Area bounded by the curves `y =sin^(4)x,x=0,x= pi/2` and `y=0` is

Options

A. `3//8`
B. `3//16`
C. `3pi//16`
D. None of these

Answer

To find the area bounded by the curves \( y = \sin^4 x \), \( x = 0 \), \( x = \frac{\pi}{2} \), and \( y = 0 \), we can follow these steps:

### Step-by-Step Solution

1. **Set Up the Integral**:
The area \( A \) can be calculated using the integral of the function \( y = \sin^4 x \) from \( x = 0 \) to \( x = \frac{\pi}{2} \).
\[ A = \int_{0}^{\frac{\pi}{2}} \sin^4 x \, dx \]

2. **Use the Power Reduction Formula**:
To integrate \( \sin^4 x \), we can use the power reduction formula:
\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \]
Therefore,
\[ \sin^4 x = \left(\sin^2 x\right)^2 = \left(\frac{1 - \cos(2x)}{2}\right)^2 = \frac{(1 - \cos(2x))^2}{4} \]

3. **Expand the Expression**:
Expanding \( (1 - \cos(2x))^2 \):
\[ (1 - \cos(2x))^2 = 1 - 2\cos(2x) + \cos^2(2x) \]
Using the power reduction again for \( \cos^2(2x) \):
\[ \cos^2(2x) = \frac{1 + \cos(4x)}{2} \]
Thus,
\[ (1 - \cos(2x))^2 = 1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2} = \frac{3}{2} - 2\cos(2x) + \frac{1}{2}\cos(4x) \]

4. **Substitute Back into the Integral**:
Now substituting back into the integral:
\[ A = \int_{0}^{\frac{\pi}{2}} \frac{1}{4} \left( \frac{3}{2} - 2\cos(2x) + \frac{1}{2}\cos(4x) \right) dx \]
Simplifying:
\[ A = \frac{1}{4} \left( \frac{3}{2} \int_{0}^{\frac{\pi}{2}} dx - 2 \int_{0}^{\frac{\pi}{2}} \cos(2x) dx + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos(4x) dx \right) \]

5. **Evaluate Each Integral**:
- The first integral:
\[ \int_{0}^{\frac{\pi}{2}} dx = \frac{\pi}{2} \]
- The second integral:
\[ \int_{0}^{\frac{\pi}{2}} \cos(2x) dx = \left[ \frac{1}{2} \sin(2x) \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} (0 - 0) = 0 \]
- The third integral:
\[ \int_{0}^{\frac{\pi}{2}} \cos(4x) dx = \left[ \frac{1}{4} \sin(4x) \right]_{0}^{\frac{\pi}{2}} = \frac{1}{4} (0 - 0) = 0 \]

6. **Combine the Results**:
Putting it all together:
\[ A = \frac{1}{4} \left( \frac{3}{2} \cdot \frac{\pi}{2} - 2 \cdot 0 + \frac{1}{2} \cdot 0 \right) = \frac{1}{4} \cdot \frac{3\pi}{4} = \frac{3\pi}{16} \]

### Final Answer:
The area bounded by the curves is:
\[ \boxed{\frac{3\pi}{16}} \]