Make a rough sketch of the graph of the function `y =2 sin x, 0 le x le (pi)/(2)` and determine the area enclosed between the curve, the x-axis and the line `x = (pi)/(2).`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Sketch the Graph of \( y = 2 \sin x \) for \( 0 \leq x \leq \frac{\pi}{2} \)

1. **Identify the function**: The function given is \( y = 2 \sin x \).
2. **Determine the range**: The sine function varies from 0 to 1 in the interval \( 0 \leq x \leq \frac{\pi}{2} \). Therefore, \( y \) will vary from \( 0 \) to \( 2 \) in this interval.
3. **Plot key points**:
- At \( x = 0 \), \( y = 2 \sin(0) = 0 \).
- At \( x = \frac{\pi}{2} \), \( y = 2 \sin\left(\frac{\pi}{2}\right) = 2 \).
4. **Draw the curve**: The graph will start at the origin (0,0), rise to the point \(\left(\frac{\pi}{2}, 2\right)\), and will be smooth and continuous.

### Step 2: Determine the Area Enclosed

1. **Identify the area to be calculated**: We need to find the area between the curve \( y = 2 \sin x \), the x-axis, and the line \( x = \frac{\pi}{2} \).
2. **Set up the integral**: The area \( A \) can be calculated using the integral:
\[ A = \int_{0}^{\frac{\pi}{2}} 2 \sin x \, dx \]

### Step 3: Calculate the Integral

1. **Factor out the constant**:
\[ A = 2 \int_{0}^{\frac{\pi}{2}} \sin x \, dx \]
2. **Integrate \( \sin x \)**:
\[ \int \sin x \, dx = -\cos x + C \]
3. **Evaluate the definite integral**:
\[ A = 2 \left[ -\cos x \right]_{0}^{\frac{\pi}{2}} = 2 \left( -\cos\left(\frac{\pi}{2}\right) + \cos(0) \right) \]
- Calculate \( -\cos\left(\frac{\pi}{2}\right) = -0 = 0 \)
- Calculate \( \cos(0) = 1 \)
- Therefore:
\[ A = 2 \left( 0 + 1 \right) = 2 \]

### Final Result

The area enclosed between the curve \( y = 2 \sin x \), the x-axis, and the line \( x = \frac{\pi}{2} \) is \( 2 \) square units.

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