Consider the intergral `A=int_(0)^(1)(e^(x)-1)/(x)dx` and `B=int_(0)^(1)(x)/(e^(2)-1)dx`. Then, which of the following is incorrect?

Options

A. `B lt 1`
B. `A gt 1`
C. `B gt A`
D. `A gt (1)/(2)`

Answer

To solve the problem, we need to analyze the two integrals \( A \) and \( B \) given by:

\[ A = \int_{0}^{1} \frac{e^x - 1}{x} \, dx \]

\[ B = \int_{0}^{1} \frac{x}{e^x - 1} \, dx \]

We want to determine which of the statements regarding these integrals is incorrect.

### Step 1: Analyze Integral \( A \)

The integral \( A \) can be rewritten using the Taylor series expansion for \( e^x \):

\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

Thus,

\[ e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

Dividing by \( x \):

\[ \frac{e^x - 1}{x} = 1 + \frac{x}{2!} + \frac{x^2}{3!} + \ldots \]

Now, we can integrate term by term from 0 to 1:

\[ A = \int_{0}^{1} \left(1 + \frac{x}{2!} + \frac{x^2}{3!} + \ldots \right) dx \]

Calculating each term:

\[ A = \left[ x + \frac{x^2}{2 \cdot 2!} + \frac{x^3}{3 \cdot 3!} + \ldots \right]_{0}^{1} = 1 + \frac{1}{4} + \frac{1}{18} + \ldots \]

This series converges to a value greater than 1.

### Step 2: Analyze Integral \( B \)

For the integral \( B \):

\[ B = \int_{0}^{1} \frac{x}{e^x - 1} \, dx \]

Using the series expansion for \( e^x - 1 \):

\[ \frac{x}{e^x - 1} = \frac{x}{x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots} = \frac{1}{1 + \frac{x}{2!} + \frac{x^2}{3!} + \ldots} \]

This function is always less than 1 for \( x \) in \( (0, 1) \). Therefore, we can conclude that:

\[ B < 1 \]

### Step 3: Compare \( A \) and \( B \)

From the analysis, we have:

- \( A > 1 \)
- \( B < 1 \)

Thus, we can conclude that:

\[ A > B \]

### Conclusion

Given that \( A > B \), any statement claiming \( B > A \) is incorrect. Therefore, if we are asked which statement is incorrect, the one stating \( B > A \) is the incorrect statement.

### Final Answer

The incorrect statement is \( C: B > A \).