If `e^-x int_3^x{3t^2+2t+4f'(t)}dt=f(x) and f'(4)=(alphae^beta-224)/(e^beta-4)^2` then value of `(alpha+beta)`

Answer

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

\[ e^{-x} \int_3^x (3t^2 + 2t + 4f'(t)) \, dt = f(x) \]

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

Using the Leibniz rule for differentiation under the integral sign, we differentiate the left-hand side:

\[ \frac{d}{dx} \left( e^{-x} \int_3^x (3t^2 + 2t + 4f'(t)) \, dt \right) \]

Using the product rule, we have:

\[ \frac{d}{dx}(e^{-x}) \cdot \int_3^x (3t^2 + 2t + 4f'(t)) \, dt + e^{-x} \cdot (3x^2 + 2x + 4f'(x)) \]

This simplifies to:

\[ -e^{-x} \int_3^x (3t^2 + 2t + 4f'(t)) \, dt + e^{-x}(3x^2 + 2x + 4f'(x)) \]

Setting this equal to \( f'(x) \):

\[ f'(x) = -e^{-x} \int_3^x (3t^2 + 2t + 4f'(t)) \, dt + e^{-x}(3x^2 + 2x + 4f'(x)) \]

### Step 2: Evaluate \( f(3) \)

To find \( f(3) \), we substitute \( x = 3 \) into the original equation:

\[ f(3) = e^{-3} \int_3^3 (3t^2 + 2t + 4f'(t)) \, dt = e^{-3} \cdot 0 = 0 \]

### Step 3: Substitute \( f(3) \) into the equation

Now substituting \( f(3) = 0 \) into the equation we derived earlier:

\[ x^3 + x^2 + 4f(x) - 36 = e^x f(x) \]

Rearranging gives:

\[ x^3 + x^2 - 36 = e^x f(x) - 4f(x) \]

Factoring out \( f(x) \):

\[ f(x) = \frac{x^3 + x^2 - 36}{e^x - 4} \]

### Step 4: Differentiate \( f(x) \)

Using the quotient rule to differentiate \( f(x) \):

Let \( u = x^3 + x^2 - 36 \) and \( v = e^x - 4 \).

\[ f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \]

Calculating \( u' \) and \( v' \):

\[ u' = 3x^2 + 2, \quad v' = e^x \]

Thus,

\[ f'(x) = \frac{(e^x - 4)(3x^2 + 2) - (x^3 + x^2 - 36)e^x}{(e^x - 4)^2} \]

### Step 5: Evaluate \( f'(4) \)

Substituting \( x = 4 \):

\[ f'(4) = \frac{(e^4 - 4)(3 \cdot 4^2 + 2) - (4^3 + 4^2 - 36)e^4}{(e^4 - 4)^2} \]

Calculating:

\[ = \frac{(e^4 - 4)(48 + 2) - (64 + 16 - 36)e^4}{(e^4 - 4)^2} \]
\[ = \frac{(e^4 - 4)(50) - 44e^4}{(e^4 - 4)^2} \]
\[ = \frac{50e^4 - 200 - 44e^4}{(e^4 - 4)^2} \]
\[ = \frac{6e^4 - 200}{(e^4 - 4)^2} \]

### Step 6: Compare with given expression

We are given:

\[ f'(4) = \frac{\alpha e^\beta - 224}{(e^\beta - 4)^2} \]

From our calculation, we have:

\[ \frac{6e^4 - 200}{(e^4 - 4)^2} \]

Setting \( \alpha = 6 \) and \( \beta = 4 \) gives:

\[ \alpha + \beta = 6 + 4 = 10 \]

### Final Answer

Thus, the value of \( \alpha + \beta \) is:

\[ \boxed{10} \]