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 between the curve \( y = \sin 2x \) and the x-axis from \( x = 0 \) to \( x = \frac{\pi}{2} \), we can follow these steps:

### Step 1: Define the Area
The area \( A' \) under the curve \( y = \sin 2x \) from \( x = 0 \) to \( x = \frac{\pi}{2} \) can be expressed as:
\[ A' = \int_{0}^{\frac{\pi}{2}} \sin 2x \, dx \]

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

### Step 3: Evaluate the Definite Integral
Now we need to evaluate the definite integral from \( 0 \) to \( \frac{\pi}{2} \):
\[ A' = \left[-\frac{1}{2} \cos 2x \right]_{0}^{\frac{\pi}{2}} \]

### Step 4: Substitute the Limits
Now substitute the upper and lower limits into the expression:
1. For the upper limit \( x = \frac{\pi}{2} \):
\[ -\frac{1}{2} \cos(2 \cdot \frac{\pi}{2}) = -\frac{1}{2} \cos(\pi) = -\frac{1}{2} \cdot (-1) = \frac{1}{2} \]
2. For the lower limit \( x = 0 \):
\[ -\frac{1}{2} \cos(2 \cdot 0) = -\frac{1}{2} \cos(0) = -\frac{1}{2} \cdot 1 = -\frac{1}{2} \]

### Step 5: Calculate the Area
Now, subtract the lower limit result from the upper limit result:
\[ A' = \frac{1}{2} - (-\frac{1}{2}) = \frac{1}{2} + \frac{1}{2} = 1 \]

### Conclusion
Thus, the area \( A' \) lying between the curve \( y = \sin 2x \) and the x-axis from \( x = 0 \) to \( x = \frac{\pi}{2} \) is:
\[ A' = 1 \]