Show that Riemann sum equals integral Suppose we have some continuous function
$\textbf{F}\colon \mathbb{R}^{n} \to \mathbb{R}^n$ (i.e. a force field) and a curve $\gamma\colon [a, b]\subseteq \mathbb{R} \to \mathbb{R}^{n}$ in $C^1$. Considering the partition $P$ of
the interval $[a, b]$
\begin{align*}
P= \left\{t_{0}=a, t_{1}=a + \frac{b-a}{n}, a + 2 \frac{b-a}{n} , \ldots,t_{n}= b\right\} \implies \Delta t = \frac{b-a}{n}
.\end{align*}
I want to show that
\begin{align*}
    \lim_{n \to\infty} \sum_{k=1}^{n}\left\langle 
    \textbf{F}\left(\gamma(t_{k}^{*})\right), \left(\frac{\gamma(t_{k})-\gamma(t_{k-1})}{t_{k}-t_{k-1}}\right)\right\rangle 
    \Delta t
&= \int_{a}^{b}\left\langle  \textbf{F}\left(\gamma(t)\right),\gamma'(t)\right\rangle \mathrm{d}t 
\\ 
&=
\lim_{n \to\infty} \sum_{k=1}^{n}\left\langle   \textbf{F}\left(\gamma(t_{k}^{*})\right),\gamma'(t_{k}^{*}) \right\rangle\Delta t
.\end{align*}
So let $\epsilon>0$ be arbitrary.
We can apply the mean value theorem componentwise, namely for all $1\le i\le m$ we have
\begin{align*}
    \forall x,y \in [a, b] \colon \gamma_{i}(y)-\gamma_{i}(x)=\gamma_{i}'(\xi_i) (y-x), \quad \xi_i \in [x, y]
.\end{align*}
By uniform continuity of $\gamma'$ there exists some $\delta$ s.t. for all $x$ and for all $y \in (x-\delta, x+\delta)$
\begin{align*}
   \left\|\gamma'(y)-\gamma'(x) \right\|_{2}   <\epsilon
.\end{align*}
Now, we choose $N>0$ s.t.
\begin{align*}
\frac{b-a}{N}<  \delta
.\end{align*}
Then, for all $n\ge N$ we find
\begin{align*}
&\left|  
\sum_{k=1}^{n}\left(\left\langle  \textbf{F}\left(\gamma(t_{k}^{*})\right),
        \left(\frac{\gamma(t_{k})-\gamma(t_{k-1})}{t_{k}-t_{k-1}}\right)\right\rangle \Delta t
-\left\langle  \textbf{F}\left(\gamma(t_{k}^{*})\right),\gamma'(t_{k}^{*}) \right\rangle\Delta t
\right)
\right | 
\\
=   
&\left|  
\sum_{k=1}^{n}\Delta t\left\langle \textbf{F} \left(\gamma(t_{k}^{*})\right), \left(\frac{\gamma(t_{k})-\gamma(t_{k-1})}{t_{k}-t_{k-1}}\right)
    -\gamma'(t_{k-1})+\gamma'(t_{k-1})
-\gamma'(t_{k}^{*})\right\rangle 
\right | 
\\
\le 
&  
\Delta t\sum_{k=1}^{n}\left\| \textbf{F}\left(\gamma(t_{k}^{*})\right)\right\|_{2}\left\| 
\left(\frac{\gamma(t_{k})-\gamma(t_{k-1})}{t_{k}-t_{k-1}}\right)-\gamma'(t_{k-1})+\gamma'(t_{k-1})
-\gamma'(t_{k}^{*})\right\|_{2}
\\
\le 
&\Delta t\sum_{k=1}^{n} \left\| \textbf{F}\left(\gamma(t_{k}^{*})\right)\right\|_{2}\left(\left\| 
\left(\frac{\gamma(t_{k})-\gamma(t_{k-1})}{t_{k}-t_{k-1}}\right)-\gamma'(t_{k-1})\right\|_2+\left\|\gamma'(t_{k-1})
-\gamma'(t_{k}^{*})\right\|_{2}\right)
\\ 
<
&\Delta t \sum_{k=1}^{n} \left\|\textbf{F}\left(\gamma(t_{k}^{*})\right)\right\|_{2} \cdot \epsilon(\sqrt{m} +1) 
.\end{align*}
The last inequality follows from the fact that
\begin{align*}
    \frac{\gamma_{i}(t_{k})-\gamma_{i}(t_{k-1})}{t_{k}-t_{k-1}} = \gamma_{i}'(\xi_{i}), \quad \xi_{i} \in (t_{k-1}, t_{k})
\end{align*}
and since $\xi_i-t_{k-1}<\delta$ we have
\begin{align*}
    |\gamma_{i}'(\xi_{i})-\gamma_{i}'(t_{k-1})| < \epsilon
\end{align*}
for all $1\le i\le m$.
