Let `F:[3,5] rarr R` be a twice differentiable function on (3,5) such that `F(x)= e^(-x) int_(3)^(x) (3t^(2)+2t+4F'(t)) dt`. If `F'(4)= (alpha e^(beta)-224)/((e^(beta)-4)^(2))`, then `alpha+beta` is equal to _____

Answer

To solve the problem, we need to analyze the given function and its properties step by step.

### Step 1: Understand the given function
We have the function defined as:
\[ F(x) = e^{-x} \int_{3}^{x} (3t^2 + 2t + 4F'(t)) dt \]

### Step 2: Differentiate both sides
To find \( F'(x) \), we will differentiate both sides using the product rule and the Fundamental Theorem of Calculus:
\[ F'(x) = \frac{d}{dx} \left( e^{-x} \int_{3}^{x} (3t^2 + 2t + 4F'(t)) dt \right) \]
Using the product rule:
\[ F'(x) = e^{-x} \cdot \left( 3x^2 + 2x + 4F'(x) \right) - e^{-x} \int_{3}^{x} (3t^2 + 2t + 4F'(t)) dt \]

### Step 3: Evaluate \( F(3) \)
To find \( F(3) \), we substitute \( x = 3 \) into the equation:
\[ F(3) = e^{-3} \int_{3}^{3} (3t^2 + 2t + 4F'(t)) dt = 0 \]
This is because the integral from 3 to 3 is zero.

### Step 4: Substitute \( F(3) \) back into the equation
Now we substitute \( F(3) = 0 \) into the equation for \( F(x) \):
\[ F(x) = e^{-x} \left( \int_{3}^{x} (3t^2 + 2t) dt + 4 \cdot 0 \right) = e^{-x} \left( \int_{3}^{x} (3t^2 + 2t) dt \right) \]

### Step 5: Calculate the integral
Now we compute the integral:
\[ \int (3t^2 + 2t) dt = t^3 + t^2 + C \]
Evaluating from 3 to \( x \):
\[ \int_{3}^{x} (3t^2 + 2t) dt = \left[ t^3 + t^2 \right]_{3}^{x} = (x^3 + x^2) - (27 + 9) = x^3 + x^2 - 36 \]

### Step 6: Substitute back into \( F(x) \)
Thus, we have:
\[ F(x) = e^{-x} (x^3 + x^2 - 36) \]

### Step 7: Differentiate \( F(x) \) to find \( F'(x) \)
Now we differentiate \( F(x) \):
\[ F'(x) = e^{-x} (3x^2 + 2x - 36) - e^{-x} (x^3 + x^2 - 36) \]
This simplifies to:
\[ F'(x) = e^{-x} \left( (3x^2 + 2x - 36) - (x^3 + x^2 - 36) \right) \]
\[ = e^{-x} \left( -x^3 + 2x^2 + 2x \right) \]

### Step 8: Evaluate \( F'(4) \)
Now we substitute \( x = 4 \):
\[ F'(4) = e^{-4} \left( -4^3 + 2 \cdot 4^2 + 2 \cdot 4 \right) \]
Calculating:
\[ = e^{-4} \left( -64 + 32 + 8 \right) = e^{-4} (-24) \]

### Step 9: Relate \( F'(4) \) to the given expression
We know:
\[ F'(4) = \frac{\alpha e^{\beta} - 224}{(e^{\beta} - 4)^2} \]
Setting this equal to our expression for \( F'(4) \):
\[ -24 e^{4} = \frac{\alpha e^{\beta} - 224}{(e^{\beta} - 4)^2} \]

### Step 10: Identify \( \alpha \) and \( \beta \)
From the equation, we can see that \( \alpha = -24 \) and \( \beta = 4 \).

### Step 11: Calculate \( \alpha + \beta \)
Thus, we find:
\[ \alpha + \beta = -24 + 4 = -20 \]

### Final Answer
The value of \( \alpha + \beta \) is:
\[ \boxed{-20} \]