Let $ f(x)$ be a function such that $ 1\minus{}f(x)\equal{}f(1\minus{}x)$ for $ 0\leq x\leq 1$.

Evaluate $ \int_0^1 f(x)\ dx$.
1. Given the functional equation \( 1 - f(x) = f(1 - x) \) for \( 0 \leq x \leq 1 \), we can rewrite it as:
   \[
   f(x) + f(1 - x) = 1
   \]
   This equation will be useful in evaluating the integral.

2. Consider the integral \( \int_0^1 f(x) \, dx \). We can use the property of the function \( f \) to split the integral:
   \[
   \int_0^1 f(x) \, dx = \int_0^{1/2} f(x) \, dx + \int_{1/2}^1 f(x) \, dx
   \]

3. Make a substitution in the second integral. Let \( u = 1 - x \). Then \( du = -dx \), and when \( x = 1/2 \), \( u = 1/2 \), and when \( x = 1 \), \( u = 0 \). Thus:
   \[
   \int_{1/2}^1 f(x) \, dx = \int_{1/2}^0 f(1 - u) (-du) = \int_0^{1/2} f(1 - u) \, du
   \]

4. Using the given functional equation \( f(1 - u) = 1 - f(u) \), we can rewrite the integral:
   \[
   \int_0^{1/2} f(1 - u) \, du = \int_0^{1/2} (1 - f(u)) \, du
   \]

5. Now, combine the integrals:
   \[
   \int_0^1 f(x) \, dx = \int_0^{1/2} f(x) \, dx + \int_0^{1/2} (1 - f(u)) \, du
   \]

6. Since \( x \) and \( u \) are dummy variables, we can combine the integrals:
   \[
   \int_0^1 f(x) \, dx = \int_0^{1/2} f(x) \, dx + \int_0^{1/2} 1 \, du - \int_0^{1/2} f(u) \, du
   \]

7. Notice that \( \int_0^{1/2} f(x) \, dx = \int_0^{1/2} f(u) \, du \), so these terms cancel out:
   \[
   \int_0^1 f(x) \, dx = \int_0^{1/2} 1 \, du
   \]

8. Evaluate the remaining integral:
   \[
   \int_0^{1/2} 1 \, du = \left. u \right|_0^{1/2} = \frac{1}{2}
   \]

Thus, the value of the integral \( \int_0^1 f(x) \, dx \) is \( \frac{1}{2} \).

The final answer is \( \boxed{\frac{1}{2}} \).