Expressing the Limit of Riemann Sums as a Definite Integral

1. Find the Riemann sums for \(f(x) =x^{2}-3\) on the closed interval \([0,6] \) for a partition of \([0,6] \) into \(n\) subintervals of length \(\Delta x_{i}=x_{i}-x_{i-1},\) \(i=1,2, \ldots , n.\)
2. Assuming that the limit of the Riemann sums exists as \(\max \Delta x_{i}\rightarrow 0\), express the limit as a definite integral.

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 \[ \lim\limits_{\max \Delta x_i \rightarrow 0}\sum\limits_{i=1}^{n}\big(u_{i}^{2}-3\big) \Delta x_{i}=\int_{0}^{6}( x^{2}-3) dx \]