The area bounded by `y = (sin x)/(x)`, x-axis and the ordinates x = 0, `x = pi//4` is equal to `pi//4` less than `pi//4` is

Options

A. equal to `pi//4`
B. less than `pi//4`
C. greater than `pi//4`
D. less than `underset(0)overset(x//4)(int) (tan x)/(x) dx`

Answer

To find the area bounded by the curve \( y = \frac{\sin x}{x} \), the x-axis, and the ordinates \( x = 0 \) and \( x = \frac{\pi}{4} \), we need to evaluate the definite integral of the function from \( 0 \) to \( \frac{\pi}{4} \).

### Step-by-Step Solution:

1. **Set up the integral**: The area \( A \) can be expressed as:
\[ A = \int_{0}^{\frac{\pi}{4}} \frac{\sin x}{x} \, dx \]

2. **Evaluate the limit as \( x \) approaches 0**: We know that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
This indicates that the function is continuous at \( x = 0 \) and the area can be calculated without any issues.

3. **Use Taylor Series Expansion**: The Taylor series expansion for \( \sin x \) around \( x = 0 \) is:
\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \]
Therefore, we can write:
\[ \frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots \]

4. **Substitute the series into the integral**:
\[ A = \int_{0}^{\frac{\pi}{4}} \left( 1 - \frac{x^2}{6} + \frac{x^4}{120} - \cdots \right) dx \]

5. **Integrate term by term**:
\[ A = \left[ x - \frac{x^3}{18} + \frac{x^5}{600} - \cdots \right]_{0}^{\frac{\pi}{4}} \]

6. **Evaluate at the upper limit \( x = \frac{\pi}{4} \)**:
\[ A = \left( \frac{\pi}{4} - \frac{(\frac{\pi}{4})^3}{18} + \frac{(\frac{\pi}{4})^5}{600} - \cdots \right) - \left( 0 \right) \]

7. **Approximate the value**: The first term \( \frac{\pi}{4} \) is dominant, and the subsequent terms will contribute less. Thus, we can conclude that:
\[ A < \frac{\pi}{4} \]

8. **Conclusion**: The area bounded by the curve \( y = \frac{\sin x}{x} \), the x-axis, and the ordinates \( x = 0 \) and \( x = \frac{\pi}{4} \) is less than \( \frac{\pi}{4} \).