The area of region betweeen `y - sin x` and x-axis in the interval `[0,(pi)/(2)]` will be :

Options

A. 1
B. 0
C. 2
D. 9

Explanation

To find the area of the region between the curve \( y = \sin x \) and the x-axis over the interval \([0, \frac{\pi}{2}]\), we can follow these steps:

### Step 1: Set up the integral
The area \( A \) can be calculated using the definite integral of the function \( y = \sin x \) from \( x = 0 \) to \( x = \frac{\pi}{2} \):
\[ A = \int_{0}^{\frac{\pi}{2}} \sin x \, dx \]

### Step 2: Calculate the integral
To evaluate the integral, we find the antiderivative of \( \sin x \):
\[ \int \sin x \, dx = -\cos x + C \]
Now, we apply the limits from \( 0 \) to \( \frac{\pi}{2} \):
\[ A = \left[-\cos x\right]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) - \left(-\cos(0)\right) \]

### Step 3: Evaluate the limits
Now we substitute the limits into the expression:
\[ A = -\cos\left(\frac{\pi}{2}\right) + \cos(0) \]
We know that:
\[ \cos\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad \cos(0) = 1 \]
Thus, substituting these values gives:
\[ A = -0 + 1 = 1 \]

### Conclusion
The area of the region between \( y = \sin x \) and the x-axis over the interval \([0, \frac{\pi}{2}]\) is:
\[ \boxed{1} \]