If A is the area lying between the curve `y=sin x and ` x-axis between x=0 `and x=pi//2` . Area of the region between the curve `y=sin 2x and x`-axis in the same interval is given by

Options

A. `A//2`
B. A
C. `2A`
D. `3//2A`

Answer

To find the area of the region between the curve \( y = \sin 2x \) and the x-axis from \( x = 0 \) to \( x = \frac{\pi}{2} \), we will follow these steps:

### Step 1: Identify the area to be calculated
We need to calculate the area under the curve \( y = \sin 2x \) from \( x = 0 \) to \( x = \frac{\pi}{2} \).

### Step 2: Set up the integral
The area \( A \) can be expressed as the integral of \( \sin 2x \) from \( 0 \) to \( \frac{\pi}{2} \):
\[ A = \int_{0}^{\frac{\pi}{2}} \sin 2x \, dx \]

### Step 3: Find the integral of \( \sin 2x \)
To integrate \( \sin 2x \), we use the substitution method. The integral of \( \sin kx \) is given by:
\[ \int \sin kx \, dx = -\frac{1}{k} \cos kx + C \]
For \( k = 2 \):
\[ \int \sin 2x \, dx = -\frac{1}{2} \cos 2x + C \]

### Step 4: Evaluate the definite integral
Now we evaluate the definite integral:
\[ A = \left[-\frac{1}{2} \cos 2x \right]_{0}^{\frac{\pi}{2}} \]

Calculating the limits:
1. At \( x = \frac{\pi}{2} \):
\[ -\frac{1}{2} \cos(2 \cdot \frac{\pi}{2}) = -\frac{1}{2} \cos(\pi) = -\frac{1}{2} \cdot (-1) = \frac{1}{2} \]
2. At \( x = 0 \):
\[ -\frac{1}{2} \cos(2 \cdot 0) = -\frac{1}{2} \cos(0) = -\frac{1}{2} \cdot 1 = -\frac{1}{2} \]

Now substituting these values back into the integral:
\[ A = \frac{1}{2} - \left(-\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{2} = 1 \]

### Final Result
The area of the region between the curve \( y = \sin 2x \) and the x-axis from \( x = 0 \) to \( x = \frac{\pi}{2} \) is:
\[ \boxed{1} \]