Let $a,b$ be constant numbers such that $0<a<b.$ If a function $f(x)$ always satisfies $f'(x) >0$ at $a<x<b,$ for $a<t<b$ find the value of $t$ for which the following the integral is minimized.

\[ \int_a^b |f(x)-f(t)|x\ dx. \]
1. Given that \( f'(x) > 0 \) for \( a < x < b \), we know that \( f(x) \) is an increasing function on the interval \( (a, b) \).

2. We need to minimize the integral:
   \[
   \int_a^b |f(x) - f(t)| x \, dx
   \]
   for \( a < t < b \).

3. We can split the integral into two parts:
   \[
   \int_a^b |f(x) - f(t)| x \, dx = \int_a^t (f(t) - f(x)) x \, dx + \int_t^b (f(x) - f(t)) x \, dx
   \]
   This is because \( f(x) < f(t) \) for \( x < t \) and \( f(x) > f(t) \) for \( x > t \) due to the increasing nature of \( f(x) \).

4. Let's denote the integral as \( F(t) \):
   \[
   F(t) = \int_a^t (f(t) - f(x)) x \, dx + \int_t^b (f(x) - f(t)) x \, dx
   \]

5. We can differentiate \( F(t) \) with respect to \( t \) to find the critical points:
   \[
   F'(t) = \frac{d}{dt} \left( \int_a^t (f(t) - f(x)) x \, dx + \int_t^b (f(x) - f(t)) x \, dx \right)
   \]

6. Using the Leibniz rule for differentiation under the integral sign, we get:
   \[
   F'(t) = (f(t) - f(t)) t + \int_a^t f'(t) x \, dx - \int_t^b f'(t) x \, dx + (f(t) - f(t)) t
   \]
   Simplifying, we have:
   \[
   F'(t) = \int_a^t f'(t) x \, dx - \int_t^b f'(t) x \, dx
   \]

7. Since \( f'(t) \) is a constant with respect to \( x \), we can factor it out:
   \[
   F'(t) = f'(t) \left( \int_a^t x \, dx - \int_t^b x \, dx \right)
   \]

8. Evaluating the integrals, we get:
   \[
   \int_a^t x \, dx = \frac{t^2 - a^2}{2}, \quad \int_t^b x \, dx = \frac{b^2 - t^2}{2}
   \]

9. Substituting these back into the expression for \( F'(t) \), we have:
   \[
   F'(t) = f'(t) \left( \frac{t^2 - a^2}{2} - \frac{b^2 - t^2}{2} \right) = f'(t) \left( \frac{2t^2 - a^2 - b^2}{2} \right)
   \]

10. Setting \( F'(t) = 0 \) to find the critical points, we get:
    \[
    f'(t) \left( \frac{2t^2 - a^2 - b^2}{2} \right) = 0
    \]

11. Since \( f'(t) \neq 0 \) (as \( f(x) \) is strictly increasing), we have:
    \[
    2t^2 - a^2 - b^2 = 0 \implies t^2 = \frac{a^2 + b^2}{2} \implies t = \sqrt{\frac{a^2 + b^2}{2}}
    \]

12. Therefore, the value of \( t \) that minimizes the integral is:
    \[
    t = \sqrt{\frac{a^2 + b^2}{2}}
    \]

The final answer is \( \boxed{ t = \sqrt{\frac{a^2 + b^2}{2}} } \)