The area bounded by the curve `y=sin^(2)x-2 sin x ` and the x-axis, where `x in [0, 2pi]`, is

Options

A. 4 sq. units
B. 8 sq. units
C. 16 sq. units
D. 20 sq. units

Explanation

To find the area bounded by the curve \( y = \sin^2 x - 2 \sin x \) and the x-axis for \( x \) in the interval \([0, 2\pi]\), we will follow these steps:

### Step 1: Find the points where the curve intersects the x-axis
We need to set \( y = 0 \):
\[ \sin^2 x - 2 \sin x = 0 \]
Factoring gives:
\[ \sin x (\sin x - 2) = 0 \]
This implies:
\[ \sin x = 0 \quad \text{or} \quad \sin x = 2 \]
Since \(\sin x\) cannot equal 2, we only consider \(\sin x = 0\). The solutions in the interval \([0, 2\pi]\) are:
\[ x = 0, \pi, 2\pi \]

### Step 2: Determine the intervals of integration
The function \( y = \sin^2 x - 2 \sin x \) will be above the x-axis between the points where it intersects the x-axis. We need to check the sign of \( y \) in the intervals:
- From \( 0 \) to \( \pi \)
- From \( \pi \) to \( 2\pi \)

### Step 3: Evaluate the function in the intervals
1. For \( x \in [0, \pi] \):
- At \( x = \frac{\pi}{2} \):
\[ y = \sin^2\left(\frac{\pi}{2}\right) - 2\sin\left(\frac{\pi}{2}\right) = 1 - 2 = -1 \quad (\text{below x-axis}) \]
So, \( y < 0 \) in this interval.

2. For \( x \in [\pi, 2\pi] \):
- At \( x = \frac{3\pi}{2} \):
\[ y = \sin^2\left(\frac{3\pi}{2}\right) - 2\sin\left(\frac{3\pi}{2}\right) = 0 + 2 = 2 \quad (\text{above x-axis}) \]
So, \( y > 0 \) in this interval.

### Step 4: Set up the integral for the area
Since the area is bounded by the x-axis, we will integrate the absolute value of the function:
\[ \text{Area} = \int_{\pi}^{2\pi} (\sin^2 x - 2 \sin x) \, dx \]

### Step 5: Simplify the integral
Using the identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \):
\[ \text{Area} = \int_{\pi}^{2\pi} \left(\frac{1 - \cos 2x}{2} - 2 \sin x\right) \, dx \]

### Step 6: Evaluate the integral
Now, we can split the integral:
\[ \text{Area} = \int_{\pi}^{2\pi} \frac{1}{2} \, dx - \int_{\pi}^{2\pi} \frac{\cos 2x}{2} \, dx - 2 \int_{\pi}^{2\pi} \sin x \, dx \]

Calculating each part:
1. \(\int_{\pi}^{2\pi} \frac{1}{2} \, dx = \frac{1}{2} [x]_{\pi}^{2\pi} = \frac{1}{2} (2\pi - \pi) = \frac{\pi}{2}\)
2. \(\int_{\pi}^{2\pi} \frac{\cos 2x}{2} \, dx = \frac{1}{4} [\sin 2x]_{\pi}^{2\pi} = \frac{1}{4} (\sin(4\pi) - \sin(2\pi)) = 0\)
3. \(\int_{\pi}^{2\pi} \sin x \, dx = [-\cos x]_{\pi}^{2\pi} = -\cos(2\pi) + \cos(\pi) = -1 + 1 = 0\)

### Step 7: Combine the results
Thus, the area is:
\[ \text{Area} = \frac{\pi}{2} - 0 - 0 = \frac{\pi}{2} \]

### Final Answer
The area bounded by the curve \( y = \sin^2 x - 2 \sin x \) and the x-axis from \( x = 0 \) to \( x = 2\pi \) is:
\[ \text{Area} = \frac{\pi}{2} \]