Let $f: [0,1]\rightarrow(0,+\infty)$ be a continuous function. 

a) Show that for any integer $n\geq 1$, there is a unique division $0=a_{0}<a_{1}<\ldots<a_{n}=1$ such that $\int_{a_{k}}^{a_{k+1}}f(x)\, dx=\frac{1}{n}\int_{0}^{1}f(x)\, dx$ holds for all $k=0,1,\ldots,n-1$.

b) For each $n$, consider the $a_{i}$ above (that depend on $n$) and define $b_{n}=\frac{a_{1}+a_{2}+\ldots+a_{n}}{n}$. Show that the sequence $(b_{n})$ is convergent and compute it's limit.
### Part (a)

1. **Define the cumulative distribution function:**
   Let \(\Phi(t) = \int_{0}^{t} f(x) \, dx\). Since \(f(x)\) is continuous and positive on \([0,1]\), \(\Phi(t)\) is strictly increasing and continuous on \([0,1]\).

2. **Determine the total integral:**
   Let \(I = \int_{0}^{1} f(x) \, dx\). Since \(f(x)\) is positive and continuous, \(I > 0\).

3. **Divide the interval \([0,1]\) into \(n\) parts:**
   We need to find points \(0 = a_0 < a_1 < \ldots < a_n = 1\) such that:
   \[
   \int_{a_k}^{a_{k+1}} f(x) \, dx = \frac{I}{n} \quad \text{for all } k = 0, 1, \ldots, n-1.
   \]

4. **Use the inverse function:**
   Since \(\Phi(t)\) is strictly increasing and continuous, it has a well-defined inverse function \(\Phi^{-1}\). Define:
   \[
   a_i = \Phi^{-1}\left(\frac{iI}{n}\right) \quad \text{for } i = 0, 1, \ldots, n.
   \]

5. **Verify the integral condition:**
   For each \(k\):
   \[
   \int_{a_k}^{a_{k+1}} f(x) \, dx = \Phi(a_{k+1}) - \Phi(a_k) = \frac{(k+1)I}{n} - \frac{kI}{n} = \frac{I}{n}.
   \]
   This shows that the division \(0 = a_0 < a_1 < \ldots < a_n = 1\) satisfies the required condition.

### Part (b)

1. **Define the sequence \(b_n\):**
   For each \(n\), let:
   \[
   b_n = \frac{a_1 + a_2 + \ldots + a_n}{n}.
   \]

2. **Express \(b_n\) in terms of \(\Phi^{-1}\):**
   Using the definition of \(a_i\):
   \[
   b_n = \frac{1}{n} \sum_{i=1}^{n} \Phi^{-1}\left(\frac{iI}{n}\right).
   \]

3. **Consider the limit of \(b_n\):**
   As \(n \to \infty\), the sum \(\frac{1}{n} \sum_{i=1}^{n} \Phi^{-1}\left(\frac{iI}{n}\right)\) approximates the integral:
   \[
   \lim_{n \to \infty} b_n = \frac{1}{I} \int_{0}^{I} \Phi^{-1}(t) \, dt.
   \]

4. **Change of variables:**
   Let \(u = \Phi(x)\), then \(du = f(x) \, dx\) and the limits change from \(0\) to \(1\) for \(x\) and from \(0\) to \(I\) for \(u\). Thus:
   \[
   \int_{0}^{I} \Phi^{-1}(t) \, dt = \int_{0}^{1} x f(x) \, dx.
   \]

5. **Compute the limit:**
   Therefore:
   \[
   \lim_{n \to \infty} b_n = \frac{1}{I} \int_{0}^{1} x f(x) \, dx.
   \]

The final answer is \( \boxed{ \frac{\int_{0}^{1} x f(x) \, dx}{\int_{0}^{1} f(x) \, dx} } \).