Consider the polynomial $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$, with degree less than or equal to 2.

When $ f$ varies with subject to the constrain $ f(0) \equal{} 0,\ f(2) \equal{} 2$, find the minimum value of $ S\equal{}\int_0^2 |f'(x)|\ dx$.
1. Given the polynomial \( f(x) = ax^2 + bx + c \) with the constraints \( f(0) = 0 \) and \( f(2) = 2 \), we need to find the minimum value of \( S = \int_0^2 |f'(x)| \, dx \).

2. From the constraint \( f(0) = 0 \), we have:
   \[
   f(0) = a(0)^2 + b(0) + c = 0 \implies c = 0
   \]
   Thus, the polynomial simplifies to:
   \[
   f(x) = ax^2 + bx
   \]

3. From the constraint \( f(2) = 2 \), we have:
   \[
   f(2) = a(2)^2 + b(2) = 2 \implies 4a + 2b = 2 \implies 2a + b = 1
   \]

4. We need to find the minimum value of \( S = \int_0^2 |f'(x)| \, dx \). First, we compute \( f'(x) \):
   \[
   f'(x) = \frac{d}{dx}(ax^2 + bx) = 2ax + b
   \]

5. Using the inequality \( \int_0^2 |f'(x)| \, dx \geq \left| \int_0^2 f'(x) \, dx \right| \), we compute:
   \[
   \int_0^2 f'(x) \, dx = \int_0^2 (2ax + b) \, dx
   \]

6. Evaluating the integral:
   \[
   \int_0^2 (2ax + b) \, dx = \left[ ax^2 + bx \right]_0^2 = (4a + 2b) - (0 + 0) = 4a + 2b
   \]

7. From the constraint \( 2a + b = 1 \), we substitute \( b = 1 - 2a \) into \( 4a + 2b \):
   \[
   4a + 2(1 - 2a) = 4a + 2 - 4a = 2
   \]

8. Therefore, we have:
   \[
   \left| \int_0^2 f'(x) \, dx \right| = |4a + 2b| = |2| = 2
   \]

9. Since \( \int_0^2 |f'(x)| \, dx \geq \left| \int_0^2 f'(x) \, dx \right| \), the minimum value of \( \int_0^2 |f'(x)| \, dx \) is:
   \[
   \int_0^2 |f'(x)| \, dx \geq 2
   \]

10. The minimum value occurs when \( f'(x) \geq 0 \) for all \( x \in [0, 2] \). Given \( f'(x) = 2ax + b \) and substituting \( b = 1 - 2a \), we get:
    \[
    f'(x) = 2ax + (1 - 2a) = 2ax + 1 - 2a
    \]
    For \( f'(x) \geq 0 \) for all \( x \in [0, 2] \), we need:
    \[
    2ax + 1 - 2a \geq 0 \implies 2ax \geq 2a - 1 \implies x \geq \frac{2a - 1}{2a}
    \]
    For \( x \in [0, 2] \), this inequality must hold for \( x = 0 \):
    \[
    0 \geq \frac{2a - 1}{2a} \implies 2a - 1 \leq 0 \implies a \leq \frac{1}{2}
    \]
    And for \( x = 2 \):
    \[
    2a(2) + 1 - 2a \geq 0 \implies 4a + 1 - 2a \geq 0 \implies 2a + 1 \geq 0 \implies a \geq -\frac{1}{2}
    \]
    Therefore, \( -\frac{1}{2} \leq a \leq \frac{1}{2} \).

11. The minimum value of \( \int_0^2 |f'(x)| \, dx \) is achieved when \( a = \frac{1}{2} \) and \( b = 0 \), giving:
    \[
    f(x) = \frac{1}{2}x^2
    \]
    \[
    f'(x) = x
    \]
    \[
    \int_0^2 |f'(x)| \, dx = \int_0^2 x \, dx = \left[ \frac{x^2}{2} \right]_0^2 = 2
    \]

Conclusion:
\[
\boxed{2}
\]