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 can follow these steps:

### Step 1: Find the points where the curve intersects the x-axis
To find the points of intersection, we set \( y = 0 \):
\[ \sin^2 x - 2 \sin x = 0 \]
Factoring gives:
\[ \sin x (\sin x - 2) = 0 \]
This gives us two cases:
1. \( \sin x = 0 \)
2. \( \sin x - 2 = 0 \) (not possible since \( \sin x \) cannot exceed 1)

From the first case, the solutions in the interval \([0, 2\pi]\) are:
\[ x = 0, \pi, 2\pi \]