\ANG{
Berechne die Fl"ache $A$, die von G$_{f}$, der $x$-Achse und den Geraden $x=a$ und
$x=b$ eingeschlossen wird. Zeichne den Grafen und veranschauliche $A$.\\[2mm]
\begin{tabular}[t]{ll}
(a) $\D f(x)=2x+1$, $a=-2$, $b=1$ & (b) $\D f(x)=x^{2}-4$, $a=-3$, $b=0$\\
(c) $\D f(x)=\left|x^{3}+1\right|$, $a=-2$, $b=1$ & 
(d) $\D f(x)=x^{-2}+x^{-3}$, $a=-2$, $b=-0{,}5$
\end{tabular}
} 
\LSG{
\begin{enumerate}
\item \EPSR{10cm}{12rr0121}{4cm}{}{
\PSFM{m2}{$-2$} \PSFM{m1}{$-1$} \PSFM{1}{$1$} \PSFM{m3}{$-3$}
\PSFM{m4}{$-4$} \PSFM{x}{$x$}  \PSFM{y}{$y$}  \PSFM{2}{$2$}  \PSFM{3}{$3$} 
}{
Nullstelle im Integrationsintervall: $x_{0}=-\frac 12$
\[f(x)<0\text{ f"ur }x<-\frac 12,\quad f(x)>0\text{ f"ur }x>-\frac 12\]
\vspace{-8mm}
\begin{align*}
  A &= \int\limits_{-2}^{1}|f(x)|\,\RMd x=
  \int\limits_{-2}^{-\frac 12}\left[-f(x)\right]\,\RMd x+
  \int\limits_{-\frac 12}^{1}f(x)\,\RMd x=\\
  &=\biggl[-x^{2}-x\biggl]_{-2}^{-\frac 12}+
    \biggl[x^{2}+x\biggl]_{-\frac 12}^{1}=\\
  &= -\frac 14+\frac{1}{2}+4-2+1+1-\frac 14+\frac{1}{2}=4{,}5
\end{align*}
elementargeometrisch: $\D A=2\cdot\frac 12\cdot 1{,}5\cdot 3=4{,}5$
}
\item \EPSR{10cm}{12rr0122}{3cm}{}{
\PSFM{-2}{$-2$} \PSFM{-1}{$-1$} \PSFM{1}{$1$} \PSFM{-3}{$-3$}
\PSFM{-4}{$-4$} \PSFM{x}{$x$}  \PSFM{y}{$y$}  \PSFM{2}{$2$}  \PSFM{4}{$4$} 
}{
Nullstelle im Integrationsintervall: $x_{0}=-2$
\[f(x)>0\text{ f"ur }x<-2,\quad f(x)<0\text{ f"ur }x>-2\]
\vspace{-8mm}
\begin{align*}
  A &= \int\limits_{-3}^{0}|f(x)|\,\RMd x=
  \int\limits_{-3}^{-2}f(x)\,\RMd x+
  \int\limits_{-2}^{0}\left[-f(x)\right]\,\RMd x=\\
  &=\biggl[\frac{x^{3}}{3}-4x\biggl]_{-3}^{-2}+
    \biggl[-\frac{x^{3}}{3}+4x\biggl]_{-2}^{0}=\\
  &= -\frac 83+8+9-12-\frac 83+8=7\frac 23
\end{align*}
}
\item \EPSR{10cm}{12rr0123}{4cm}{}{
\PSFM{-2}{$-2$} \PSFM{-1}{$-1$} \PSFM{1}{$1$} \PSFM{5}{$5$}
\PSFM{7}{$7$} \PSFM{x}{$x$}  \PSFM{y}{$y$}  \PSFM{2}{$2$}  \PSFM{4}{$4$} 
}{
Nullstelle von $x^{3}+1$: $x_{0}=-1\QDQ$
\[f(x)=
\begin{cases}
  -x^{3}-1 &\text{f"ur $x<-1$}\\
   x^{3}+1 &\text{f"ur $x\geqq -1$}
\end{cases}
\]
\vspace{-8mm}
\begin{align*}
  A &= 
  \int\limits_{-2}^{-1}(-x^{3}-1)\,\RMd x+
  \int\limits_{-1}^{1}(x^{3}+1)\,\RMd x=\\
  &=\biggl[-\frac{x^{4}}{4}-x\biggl]_{-2}^{-1}+
    \biggl[\frac{x^{4}}{4}+x\biggl]_{-1}^{1}=\\
  &= -\frac 14+1+4-2+\frac 14+1-\frac 14+1=4{,}75
\end{align*}
}
\item \EPSR{10cm}{12rr0124}{4cm}{}{
\PSFM{-2}{$-2$} \PSFM{-1}{$-1$} \PSFM{-0.5}{$-0{,}5$} \PSFM{-3}{$-3$}
\PSFM{-4}{$-4$} \PSFM{x}{$x$}  \PSFM{y}{$y$}
}{
Nullstelle im Integrationsintervall: $x_{0}=-1$
\[f(x)>0\text{ f"ur }x<-1,\quad f(x)<0\text{ f"ur }-1<x<0\]
\vspace{-8mm}
\begin{align*}
  A &= \int\limits_{-2}^{-0{,}5}|f(x)|\,\RMd x=
  \int\limits_{-2}^{-1}f(x)\,\RMd x-\int\limits_{-1}^{-0{,}5}f(x)\,\RMd x=\\
  &=\biggl[-\frac{1}{x}-\frac{1}{2x^{2}}\biggl]_{-2}^{-1}+
    \biggl[\frac{1}{x}+\frac{1}{2x^{2}}\biggl]_{-1}^{-0{,}5}=\\
  &= 1-\frac{1}{2}-\frac{1}{2}+\frac{1}{8}-2+2+1-\frac{1}{2}=\frac 58
\end{align*}
}
\end{enumerate}

}
