If `f(x)=x+int_0^1t(x+t)f(t)`dt ,then the value of `23/2f(0)` is equal to _________

Explanation

To solve the problem, we need to evaluate the function defined by the integral equation and then find the value of \( \frac{23}{2} f(0) \).

### Step 1: Define the function
We start with the given function:
\[ f(x) = x + \int_0^1 t(x + t)f(t) \, dt \]

### Step 2: Substitute \( x = 0 \)
To find \( f(0) \), we substitute \( x = 0 \) into the equation:
\[ f(0) = 0 + \int_0^1 t(0 + t)f(t) \, dt = \int_0^1 t^2 f(t) \, dt \]
This simplifies to:
\[ f(0) = \int_0^1 t^2 f(t) \, dt \]

### Step 3: Express \( f(t) \)
We can express \( f(t) \) in terms of \( t \):
\[ f(t) = t + \int_0^1 s(t + s)f(s) \, ds \]
This means we need to evaluate \( f(t) \) in a similar manner.

### Step 4: Substitute \( f(t) \) into the integral
Substituting \( f(t) \) into the integral:
\[ f(t) = t + \int_0^1 t^2 f(t) \, dt + \int_0^1 t(s^2 + ts)f(s) \, ds \]
This gives us a recursive relationship.

### Step 5: Define constants
Let:
\[ A = \int_0^1 t^2 f(t) \, dt \]
Then we can express \( f(0) \) as:
\[ f(0) = A \]

### Step 6: Set up equations
From the previous steps, we have:
\[ f(0) = A \quad \text{and} \quad f(t) = t + A \]
Substituting \( f(t) \) back into the integral:
\[ A = \int_0^1 t^2 (t + A) \, dt = \int_0^1 (t^3 + At^2) \, dt \]
Calculating the integrals:
\[ A = \left[ \frac{t^4}{4} \right]_0^1 + A \left[ \frac{t^3}{3} \right]_0^1 = \frac{1}{4} + \frac{A}{3} \]

### Step 7: Solve for \( A \)
Rearranging gives:
\[ A - \frac{A}{3} = \frac{1}{4} \]
This simplifies to:
\[ \frac{2A}{3} = \frac{1}{4} \]
Multiplying both sides by \( \frac{3}{2} \):
\[ A = \frac{3}{8} \]

### Step 8: Find \( f(0) \)
Thus, we have:
\[ f(0) = \frac{3}{8} \]

### Step 9: Calculate \( \frac{23}{2} f(0) \)
Now, we can find \( \frac{23}{2} f(0) \):
\[ \frac{23}{2} f(0) = \frac{23}{2} \cdot \frac{3}{8} = \frac{69}{16} \]

### Final Answer
The value of \( \frac{23}{2} f(0) \) is:
\[ \frac{69}{16} \]