Answer
They are connected by the limit:
$$\int_a^b f(x)dx = \lim_{\Delta\to0}\sum_{k=1}^nf(x_k^*)\Delta x_k.$$
Work Step by Step
The definite integral is the limit when the width $\Delta = \max\{\Delta x_i\}$ of the arbitrary partition of the interval $[a,b]$ tends to zero of the Riemann sum $\sum_{k=1}^nf(x_k^*)\Delta x_k$ for arbitrary choice $x_k^*\in[x_{i},x_{i+1}]$. Thus the Riemann sum approximates the value of the definite integral and it is a better approximation if the partition contains more points and its width is smaller. Precisely:
$$\int_a^b f(x)dx = \lim_{\Delta\to0}\sum_{k=1}^nf(x_k^*)\Delta x_k $$
