(i) Make a rough sketch of the graph of the function `y = 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).`
(ii) Find the area bounded by the curve:
`(I) y = sin x" "(II) y = cos x`
bewtween `x =0 and x =2pi.`

Answer

To solve the given problem, we will break it down into two parts as specified in the question.

### Part (i):
**Make a rough sketch of the graph of the function \( y = \sin x \) for \( 0 \leq x \leq \frac{\pi}{2} \) and determine the area enclosed between the curve, the x-axis, and the line \( x = \frac{\pi}{2} \).**

1. **Sketch the graph of \( y = \sin x \)**:
- The graph of \( y = \sin x \) starts at the origin (0,0) when \( x = 0 \).
- At \( x = \frac{\pi}{2} \), \( y = \sin\left(\frac{\pi}{2}\right) = 1 \).
- The graph is a smooth curve that rises from (0,0) to (\(\frac{\pi}{2}\), 1).

2. **Identify the area to be calculated**:
- The area we need to find is bounded by the curve \( y = \sin x \), the x-axis, and the vertical line \( x = \frac{\pi}{2} \).

3. **Set up the integral to find the area**:
- The area \( A \) can be calculated using the integral:
\[ A = \int_{0}^{\frac{\pi}{2}} \sin x \, dx \]

4. **Calculate the integral**:
- The integral of \( \sin x \) is \( -\cos x \).
- Thus, we compute:
\[ A = \left[-\cos x\right]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) = 0 - (-1) = 1 \]

5. **Conclusion**:
- The area enclosed between the curve, the x-axis, and the line \( x = \frac{\pi}{2} \) is \( 1 \) square unit.

### Part (ii):
**Find the area bounded by the curves \( y = \sin x \) and \( y = \cos x \) between \( x = 0 \) and \( x = 2\pi \).**

1. **Sketch the graphs of \( y = \sin x \) and \( y = \cos x \)**:
- The graph of \( y = \sin x \) oscillates between -1 and 1, starting at (0,0).
- The graph of \( y = \cos x \) also oscillates between -1 and 1, starting at (0,1).

2. **Find the points of intersection**:
- Set \( \sin x = \cos x \).
- This occurs when \( x = \frac{\pi}{4} \) and \( x = \frac{5\pi}{4} \) in the interval \( [0, 2\pi] \).

3. **Set up the integral for the area**:
- The area between \( 0 \) and \( \frac{\pi}{4} \) is given by \( \cos x - \sin x \).
- The area between \( \frac{\pi}{4} \) and \( \frac{5\pi}{4} \) is given by \( \sin x - \cos x \).
- The area between \( \frac{5\pi}{4} \) and \( 2\pi \) is again given by \( \cos x - \sin x \).

4. **Calculate the total area**:
- The total area \( A \) is:
\[ A = \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin x - \cos x) \, dx + \int_{\frac{5\pi}{4}}^{2\pi} (\cos x - \sin x) \, dx \]

5. **Evaluate each integral**:
- For \( \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx \):
\[ = \left[\sin x + \cos x\right]_{0}^{\frac{\pi}{4}} = \left(\sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right)\right) - (0 + 1) = \sqrt{2} - 1 \]
- For \( \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin x - \cos x) \, dx \):
\[ = \left[-\cos x - \sin x\right]_{\frac{\pi}{4}}^{\frac{5\pi}{4}} = \left(-\cos\left(\frac{5\pi}{4}\right) - \sin\left(\frac{5\pi}{4}\right)\right) - \left(-\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right)\right) = -(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}) - (-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}) = 2\sqrt{2} \]
- For \( \int_{\frac{5\pi}{4}}^{2\pi} (\cos x - \sin x) \, dx \):
\[ = \left[\sin x + \cos x\right]_{\frac{5\pi}{4}}^{2\pi} = \left(\sin(2\pi) + \cos(2\pi)\right) - \left(\sin\left(\frac{5\pi}{4}\right) + \cos\left(\frac{5\pi}{4}\right)\right) = 1 - (-\sqrt{2}) = 1 + \sqrt{2} \]

6. **Combine the areas**:
- Total area \( A = (\sqrt{2} - 1) + 2\sqrt{2} + (1 + \sqrt{2}) = 4\sqrt{2} \).

### Final Answers:
1. The area enclosed in part (i) is \( 1 \) square unit.
2. The area bounded by the curves in part (ii) is \( 4\sqrt{2} \) square units.