The area bounded by the curve `y = sin x , x in [0,2pi]` and the `x`-axis is equal to:

Options

A. (a) `4` sq. unit
B. (b) `0` sq. unit
C. (c) `2` sq. unit
D. (d) `6` sq. unit

Answer

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

### Step 1: Understand the Graph of \( y = \sin x \)
The sine function oscillates between -1 and 1. Over the interval \( [0, 2\pi] \), it has positive values from \( 0 \) to \( \pi \) and negative values from \( \pi \) to \( 2\pi \).

### Step 2: Identify the Areas
We can divide the area into two parts:
- Area \( A_1 \) from \( x = 0 \) to \( x = \pi \) (where \( \sin x \) is positive)
- Area \( A_2 \) from \( x = \pi \) to \( x = 2\pi \) (where \( \sin x \) is negative)

### Step 3: Calculate Area \( A_1 \)
The area \( A_1 \) can be calculated using the definite integral:
\[ A_1 = \int_0^{\pi} \sin x \, dx \]

### Step 4: Perform the Integration
The integral of \( \sin x \) is \( -\cos x \). Thus,
\[ A_1 = \left[-\cos x\right]_0^{\pi} \]
Calculating the limits:
\[ A_1 = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \]

### Step 5: Calculate Area \( A_2 \)
For area \( A_2 \), since \( \sin x \) is negative, we take the absolute value:
\[ A_2 = \int_{\pi}^{2\pi} -\sin x \, dx = \int_{\pi}^{2\pi} \sin x \, dx \]
Using the same integration:
\[ A_2 = \left[-\cos x\right]_{\pi}^{2\pi} \]
Calculating the limits:
\[ A_2 = -\cos(2\pi) - (-\cos(\pi)) = -1 - 1 = 2 \]

### Step 6: Total Area
The total area \( A \) bounded by the curve and the x-axis from \( x = 0 \) to \( x = 2\pi \) is:
\[ A = A_1 + A_2 = 2 + 2 = 4 \]

### Final Answer
The area bounded by the curve \( y = \sin x \) from \( x = 0 \) to \( x = 2\pi \) is \( 4 \) square units.

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