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\begin{frame}
  \frametitle{The Area below a Curve}

  \begin{exampleblock}{}
    Estimate the area below the curve $f(x) = x^2$ from $0$ to $1$.
  \end{exampleblock}
  
  We now let the number of strips go to infinity: \alert{$\lim_{n\to \infty} R_n$}
  \pause\smallskip
  
  The formula for the area \alert{$R_n$ with $n$ strips} is:\vspace{-.5ex}
  \begin{talign}
    R_n &= \mpause[1]{\frac{1}{n}\cdot \left(\frac{1}{n}\right)^2}
    \mpause{+ \frac{1}{n}\cdot \left(\frac{2}{n}\right)^2}
    \mpause{+ \frac{1}{n}\cdot \left(\frac{3}{n}\right)^2}
    \mpause{+ \ldots + \frac{1}{n}\cdot \left(\frac{n}{n}\right)^2}
    \\
    \mpause{& = \frac{1}{n}\cdot \frac{1}{n^2} (1^2 + 2^2 + 3^2 + \ldots + n^2)}
  \end{talign}
  \pause\pause\pause\pause\pause\pause\vspace{-2ex}
  \begin{block}{}
    \begin{malign}
      1^2 + 2^2 + 3^2 + \ldots + n^2 \;\;=\;\; \frac{n(n+1)(2n+1)}{6} 
    \end{malign}
  \end{block}
  \pause\vspace{-2ex}
  \begin{talign}
    R_n &= \frac{1}{n^3}\cdot \frac{n(n+1)(2n+1)}{6} 
    \mpause[1]{= \frac{(n+1)(2n+1)}{6n^2}}
  \end{talign}
  \pause\pause
  Hence, we have\vspace{-1ex}
  \begin{talign}
    \alert{\lim_{n\to \infty} R_n} = %\lim_{n\to \infty} \frac{(n+1)(2n+1)}{6n^2}
    \mpause[1]{\lim_{n\to \infty} \frac{1}{6}\left(\frac{n+1}{n}\right)\left(\frac{2n+1}{n}\right) }
    \mpause{= \frac{2}{6}}
    \mpause{= \alert{\frac{1}{3}}}
  \end{talign}    

  \vspace{10cm}
\end{frame}