\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} \begin{center} \scalebox{.65}{ \begin{tikzpicture}[default] \def\mfun{(4*((\x+\mfunshift)/4)^2)} \diagram[1]{-.5}{4.5}{-.4}{4.5}{1} \diagramannotatez \diagramannotatexx{1/.25,2/.5,3/.75,4/1} \diagramannotateyy{1/.25,2/.5,3/.75,4/1} \def\mfunshift{0} \begin{scope}[ultra thick] \draw[cred] plot[smooth,domain=0:4,samples=20] (\x,{\mfun}); \def\mwidth{4} \setcounter{slide}{2} \foreach \nrsteps/\mcolor in {3/cred} { \setcounter{roundcounter}{\arabic{slide}} \def\mstep{\mwidth/(\nrsteps+1)} \def\mfunshift{0} \foreach \xx in {0,...,\nrsteps} { \def\x{\xx*\mstep} \setcounter{tmpcount}{\arabic{roundcounter}} \addtocounter{tmpcount}{\nrsteps} \onslide<\arabic{slide}->{ \draw[thick,draw=\mcolor!60!black,fill=\mcolor,opacity=.5] ({\x},0) rectangle ({\x+\mstep},{\mfun}); \node[include=\mcolor] at ({\x+\mfunshift},{\mfun}) {}; } \addtocounter{slide}{1} } } \end{scope} \end{tikzpicture} } \end{center} \vspace{-1ex} Let's split the area in $4$ vertical strips: \begin{itemize} \item height of each rectangle = value at \alert{left} endpoint \end{itemize} \pause\pause\pause\pause\pause\smallskip The sum of the area of these rectangles is:\vspace{-1ex} \begin{talign} &\alert{L_4} = \mpause[1]{ \frac{1}{4} \cdot 0^2 } \mpause{ + \frac{1}{4} \cdot \left(\frac{1}{4}\right)^2 } \mpause{ + \frac{1}{4} \cdot \left(\frac{2}{4}\right)^2 } \mpause{ + \frac{1}{4} \cdot \left(\frac{3}{4}\right)^2 } \mpause{ = \alert{0.21875}} \end{talign} \pause\pause\pause\pause\pause\pause The area $A$ below the curve is \pause larger then $L_4$, that is, \alert{$L_4 < A$}. \vspace{10cm} \end{frame}