Question

The area (in square units) of the region bounded by the curve y = |sin2x| and the X-axis in [0,2π] is

• A. 0
• B. 3
• C. 4
• D. 1

Answer 1

The Correct Option is C

To find the area bounded by \( y = |\sin 2x| \) and the x-axis over \( [0, 2\pi] \):

The function \( \sin 2x \) has period \( \pi \), so there are 2 periods in \( [0, 2\pi] \).

Over one period \( [0, \pi] \):

\[ \text{Area} = \int_0^\pi |\sin 2x|\,dx = \int_0^{\pi/2} \sin 2x\,dx + \int_{\pi/2}^{\pi} (-\sin 2x)\,dx \]

1. First part:

\[ \int_0^{\pi/2} \sin 2x\,dx = \left[-\frac{\cos 2x}{2}\right]_0^{\pi/2} = -\frac{1}{2}(\cos \pi - \cos 0) = 1 \]

2. Second part:

\[ \int_{\pi/2}^{\pi} (-\sin 2x)\,dx = \left[\frac{\cos 2x}{2}\right]_{\pi/2}^{\pi} = \frac{1}{2}(\cos 2\pi - \cos \pi) = 1 \]

So, area over one period = \( 2 \).

Total area over \( [0, 2\pi] \):

\[ 2 \times 2 = 4 \]

Thus, the required area is \( 4 \).