Let `f : R rarr R` be a differentiable function satisfying `f(x) = f(y) f(x - y), AA x, y in R and f'(0) = int_(0)^(4) {2x}dx`, where {.} denotes the fractional part function and `f'(-3) = alpha e^(beta)`. Then, `|alpha + beta|` is equal to.......

Answer

To solve the problem step by step, we will follow the reasoning and calculations presented in the video transcript.

### Step 1: Calculate \( f'(0) \)
We are given that:

\[ f'(0) = \int_{0}^{4} \{2x\} \, dx \]

The fractional part function \(\{x\}\) is periodic with a period of 1. Therefore, we can break the integral into intervals of 1:

\[ \int_{0}^{4} \{2x\} \, dx = \int_{0}^{0.5} \{2x\} \, dx + \int_{0.5}^{1.5} \{2x\} \, dx + \int_{1.5}^{2.5} \{2x\} \, dx + \int_{2.5}^{3.5} \{2x\} \, dx + \int_{3.5}^{4} \{2x\} \, dx \]

Calculating each part:

1. For \(x \in [0, 0.5]\), \(\{2x\} = 2x\):
\[ \int_{0}^{0.5} 2x \, dx = 2 \cdot \frac{x^2}{2} \bigg|_{0}^{0.5} = \frac{1}{2} \]

2. For \(x \in [0.5, 1.5]\), \(\{2x\} = 2x - 1\):
\[ \int_{0.5}^{1.5} (2x - 1) \, dx = \left[ x^2 - x \right]_{0.5}^{1.5} = (2.25 - 1.5) - (0.25 - 0.5) = 0.75 + 0.25 = 1 \]

3. For \(x \in [1.5, 2.5]\), \(\{2x\} = 2x - 2\):
\[ \int_{1.5}^{2.5} (2x - 2) \, dx = \left[ x^2 - 2x \right]_{1.5}^{2.5} = (6.25 - 5) - (2.25 - 3) = 1.25 + 0.75 = 2 \]

4. For \(x \in [2.5, 3.5]\), \(\{2x\} = 2x - 3\):
\[ \int_{2.5}^{3.5} (2x - 3) \, dx = \left[ x^2 - 3x \right]_{2.5}^{3.5} = (12.25 - 10.5) - (6.25 - 7.5) = 1.75 + 1.25 = 3 \]

5. For \(x \in [3.5, 4]\), \(\{2x\} = 2x - 7\):
\[ \int_{3.5}^{4} (2x - 7) \, dx = \left[ x^2 - 7x \right]_{3.5}^{4} = (16 - 28) - (12.25 - 24.5) = -12 + 12.25 = 0.25 \]

Now, summing all these contributions:
\[ f'(0) = \frac{1}{2} + 1 + 2 + 3 + 0.25 = 6.75 \]

### Step 2: Analyzing the functional equation
Given that:

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

Let’s set \(y = 0\):
\[ f(x) = f(0) f(x) \implies f(0) = 1 \]

### Step 3: Deriving the function
From the functional equation, we can differentiate both sides with respect to \(x\):
\[ f'(x) = f'(y) f(x - y) + f(y) f'(x - y)(1) \]

Setting \(y = 0\):
\[ f'(x) = f'(0) f(x) + f(0) f'(x) \implies f'(x) = 2f'(0) f(x) \]

This leads us to the differential equation:
\[ \frac{f'(x)}{f(x)} = 2f'(0) \]

Integrating both sides gives:
\[ \ln |f(x)| = 2f'(0)x + C \implies f(x) = e^{2f'(0)x + C} = Ae^{2f'(0)x} \]

Where \(A = e^C\).

### Step 4: Finding \(f'(-3)\)
Differentiating \(f(x)\):
\[ f'(x) = 2f'(0)Ae^{2f'(0)x} \]

Now substituting \(x = -3\):
\[ f'(-3) = 2f'(0)Ae^{-6f'(0)} \]

Given \(f'(-3) = \alpha e^{\beta}\), we can equate:
\[ \alpha = 2f'(0)A, \quad \beta = -6f'(0) \]

### Step 5: Calculate \(|\alpha + \beta|\)
We know \(f'(0) = 6.75\):
\[ \alpha = 2 \cdot 6.75 \cdot A = 13.5A, \quad \beta = -6 \cdot 6.75 = -40.5 \]

Thus:
\[ |\alpha + \beta| = |13.5A - 40.5| \]

Assuming \(A = 1\) for simplicity:
\[ |\alpha + \beta| = |13.5 - 40.5| = | -27 | = 27 \]

### Final Answer
The value of \(|\alpha + \beta|\) is:

\[ \boxed{27} \]