Draw a rough sketch of `y=sin 2x` and determine the area enclosed by the curve. X-axis and the lines `x=pi//4` and `x=3pi//4`.

Answer

To solve the problem of finding the area enclosed by the curve \( y = \sin(2x) \), the x-axis, and the lines \( x = \frac{\pi}{4} \) and \( x = \frac{3\pi}{4} \), we will follow these steps:

### Step 1: Sketch the graph of \( y = \sin(2x) \)

1. **Identify key points**: The function \( y = \sin(2x) \) oscillates between -1 and 1. It completes one full cycle from \( 0 \) to \( \pi \).
- At \( x = 0 \), \( y = \sin(0) = 0 \)
- At \( x = \frac{\pi}{4} \), \( y = \sin\left(\frac{\pi}{2}\right) = 1 \)
- At \( x = \frac{\pi}{2} \), \( y = \sin(\pi) = 0 \)
- At \( x = \frac{3\pi}{4} \), \( y = \sin\left(\frac{3\pi}{2}\right) = -1 \)
- At \( x = \pi \), \( y = \sin(2\pi) = 0 \)

2. **Draw the graph**: Plot these points on the coordinate system and connect them smoothly to represent the sine wave.

### Step 2: Set up the integral for the area

The area \( A \) enclosed by the curve and the x-axis from \( x = \frac{\pi}{4} \) to \( x = \frac{3\pi}{4} \) can be calculated using the integral:

\[ A = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} |\sin(2x)| \, dx \]

### Step 3: Determine the intervals for integration

1. **Identify the behavior of \( \sin(2x) \)**:
- From \( x = \frac{\pi}{4} \) to \( x = \frac{\pi}{2} \), \( \sin(2x) \) is positive.
- From \( x = \frac{\pi}{2} \) to \( x = \frac{3\pi}{4} \), \( \sin(2x) \) is negative.

2. **Split the integral**:
\[ A = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(2x) \, dx - \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \sin(2x) \, dx \]

### Step 4: Calculate the integrals

1. **First integral**:
\[ \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C \]
Thus,
\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(2x) \, dx = \left[-\frac{1}{2} \cos(2x)\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = -\frac{1}{2} \left(\cos(\pi) - \cos\left(\frac{\pi}{2}\right)\right) = -\frac{1}{2} \left(-1 - 0\right) = \frac{1}{2} \]

2. **Second integral**:
\[ \int_{\frac{\pi}{2}}^{\frac{3\pi}{4}} \sin(2x) \, dx = \left[-\frac{1}{2} \cos(2x)\right]_{\frac{\pi}{2}}^{\frac{3\pi}{4}} = -\frac{1}{2} \left(\cos\left(\frac{3\pi}{2}\right) - \cos(\pi)\right) = -\frac{1}{2} \left(0 - (-1)\right) = -\frac{1}{2} \cdot 1 = -\frac{1}{2} \]

### Step 5: Combine the areas

Now, substituting back into the area formula:
\[ A = \frac{1}{2} - \left(-\frac{1}{2}\right) = \frac{1}{2} + \frac{1}{2} = 1 \]

### Conclusion

The area enclosed by the curve \( y = \sin(2x) \), the x-axis, and the lines \( x = \frac{\pi}{4} \) and \( x = \frac{3\pi}{4} \) is \( 1 \) square unit.

---