For a function satisfying $ f'(x) > 0$ for $ a\leq x\leq b$, let $ F(x) \equal{} \int_a^b |f(t) \minus{} f(x)|\ dt$. For what value of $ x$ is $ F(x)$ is minimized?
1. Given that \( f'(x) > 0 \) for \( a \leq x \leq b \), the function \( f(x) \) is strictly increasing on the interval \([a, b]\).
2. We need to find the value of \( x \) that minimizes the function \( F(x) = \int_a^b |f(t) - f(x)| \, dt \).

3. To find the minimum of \( F(x) \), we first differentiate \( F(x) \) with respect to \( x \). Using the Leibniz rule for differentiation under the integral sign, we have:
   \[
   \frac{dF}{dx} = \frac{d}{dx} \left( \int_a^b |f(t) - f(x)| \, dt \right)
   \]

4. We split the integral into two parts:
   \[
   F(x) = \int_a^x |f(t) - f(x)| \, dt + \int_x^b |f(t) - f(x)| \, dt
   \]

5. Since \( f(x) \) is strictly increasing, for \( t \leq x \), \( f(t) \leq f(x) \) and for \( t \geq x \), \( f(t) \geq f(x) \). Therefore, the absolute value can be simplified:
   \[
   F(x) = \int_a^x (f(x) - f(t)) \, dt + \int_x^b (f(t) - f(x)) \, dt
   \]

6. Differentiating \( F(x) \) with respect to \( x \):
   \[
   \frac{dF}{dx} = \frac{d}{dx} \left( \int_a^x (f(x) - f(t)) \, dt + \int_x^b (f(t) - f(x)) \, dt \right)
   \]

7. Applying the Leibniz rule:
   \[
   \frac{dF}{dx} = \left[ (f(x) - f(t)) \bigg|_{t=a}^{t=x} + \int_a^x \frac{\partial}{\partial x} (f(x) - f(t)) \, dt \right] + \left[ (f(t) - f(x)) \bigg|_{t=x}^{t=b} + \int_x^b \frac{\partial}{\partial x} (f(t) - f(x)) \, dt \right]
   \]

8. Simplifying the boundary terms and the integrals:
   \[
   \frac{dF}{dx} = (f(x) - f(x)) + \int_a^x f'(x) \, dt - \int_x^b f'(x) \, dt
   \]
   \[
   \frac{dF}{dx} = \int_a^x f'(x) \, dt - \int_x^b f'(x) \, dt
   \]

9. Since \( f'(x) \) is constant with respect to \( t \):
   \[
   \frac{dF}{dx} = f'(x) \left( \int_a^x dt - \int_x^b dt \right)
   \]
   \[
   \frac{dF}{dx} = f'(x) \left( x - a - (b - x) \right)
   \]
   \[
   \frac{dF}{dx} = f'(x) (2x - a - b)
   \]

10. Setting the derivative to zero to find the critical points:
    \[
    f'(x) (2x - a - b) = 0
    \]

11. Since \( f'(x) > 0 \), we have:
    \[
    2x - a - b = 0
    \]
    \[
    x = \frac{a + b}{2}
    \]

Conclusion:
The value of \( x \) that minimizes \( F(x) \) is \( x = \frac{a + b}{2} \).

The final answer is \( \boxed{ x = \frac{a + b}{2} } \).