Solution (a) The Riemann sums for $f(x)=x^{2}-3$on the closed interval $[0,6]$ are $$\sum\limits_{i=1}^{n}f(u_{i}) \Delta x_{i}=\sum\limits_{i=1}^{n}\big( u_{i}^{2}-3\big) \Delta x_{i}$$

where $[0,6]$ is partitioned into $n$ subintervals $[x_{i-1}, x_{i}]$, and $u_{i}$ is some number in the subinterval $[x_{i-1}, x_{i}]$, $i=1,2, \ldots, n$.

(b) Since $\lim\limits_{\max \Delta x_{i}\rightarrow 0}\sum\limits_{i=1}^{n}\big( u_{i}^{2}-3\big) \Delta x_{i}$ exists, then $$\lim\limits_{\max \Delta x_i \rightarrow 0}\sum\limits_{i=1}^{n}\big(u_{i}^{2}-3\big) \Delta x_{i}=\int_{0}^{6}( x^{2}-3) dx$$