Consider the polynomial `f(x)=a x^2+b x+cdot` If `f(0),f(2)=2,` then the minimum value of `int_0^2|f^(prime)(x)dxi s___`

Explanation

To solve the problem, we need to find the minimum value of the integral \( \int_0^2 |f'(x)| \, dx \) given that \( f(0) = 0 \) and \( f(2) = 2 \) for the polynomial \( f(x) = ax^2 + bx + c \).

### Step-by-Step Solution:

1. **Define the Polynomial**:
The polynomial is given as:
\[ f(x) = ax^2 + bx + c \]

2. **Use the Conditions**:
We know:
- \( f(0) = c = 0 \)
- \( f(2) = 4a + 2b + c = 2 \)

Since \( c = 0 \), we can simplify the second condition:
\[ 4a + 2b = 2 \]
This can be rewritten as:
\[ 2a + b = 1 \quad \text{(Equation 1)} \]

3. **Find the Derivative**:
The derivative of the polynomial is:
\[ f'(x) = 2ax + b \]

4. **Set Up the Integral**:
We need to evaluate:
\[ \int_0^2 |f'(x)| \, dx = \int_0^2 |2ax + b| \, dx \]

5. **Determine Where \( f'(x) \) Changes Sign**:
To find where \( f'(x) = 0 \):
\[ 2ax + b = 0 \implies x = -\frac{b}{2a} \]

We need to check if this point lies within the interval \([0, 2]\).

6. **Evaluate the Integral**:
Depending on the sign of \( a \):
- If \( a > 0 \), \( f'(x) \) is increasing, and we evaluate the integral based on whether \( -\frac{b}{2a} \) is in \([0, 2]\).
- If \( a < 0 \), \( f'(x) \) is decreasing.

For simplicity, let's assume \( a = 0 \) (linear case):
\[ f'(x) = b \]
Then:
\[ \int_0^2 |b| \, dx = 2|b| \]

7. **Using Equation 1**:
From Equation 1, we have \( b = 1 - 2a \). Thus:
\[ 2|b| = 2|1 - 2a| \]

8. **Minimize the Integral**:
To minimize \( 2|1 - 2a| \), we find the value of \( a \) that makes \( 1 - 2a = 0 \):
\[ 1 - 2a = 0 \implies a = \frac{1}{2} \]
Substituting \( a = \frac{1}{2} \) into Equation 1 gives:
\[ b = 1 - 2 \cdot \frac{1}{2} = 0 \]

9. **Final Evaluation**:
Thus, substituting \( a \) and \( b \) back, we find:
\[ \int_0^2 |f'(x)| \, dx = 2|0| = 0 \]
However, since we need \( f(2) = 2 \), we find that the minimum value of the integral is actually:
\[ \int_0^2 |f'(x)| \, dx \geq 2 \]

### Conclusion:
The minimum value of \( \int_0^2 |f'(x)| \, dx \) is:
\[ \boxed{2} \]