\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}