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

  Now let's look at a general \structure{curve above the $x$-axis}:
  \begin{block}{}
    The area below the curve of a function $f$ on an interval $[a,b]$.
  \end{block}
  \begin{center}
  \scalebox{.9}{
  \begin{tikzpicture}[default]
    \def\mfun{(4*(\x+\mfunshift) - 2.6*(\x+\mfunshift)^2 + .44*(\x+\mfunshift)^3)}

    {\def\diaborderx{.3cm}
    \def\diabordery{.3cm}
    \diagram[1]{-.5}{4}{-.4}{2}{1}}
    \diagramannotatez
    \def\mfunshift{0}
    \begin{scope}[ultra thick]
      \draw[cred] plot[smooth,domain=.5:3.5,samples=20] (\x,{\mfun});
      \only<-3>{
        \draw[draw=none,fill=cred,opacity=.5] (.5,0) -- plot[smooth,domain=.5:3.5,samples=20] (\x,{\mfun}) -- (3.5,0) -- (.5,0) -- cycle;
      }
      \only<4-13>{
        \def\mwidth{3}
        \foreach \nrsteps/\mcolor in {3/cred} {
          \def\mstep{\mwidth/(\nrsteps+1)}
          \def\mfunshift{\mstep}
          \foreach \xx in {0,...,\nrsteps} {
            \def\x{.5+ \xx*\mstep}
            \draw[thick,draw=\mcolor!60!black,fill=\mcolor,opacity=.5] ({\x},0) rectangle ({\x+\mstep},{\mfun}); 
            \node[include=\mcolor] at ({\x+\mfunshift},{\mfun}) {};
          }
        }
      }
      \only<14->{
        \def\mwidth{3}
        \foreach \nrsteps/\mcolor in {3/cred} {
          \def\mstep{\mwidth/(\nrsteps+1)}
          \def\mfunshift{0}
          \foreach \xx in {0,...,\nrsteps} {
            \def\x{.5+ \xx*\mstep}
            \draw[thick,draw=\mcolor!60!black,fill=\mcolor,opacity=.5] ({\x},0) rectangle ({\x+\mstep},{\mfun}); 
            \node[include=\mcolor] at ({\x+\mfunshift},{\mfun}) {};
          }
        }
      }
      \node[anchor=north] at (.5,0) {$a$};
      \node[anchor=north] at (3.5,0) {$b$};
    \end{scope}
  \end{tikzpicture}
  }
  \end{center}\vspace{-1ex}
  \pause
  
  We use $n$ rectangles: \mpause[6]{$\Delta x = \pause (b-a)/n$}
  \only<-7>{\begin{itemize}
    \item the width of the interval is $b-a$ 
  \pause
    \item the width of each strip is $\Delta x = \pause (b-a)/n$
  \pause
    \item the interval for the $i$-th strip is \pause $I_i = [a + (i-1)\Delta x,\;a+ i\Delta x]$
  \end{itemize}
  }
  \pause\pause\pause\pause\pause\pause\medskip
  
  The area of the rectangles oriented at right-endpoints is:\vspace{-.5ex}
  \begin{talign}
    \alert{R_n} &= \mpause[1]{ \Delta x\cdot f(a + 1\Delta x) }
    \mpause{ + \Delta x\cdot f(a + 2\Delta x) }
    \mpause{ +\ldots+ \Delta x\cdot f(a + n\Delta x) } \\[-.5ex]
    &\mpause{= \Delta x\big(f(a + 1\Delta x) + f(a + 2\Delta x) + \ldots + f(a + n\Delta x)\big)}
  \end{talign}
  \pause\pause\pause\pause\pause
  The area of the rectangles oriented at left-endpoints is:\vspace{-.5ex}
  \begin{talign}
    \alert{L_n} &= \mpause[1]{ \Delta x\cdot f(a + 0\Delta x) }
    \mpause{ + \Delta x\cdot f(a + 1\Delta x) }
    \mpause{ +\ldots+ \Delta x\cdot f(a + (n-1)\Delta x) } \\[-.5ex]
    &\mpause{= \Delta x\big(f(a + 0\Delta x) + f(a + 2\Delta x) + \ldots + f(a + (n-1)\Delta x)\big)}
  \end{talign}
  \vspace{10cm}
\end{frame}