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

Is the former just popular preference or is there something I am not conceptually understanding here?

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. 
