Find the sum of all possible integral values of in [1,100] for which the function `f(x)=2x^(3)-3(2+alpha)x^(2)` has exactly one local maximum and one local minimum.

Answer

To find the sum of all possible integral values of \(\alpha\) in the interval \([1, 100]\) for which the function \(f(x) = 2x^3 - 3(2 + \alpha)x^2\) has exactly one local maximum and one local minimum, we can follow these steps:

### Step 1: Find the first derivative of the function
The first step is to differentiate the function \(f(x)\):

\[ f'(x) = \frac{d}{dx}(2x^3 - 3(2 + \alpha)x^2) \]

Using the power rule, we get:

\[ f'(x) = 6x^2 - 6(2 + \alpha)x \]

### Step 2: Set the first derivative to zero
To find the critical points, we set the first derivative equal to zero:

\[ 6x^2 - 6(2 + \alpha)x = 0 \]

Factoring out \(6x\):

\[ 6x(x - (2 + \alpha)) = 0 \]

This gives us two roots:

1. \(x = 0\)
2. \(x = 2 + \alpha\)

### Step 3: Determine conditions for local extrema
For the function to have exactly one local maximum and one local minimum, the quadratic \(f'(x)\) must have two distinct roots. This means that the root \(x = 2 + \alpha\) must not be equal to zero:

\[ 2 + \alpha \neq 0 \]

This simplifies to:

\[ \alpha \neq -2 \]

### Step 4: Analyze the interval for \(\alpha\)
Given that \(\alpha\) must be in the interval \([1, 100]\), we see that \(-2\) is not in this interval. Therefore, \(\alpha\) can take any integral value from \(1\) to \(100\).

### Step 5: Calculate the sum of integral values of \(\alpha\)
The integral values of \(\alpha\) are \(1, 2, 3, \ldots, 100\). We need to find the sum of these values:

\[ \text{Sum} = 1 + 2 + 3 + \ldots + 100 \]

This is the sum of the first \(n\) natural numbers, which can be calculated using the formula:

\[ S_n = \frac{n(n + 1)}{2} \]

For \(n = 100\):

\[ S_{100} = \frac{100 \times (100 + 1)}{2} = \frac{100 \times 101}{2} = 5050 \]

### Final Answer
The sum of all possible integral values of \(\alpha\) in the interval \([1, 100]\) is:

\[ \boxed{5050} \]