If `int_(0)^(x)f(t)dt=x+int_(x)^(1)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:

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

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

Using the Leibniz rule for differentiation under the integral sign, we differentiate both sides:

\[ \frac{d}{dx} \left( \int_{0}^{x} f(t) \, dt \right) = \frac{d}{dx} \left( x + \int_{x}^{1} f(t) \, dt \right) \]

### Step 2: Apply Leibniz rule

The left-hand side becomes:

\[ f(x) \]

The right-hand side can be differentiated as follows:

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

### Step 3: Set the derivatives equal to each other

Now we have:

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

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

Adding \(f(x)\) to both sides gives:

\[ 2f(x) = 1 \]

Dividing both sides by 2, we find:

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

### Step 5: Find \(f(1)\)

Since \(f(x)\) is a constant function, we can directly find:

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

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

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

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