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

Options

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

Answer

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

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

### Step 1: Differentiate both sides with respect to \(x\)
We will apply the Fundamental Theorem of Calculus on the left side and use the Leibniz rule on the right side.

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

For the right 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 the Leibniz rule, we have:

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

Thus, the right side becomes:

\[ 1 - x f(x) \]

Setting both sides equal gives us:

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

### Step 2: Rearranging the equation
Now we rearrange the equation to isolate \(f(x)\):

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

Factoring out \(f(x)\):

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

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

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

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