Is $\frac{1}{n}\sum_{k=1}^nf(x_k)f'(c_k)$ a Riemann Sum? 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 $n \to \infty ?$
 A: 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.
