If `f(x)` and `g(x)` are continuous functions, then `int_(In lamda)^(In (1//lamda))(f(x^(2)//4)[f(x)-f(-x)])/(g(x^(2)//4)[g(x)+g(-x)])dx` is

Options

A. depenent on `lamda`
B. a non zero constant
C. zero
D. none of these

Explanation

To solve the integral

\[ I = \int_{\ln \lambda}^{\ln \frac{1}{\lambda}} \frac{f\left(\frac{x^2}{4}\right) \left[f(x) - f(-x)\right]}{g\left(\frac{x^2}{4}\right) \left[g(x) + g(-x)\right]} \, dx, \]

we will analyze the properties of the integrand and use the properties of definite integrals.

### Step 1: Change of Variables
First, we will perform a change of variables. Let \( u = -x \). Then, \( du = -dx \). The limits change as follows:
- When \( x = \ln \lambda \), \( u = -\ln \lambda \).
- When \( x = \ln \frac{1}{\lambda} \), \( u = -\ln \frac{1}{\lambda} = \ln \lambda \).

Thus, the integral becomes:

\[ I = \int_{-\ln \lambda}^{-\ln \frac{1}{\lambda}} \frac{f\left(\frac{(-u)^2}{4}\right) \left[f(-u) - f(u)\right]}{g\left(\frac{(-u)^2}{4}\right) \left[g(-u) + g(u)\right]} (-du). \]

This simplifies to:

\[ I = \int_{-\ln \lambda}^{\ln \lambda} \frac{f\left(\frac{u^2}{4}\right) \left[f(-u) - f(u)\right]}{g\left(\frac{u^2}{4}\right) \left[g(-u) + g(u)\right]} du. \]

### Step 2: Analyze the Integrand
Now, we analyze the integrand. Notice that:
- \( f\left(\frac{u^2}{4}\right) \) is an even function since \( \frac{u^2}{4} \) is even.
- \( f(-u) - f(u) \) is an odd function.
- \( g\left(\frac{u^2}{4}\right) \) is also an even function.
- \( g(-u) + g(u) \) is an even function.

Thus, the entire integrand can be expressed as:

\[ \frac{f\left(\frac{u^2}{4}\right) \cdot \text{(odd function)}}{g\left(\frac{u^2}{4}\right) \cdot \text{(even function)}} \]

This shows that the integrand is an odd function because the product of an even function and an odd function is odd.

### Step 3: Evaluate the Integral
Since \( I \) is the integral of an odd function over a symmetric interval \([-a, a]\), we have:

\[ I = 0. \]

### Conclusion
Thus, the value of the integral is

\[ \boxed{0}. \]