What is the area of the portion of the curve `y=sinx,` lying between x=0,y=0 and `x=2pi`?

Options

A. 1 sq. unit
B. 2 sq. units
C. 4 sq. units
D. 8 sq. units

Answer

To find the area of the portion of the curve \( y = \sin x \) lying between \( x = 0 \), \( y = 0 \), and \( x = 2\pi \), we can follow these steps:

### Step 1: Understand the curve and the boundaries
The curve \( y = \sin x \) oscillates between 0 and 1 for \( x \) in the interval \( [0, 2\pi] \). The points where \( y = 0 \) are at \( x = 0 \), \( x = \pi \), and \( x = 2\pi \).

### Step 2: Identify the areas to calculate
The area under the curve from \( x = 0 \) to \( x = 2\pi \) consists of two parts:
1. From \( x = 0 \) to \( x = \pi \) where \( y = \sin x \) is above the x-axis.
2. From \( x = \pi \) to \( x = 2\pi \) where \( y = \sin x \) is below the x-axis.

### Step 3: Calculate the area from \( x = 0 \) to \( x = \pi \)
The area \( A_1 \) can be calculated using the integral:
\[ A_1 = \int_0^{\pi} \sin x \, dx \]

### Step 4: Evaluate the integral
To evaluate \( A_1 \):
\[ A_1 = \int_0^{\pi} \sin x \, dx = -\cos x \bigg|_0^{\pi} \]
Calculating this gives:
\[ A_1 = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2 \]

### Step 5: Calculate the area from \( x = \pi \) to \( x = 2\pi \)
The area \( A_2 \) can be calculated using the integral:
\[ A_2 = \int_{\pi}^{2\pi} \sin x \, dx \]
Since \( \sin x \) is negative in this interval, we take the absolute value:
\[ A_2 = -\int_{\pi}^{2\pi} \sin x \, dx = -(-\cos x \bigg|_{\pi}^{2\pi}) = \cos(\pi) - \cos(2\pi) \]
Calculating this gives:
\[ A_2 = -(-1) - 1 = 1 - (-1) = 2 \]

### Step 6: Total area
The total area \( A \) is the sum of the two areas:
\[ A = A_1 + A_2 = 2 + 2 = 4 \]

### Conclusion
The area of the portion of the curve \( y = \sin x \) lying between \( x = 0 \), \( y = 0 \), and \( x = 2\pi \) is \( 4 \) square units.