If F(x)`=(1)/(x^(2))int_(4)^(x) [4t^(2)-2F'(t)]dt` then F'(4) equals

Options

A. 32
B. `32//3`
C. `32//9`
D. none of these

Answer

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

\[ F'(x) = \frac{1}{x^2} \int_{4}^{x} [4t^2 - 2F'(t)] dt \]

### Step 1: Separate the integral
We can separate the integral into two parts:

\[ F'(x) = \frac{1}{x^2} \left( \int_{4}^{x} 4t^2 dt - 2 \int_{4}^{x} F'(t) dt \right) \]

### Step 2: Calculate the first integral
The first integral can be calculated as follows:

\[ \int_{4}^{x} 4t^2 dt = 4 \left[ \frac{t^3}{3} \right]_{4}^{x} = 4 \left( \frac{x^3}{3} - \frac{4^3}{3} \right) = \frac{4}{3} (x^3 - 64) \]

### Step 3: Calculate the second integral
The second integral is:

\[ \int_{4}^{x} F'(t) dt = F(x) - F(4) \]

### Step 4: Substitute back into the equation
Substituting both integrals back into the equation gives us:

\[ F'(x) = \frac{1}{x^2} \left( \frac{4}{3} (x^3 - 64) - 2(F(x) - F(4)) \right) \]

### Step 5: Simplify the equation
This simplifies to:

\[ F'(x) = \frac{4}{3x^2} (x^3 - 64) - \frac{2}{x^2} (F(x) - F(4)) \]

### Step 6: Evaluate at \( x = 4 \)
To find \( F'(4) \), we substitute \( x = 4 \):

\[ F'(4) = \frac{4}{3 \cdot 4^2} (4^3 - 64) - \frac{2}{4^2} (F(4) - F(4)) \]

Since \( F(4) - F(4) = 0 \), the second term vanishes:

\[ F'(4) = \frac{4}{3 \cdot 16} (64 - 64) = 0 \]

### Step 7: Evaluate the first term
Now calculating the first term:

\[ F'(4) = \frac{4}{48} \cdot 0 = 0 \]

### Step 8: Solve for \( F'(4) \)
Thus, we find:

\[ F'(4) = 0 \]

### Final Answer
The value of \( F'(4) \) is:

\[ \boxed{2} \]