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 solve the problem, we need to find the area under the curve $y = |\sin 2x|$ over the interval $[0, 2\pi]$.

1. Set Up the Integral:
Since $y = |\sin 2x|$ is non-negative, the area is:
$\int_0^{2\pi} |\sin 2x| dx$.

2. Use Periodicity:
The period of $\sin 2x$ is $\pi$, so $|\sin 2x|$ has period $\pi/2$.
Thus, the integral over $[0, 2\pi]$ (four periods) is:
$\int_0^{2\pi} |\sin 2x| dx = 4 \int_0^{\pi/2} |\sin 2x| dx$.
Since $\sin 2x \ge 0$ on $[0, \pi/2]$, $|\sin 2x| = \sin 2x$ there.

3. Evaluate the Integral:
Compute:
$4 \int_0^{\pi/2} \sin 2x dx = 4 \left[ -\frac{\cos 2x}{2} \right]_0^{\pi/2} = 4 \left( -\frac{\cos \pi}{2} - \left( -\frac{\cos 0}{2} \right) \right)$.
Simplify:
$= 4 \left( \frac{1}{2} + \frac{1}{2} \right) = 4 \cdot 1 = 4$.

Final Answer:
The area is $4$.