\begin{frame} \frametitle{Fundamental Theorem of Calculus} \fundamental \medskip \begin{exampleblock}{} Evaluate\vspace{-1ex} \begin{talign} \int_{-1}^3 \frac{1}{x^2}dx \quad\mpause[10]{\text{\alert{does not exist}}} \end{talign} \pause An antiderivative of $f(x) = \frac{1}{x^2}$ is $F(x) = \pause -\frac{1}{x}$. \pause Then \begin{talign} \int_{-1}^3 \frac{1}{x^2}dx = \mpause[1]{-\frac{1}{x} \Big]_{-1}^3} \mpause{= -\frac{1}{3} - (-\frac{1}{-1})} \mpause{= -\frac{4}{3}} \end{talign} \pause\pause\pause\pause Does this make sense? \pause Note that $\frac{1}{x^2}$ is above the $x$-axis! \pause\medskip \alert{The calculation is wrong since $\frac{1}{x^2}$ is not continuous on $[-1,3]$!} \end{exampleblock} \vspace{10cm} \end{frame}