos x \, dx = [\sin x]_0^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1 \]
- **Second Integral**:
\[ A_2 = \int_{\pi/2}^{3\pi/2} -\cos x \, dx = -[\sin x]_{\pi/2}^{3\pi/2} = -(\sin(3\pi/2) - \sin(\pi/2)) = -(-1 - 1) = 2 \]
- **Third Integral**:
\[ A_3 = \int_{3\pi/2}^{2\pi} \cos x \, dx = [\sin x]_{3\pi/2}^{2\pi} = \sin(2\pi) - \sin(3\pi/2) = 0 - (-1) = 1 \]

5. **Total Area**:
Now, sum the areas:
\[ \text{Total Area} = A_1 + A_2 + A_3 = 1 + 2 + 1 = 4 \]

### Final Answer:
The area of the region bounded by the curve \( y = \cos x \), the x-axis, and the lines \( x = 0 \) and \( x = 2\pi \) is \( 4 \).