Why is the integral defined as the limit of the sum $\int_a^b f(x) dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x$?

Let's cut $[a, b]$ up into infinitely many equally spaced slices. So $x_1 = a$, obviously. What's $x_2$? What's $\Delta x$, if not zero?

---

In general, there isn't a "nice" way to cut a finite interval into infinitely many sampling intervals. Ultimately, the Riemann integral samples a bunch of function values in a fairly uniform way (meaning one from each interval of length $\Delta x$) and averages them. It doesn't make sense to make an infinite uniform sampling of a bounded interval.

A short answer is that $x_i^\ast$ and $\Delta x$ depend on $n$, so the expression $$\sum_{i=1}^\infty f(x_i^\ast)\Delta x$$ does not make sense.

Actually, the definition is:

Let $\mathcal{P}=\{a, x_1, \ldots, x_k, b\}$ be a partition of $[a, b]$ and denote $\Delta \mathcal{P} = \max_i |x_i-x_{i+1}|$. Then we say a bounded function $f$ is Riemann integrable provided for any $\varepsilon>0$ there exists $\delta>0$ such that if $\Delta \mathcal{P}<\delta$ implies \begin{align} \left|\sum_{\mathcal{P}} \left\{M_i-m_i \right\}(x_{i+1}-x_i)\right|<\varepsilon \end{align} where \begin{align} M_i:=\sup_{t \in[x_i, x_{i+1}]}f(t) \ \ \text{ and } \ \ m_i:=\inf_{t \in[x_i, x_{i+1}]}f(t). \end{align}

The partition is finite (just like how partial sum is finite).

Once you know that $f$ is Riemann integrable, then you can define the integral of $f$ to be \begin{align} \int^b_a f(t)\ dt:= \lim_{\Delta\mathcal{P}\rightarrow 0}\sum_{\mathcal{P}} M_i (x_{i+1}-x_i). \end{align}

Once you know $f$ is Riemann integrable, then you can specialize a nested sequence of partition $\mathcal{P}_n \subset \mathcal{P}_{n+1}$ with $\Delta \mathcal{P}_n \geq \Delta \mathcal{P}_{n+1}\rightarrow 0$ so that \begin{align} \int^b_a f(t)\ dt = \lim_{n\rightarrow \infty} \sum^n_{i=1} f(x^\ast_i)\Delta x_i \end{align} is actually meaningful.