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\begin{frame}
  \frametitle{Antiderivatives / Integrals}

  \begin{exampleblock}{}
    Find $f$ if
    \begin{talign}
      f''(x) = 12x^2 + 6x - 4
    \end{talign}
    and $f(0) = 4$ and $f(1) = 1$.
    \pause\medskip
    
    The general antiderivative of $f''$ is:
    \begin{talign}
      f'(x) = \mpause[1]{4x^3}\mpause{ + 3x^2}\mpause{- 4x}\mpause{+ C}
    \end{talign}
    \pause\pause\pause\pause\pause
    The general antiderivative of $f'$ is:
    \begin{talign}
      f(x) = \mpause[1]{x^4}\mpause{+ x^3}\mpause{- 2x^2}\mpause{+ Cx} \mpause{+ D}
    \end{talign}
    \pause\pause\pause\pause\pause\pause
    To ensure  $f(0) = 4$ and $f(1) = 1$, we need to find $C$ and $D$\pause:
    \begin{talign}
      f(0) &= \mpause[1]{D} \mpause{ = 4} \\
      \mpause{f(1) &= }\mpause{1 + 1 - 2 + C + 4}\mpause{ = C + 4}\mpause{= 1} \mpause{\quad\implies\quad C = -3} 
    \end{talign}
    \pause\pause\pause\pause\pause\pause\pause\pause
    Therefore the function $f$ we are looking for is:
    \begin{talign}
      f(x) = x^4 + x^3 - 2x^2 - 3x + 4
    \end{talign}
  \end{exampleblock}
\end{frame}