Since $\left\|\textbf{F}\left(\gamma(t)\right)\right\|_{2} $ is a continuous function on a compact set it takes a maximal value, let
this be $M$.
It follows that
\begin{align*}
    \Delta t \sum_{k=1}^{n} \left\|\textbf{F}\left(\gamma(t_{k}^{*})\right)\right\|_{2} \cdot \epsilon(\sqrt{m} +1)
    \le (b-a)^{2} m\cdot M\cdot \epsilon(\sqrt{m} +1)
.\end{align*}
Choosing $\epsilon$ in the beginning accordingly concludes the proof.
 A: The overall structure of your proof is good but there is a problem with applying the MVT since $\gamma$ is a vector valued function.
So I just write what you need to fix avoiding some detail also.
Then $$\left |\sum_{k=1}^n \left <\mathbf F(\gamma(t_k^*)),\gamma(t_k)-\gamma(t_{k-1})\right> - \sum_{k=1}^n \left <\mathbf F(\gamma(t_k^*)),\gamma' (t_k^*) \right> \Delta t\right|$$$$=\left |\sum_{k=1}^n \left <\mathbf F(\gamma(t_k^*)),\gamma(t_k)-\gamma(t_{k-1})-\gamma' (t_k^*) \Delta t\right> \right|$$$$\le\sum_{k=1}^n \|\mathbf F(\gamma(t_k^*))\|_2\cdot \|\gamma(t_k)-\gamma(t_{k-1})-\gamma' (t_k^*) \Delta t\|_2$$$$=\sum_{k=1}^n \|\mathbf F(\gamma(t_k^*))\|_2\cdot \|\theta_k(t_k)-\theta_k(t_{k-1})\|_2$$where $\,\theta_k (t)=\gamma (t)-\gamma'(t_k^*)\;t$ .
Apply now to $\,\theta_k\,$ a weakened version of the MVT valid for a vector valued function.
See Baby Rudin, third ed., p. 113 .
Then for every $k$ there exists $\xi_k \in [t_{k-1},t_k]$ such that$$\|\theta_k(t_k)-\theta_k(t_{k-1})\|_2\le \|\theta_k'(\xi_k)\|_2\,\Delta t$$
Since $$\theta_k'(\xi_k)=\gamma' (\xi_k)-\gamma' (t_k^*)$$you get finally$$\left |\sum_{k=1}^n \left <\mathbf F(\gamma(t_k^*)),\gamma(t_k)-\gamma(t_{k-1})\right> - \sum_{k=1}^n \left <\mathbf F(\gamma(t_k^*)),\gamma' (t_k^*) \right>\Delta t \right|$$$$\le \sum_{k=1}^n \|\mathbf F(\gamma(t_k^*))\|_2\cdot \|\gamma' (\xi_k)-\gamma' (t_k^*)\|_2\,\Delta t$$Anyway know that you can succeed even using just the classic monodimensional MVT: this means you must work componentwise of course
Edit
By the uniform continuity of $\gamma'$, for every $\varepsilon>0$ there exists $\delta_i>0$ such that $$|\gamma_i'(x)-\gamma_i'(y)|<\varepsilon\quad\text{if}\quad |x-y|<\delta_i$$Let $\,\delta=\min\, {\delta_i}$ .
Then, by the monodimensional MVT, there exists $\,\xi_{ki}\in[t_{k-1},t_k]\,$ such that $$\|\gamma(t_k)-\gamma(t_{k-1})-\gamma' (t_k^*) \Delta t\|_2$$$$= \sqrt {\sum_{i=1}^m (\gamma_i' (\xi_{ki})-\gamma_i' (t_k^*))^2}\,\Delta t < \sqrt m\, \varepsilon\, \Delta t\quad \text{if}\quad \Delta t < \delta$$
A: The Edit is wrong. If $\delta_1$ depends on $n$ then so does $N$ and you have a problem.
Formula (2) is also not correct. You are subtracting a scalar and a vector. You should have
$$\|\frac{\gamma(x)-\gamma(y)}{x-y}-\gamma’(x)\| \le\epsilon$$
It’s not clear how you get (2) using uniform continuity of the derivative. I think you need to use either the fundamental theorem of calculus combined with the uniform continuity of the derivative or the mean value theorem applied to each component combined with the uniform continuity of the derivative… Something like that.
