`y=f(x)` satisfies the relation `int_(2)^(x)f(t)dt=(x^(2))/2+int_(x)^(2)t^(2)f(t)dt`
The value of `int_(-2)^(2)f(x)dx` is

Options

A. `0`
B. `-2`
C. `2log_(e)2`
D. none of these

Explanation

To solve the problem step by step, we start with the given relation:

\[ \int_{2}^{x} f(t) \, dt = \frac{x^2}{2} + \int_{x}^{2} t^2 f(t) \, dt \]

### Step 1: Differentiate both sides with respect to \(x\)

Using the Leibniz rule for differentiation under the integral sign, we differentiate the left-hand side and the right-hand side:

\[ \frac{d}{dx} \left( \int_{2}^{x} f(t) \, dt \right) = f(x) \]

For the right-hand side, we differentiate:

\[ \frac{d}{dx} \left( \frac{x^2}{2} + \int_{x}^{2} t^2 f(t) \, dt \right) = x - x^2 f(x) \]

### Step 2: Set the derivatives equal to each other

Equating the derivatives from both sides gives us:

\[ f(x) = x - x^2 f(x) \]

### Step 3: Rearranging the equation

Now, we can rearrange this equation to isolate \(f(x)\):

\[ f(x) + x^2 f(x) = x \]

Factoring out \(f(x)\):

\[ f(x)(1 + x^2) = x \]

### Step 4: Solve for \(f(x)\)

Now, we can solve for \(f(x)\):

\[ f(x) = \frac{x}{1 + x^2} \]

### Step 5: Evaluate the integral \(\int_{-2}^{2} f(x) \, dx\)

Now we need to evaluate:

\[ \int_{-2}^{2} f(x) \, dx = \int_{-2}^{2} \frac{x}{1 + x^2} \, dx \]

### Step 6: Check if \(f(x)\) is an odd function

To determine if the integral from \(-2\) to \(2\) is zero, we check if \(f(x)\) is an odd function:

\[ f(-x) = \frac{-x}{1 + (-x)^2} = \frac{-x}{1 + x^2} = -f(x) \]

Since \(f(-x) = -f(x)\), \(f(x)\) is indeed an odd function.

### Step 7: Conclusion

The integral of an odd function over a symmetric interval around zero is zero:

\[ \int_{-2}^{2} f(x) \, dx = 0 \]

Thus, the value of \(\int_{-2}^{2} f(x) \, dx\) is:

\[ \boxed{0} \]