\documentclass[a4]{article}
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\begin{document}
\section* {S.146 Nr.11}
\subsection* {x-Komponente des Schnittpunktes berechnen}

\begin{math}
m = 0{,}5 \newline
f_1 = -(x-2)^2+4 = -x^2+4x \newline
f_2 = m x
\end{math}


\begin{displaymath}
\begin{array}{rcl}
f_1 & = & f_2 \\
-x^2 -4x & = & 0{,}5x \\
-x-4 & = & 0{,}5 \\
x & = & 3{,}5
\end{array}
\end{displaymath}
\begin{math}
\Rightarrow z = 3,5
\end{math}

\subsection* {Errechnen der Flächeninhalte \(A_1\) und \(A_2\)}

\begin{displaymath}
\begin{array}{rcl}
A_1 & = & \int\limits_{0}^z (-x^2+4x)dx - \int\limits_{0}^z (mx)dx \\
& = & \lbrack - \frac{1}{3} x^3 + 2x^2 \rbrack_{0}^z -\lbrack \frac{m}{2}x^2 \rbrack_{0}^z \\
& = & -\frac{1}{3} z^3 + 2z^2 - \frac{m}{2}z^2 \\
& \approx & 7{,}146
\end{array}
\end{displaymath}

\begin{displaymath}
\begin{array}{rcl}
A_2 & = & \int\limits_{z}^4 (-x^2+4x)dx - \int\limits_{z}^4 (mx)dx \\
& = & \lbrack - \frac{1}{3} x^3 + 2x^2 \rbrack_{z}^4 -\lbrack \frac{m}{2}x^2 \rbrack_{z}^4 \\
& = & \frac{32}{3} + \frac{1}{3} z^3 - 2z^2  - 8m + \frac{m}{2}z^2 \\
& \approx & 0{,}479
\end{array}
\end{displaymath}



\paragraph* {}

ENDE

\end{document}