If `int_(0)^(x) f(t)dt=x^(2)+2x-int_(0)^(x) tf(t)dt, x in (0,oo)`. Then, f(x) is

Options

A. Periodic
B. Periodic but fundamental does not exist
C. Periodic but fundamental period exits
D. Nothing can be said

Answer

To solve the problem, we start with the given equation:

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

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

Using the Fundamental Theorem of Calculus, we differentiate the left-hand side and the right-hand side:

\[ \frac{d}{dx} \left( \int_{0}^{x} f(t) \, dt \right) = f(x) \]

For the right-hand side, we differentiate each term:

1. The derivative of \(x^2\) is \(2x\).
2. The derivative of \(2x\) is \(2\).
3. For the term \(-\int_{0}^{x} t f(t) \, dt\), we apply the Leibniz rule:

\[ -\frac{d}{dx} \left( \int_{0}^{x} t f(t) \, dt \right) = -x f(x) \]

Thus, the differentiation of the right-hand side gives:

\[ 2x + 2 - x f(x) \]

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

Now we equate both sides:

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

### Step 3: Rearrange the equation

We can rearrange this equation to isolate \(f(x)\):

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

Factoring out \(f(x)\) gives:

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

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

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

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

### Step 5: Simplify \(f(x)\)

We can simplify \(f(x)\):

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

Thus, we find that:

\[ f(x) = 2 \]

### Conclusion

The function \(f(x)\) is a constant function equal to 2.