Let $f$ be a differentiable function. If we have $x_k \in \left[\frac{k-1}{n},\frac{k}{n}\right]$ and $c_k \in \left[\frac{k-1}{n},\frac{k}{n}\right]$.
Does it mean that the sum $$\frac{1}{n}\sum_{k=1}^nf(x_k)f'(c_k)\rightarrow \int_0^1f(x)f'(x)dx$$ as $n \to \infty ?$
Intuitively speaking, the difference between $f(x_k)$ and $f(c_k)$ should be neglible as $n\to\infty$, so these sums should have the same limit: $$ \lim_{n\to\infty}\sum_{k=1}^nf(x_k) f'(c_k) \frac{1}{n} = \lim_{n\to\infty}\sum_{k=1}^nf(c_k) f'(c_k) \frac{1}{n} = \int_0^1 f(x)f'(x)\,dx $$ More formally, for each $k$ there exists a point $t_k$ between $x_k$ and $c_k$ such that $$ f(c_k) - f(x_k) = f'(t_k)(c_k-x_k) $$ this is by the Mean Value Theorem. If we assume additionally that $f'$ is continuous (and therefore bounded) on $[a,b]$, then we can say $$ |f(c_k) - f(x_k)| \leq M |c_k - x_k| \leq M \frac{1}{n} $$ for some $M$. This means that the difference between $\sum_{k=1}^n f(x_k) f'(c_k) \frac{1}{n}$ and $\sum_{k=1}^n f(c_k) f'(c_k) \frac{1}{n}$ is bounded by a multiple of $\frac{1}{n}$: \begin{align*} \left|\sum_{k=1}^n f(x_k) f'(c_k) \frac{1}{n}-\sum_{k=1}^n f(c_k) f'(c_k) \frac{1}{n}\right| &\leq \sum_{k=1}^n\left|f(x_k) - f(c_k)\right|f'(c_k)\frac{1}{n} \\ &\leq\sum_{k=1}^n\frac{M}{n}\frac{1}{n} = \frac{M}{n} \\ \end{align*} and this tends to zero as $n\to\infty$.
You might be able to dispense with the assumption that $f'$ is continuous, if you use Duhamel's Theorem.