The area bounded by `y=sinx` in `[0, 2pi]`

Options

A. 1
B. 2
C. 4
D. 3

Answer

To find the area bounded by the curve \( y = \sin x \) over the interval \( [0, 2\pi] \), we need to follow these steps:

### Step 1: Understand the Function and Interval
The function \( y = \sin x \) oscillates between 0 and 1, and between 0 and -1 over the interval \( [0, 2\pi] \). We need to identify where the function is above and below the x-axis.

### Step 2: Identify the Points of Intersection with the x-axis
The function \( y = \sin x \) intersects the x-axis at:
- \( x = 0 \)
- \( x = \pi \)
- \( x = 2\pi \)

### Step 3: Set Up the Integral
To find the area bounded by the curve, we need to compute the integral of \( \sin x \) from \( 0 \) to \( 2\pi \). However, since the area below the x-axis (between \( \pi \) and \( 2\pi \)) will contribute negatively to the integral, we will take the absolute value of the area above and below the x-axis separately.

### Step 4: Calculate the Area Above the x-axis
The area above the x-axis from \( 0 \) to \( \pi \) is given by:
\[ \text{Area}_{\text{above}} = \int_0^{\pi} \sin x \, dx \]

### Step 5: Calculate the Area Below the x-axis
The area below the x-axis from \( \pi \) to \( 2\pi \) is given by:
\[ \text{Area}_{\text{below}} = \int_{\pi}^{2\pi} -\sin x \, dx \]
(The negative sign is used because we want the area to be positive.)

### Step 6: Evaluate the Integrals
1. **For the area above the x-axis:**
\[ \int_0^{\pi} \sin x \, dx = -\cos x \bigg|_0^{\pi} = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \]

2. **For the area below the x-axis:**
\[ \int_{\pi}^{2\pi} -\sin x \, dx = -\left(-\cos x \bigg|_{\pi}^{2\pi}\right) = \cos(2\pi) - \cos(\pi) = 1 - (-1) = 1 + 1 = 2 \]

### Step 7: Total Area
Now, we add the areas above and below the x-axis:
\[ \text{Total Area} = \text{Area}_{\text{above}} + \text{Area}_{\text{below}} = 2 + 2 = 4 \]

### Final Answer
The area bounded by \( y = \sin x \) in the interval \( [0, 2\pi] \) is \( 4 \).

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