\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}