\begin{frame} \frametitle{The Definite Integral} \begin{exampleblock}{} Evaluate the definite integral of $f$ from $0$ to $6$ is \begin{talign} \int_{a}^{b} f(x)dx = \lim_{n\to \infty} \sum_{i = 1}^n f(x_i) \Delta x &&&&\text{where $f(x) = 2x-5$} \end{talign} using right endpoints for the sample points $x_i$. \pause\medskip Let $n > 0$. Then \mpause[10]{\alert{$\Delta x = 6/n$} and \alert{$f(x_i) = 2(i\Delta x) - 5$}.} \only<-11>{ \begin{itemize} \pause \item $\Delta x = \pause (6-0) / n \pause = 6/n$ \pause \item the $i$-th interval is \pause $[0 + (i-1)\Delta x,\; 0+ i\Delta x]$ \pause \item the right endpoints are $x_i = \pause i\Delta x$ \pause \item the values at $x_i$'s are $f(x_i) = \pause 2(i\Delta x) - 5$ \end{itemize} } \pause[13]\medskip Thus the Riemann sum is: \only<-18>{ \begin{talign} R_n &= \sum_{i=1}^n f(x_i)\cdot \Delta x = \mpause[1]{\sum_{i=1}^n \left(2(i\Delta x) - 5\right)\Delta x} \\ &\mpause{= \Delta x \sum_{i=1}^n \left(2i\Delta x - 5\right)} \mpause{= \Delta x \left( \sum_{i=1}^n 2i\Delta x - \sum_{i=1}^n 5 \right)} \\ &\mpause{= \Delta x \left( 2\Delta x \left(\sum_{i=1}^n i\right) - 5n \right)} \mpause{= \Delta x \left( 2\Delta x \frac{n(n+1)}{2} - 5n \right)} \end{talign} } \pause[19] \begin{talign} R_n &= \Delta x \left( 2\Delta x \frac{n(n+1)}{2} - 5n \right) \mpause[1]{= \Delta x \left( \Delta x n(n+1) - 5n \right)} \\ &\mpause{= \frac{6}{n} \left( \frac{6}{n} n(n+1) - 5n \right)} \mpause{= \frac{6}{n} \left( 6(n+1) - 5n \right)} \\ &\mpause{= \frac{6}{n} \left( n+1 \right)} \mpause{= \frac{6n + 6}{n} } \\ \mpause{ \alert{\int_{a}^{b}}& \alert{f(x)dx} = \lim_{n\to \infty} R_n = } \mpause{\lim_{n\to \infty} \frac{6n + 6}{n}} \mpause{\alert{= 6}} \end{talign} \end{exampleblock} \vspace{10cm} \end{frame}