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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}}