If` int_(0) ^(x) f (t) dt = x + int _(x ) ^(1) t f (t) dt, ` then the value of ` f (1)` , is

Options

A. `1//2`
B. `0`
C. `1`
D. `-1//2`

Explanation

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

\[ \int_{0}^{x} f(t) \, dt = x + \int_{x}^{1} t f(t) \, dt \]

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

Using Leibniz's rule for differentiating under the integral sign, 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:

\[ \frac{d}{dx} \left( x + \int_{x}^{1} t f(t) \, dt \right) = 1 + \frac{d}{dx} \left( \int_{x}^{1} t f(t) \, dt \right) \]

Using Leibniz's rule again for the integral:

\[ \frac{d}{dx} \left( \int_{x}^{1} t f(t) \, dt \right) = -x f(x) \]

Thus, the right-hand side becomes:

\[ 1 - x f(x) \]

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

Now we equate the derivatives from both sides:

\[ f(x) = 1 - x f(x) \]

### Step 3: Solve for \( f(x) \)

Rearranging the equation gives:

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

Factoring out \( f(x) \):

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

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

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

### Step 4: Find \( f(1) \)

Now we need to find \( f(1) \):

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

Thus, the value of \( f(1) \) is:

\[ \boxed{\frac{1}{2}} \]