HMF 6 - Lösung

---

\(\\\)

Aufgabe 1 – Punkt P(0|1)

\(P\) einsetzten in \(f_k\) :

\( \quad \begin{align} 1 & = k \cdot e^{k \cdot 0} - 0^3 \\[6pt] 1 & = k \cdot e^0 \\[6pt] 1 & = k \cdot 1 \\[6pt] 1 & = k \\[6pt] k & = 1 \end{align} \\[1em] \)

Aufgabe 2 – Integral

\( \quad \begin{align} \displaystyle{\int_0^2}f_k(x) dx & = e - 5 ; \\[12pt] \displaystyle{\int_0^2} (k \cdot e^{kx} - x^3) dx & = e - 5 \\[12pt] \Big[ k \cdot \tfrac{1}{k}\cdot e^{kx} - \tfrac{1}{4} x^4 \Big]_0^2 & = e - 5 \\[12pt] \Big[ e^{kx} - \tfrac{1}{4} x^4 \Big]_0^2 & = e - 5 \\[12pt] e^{2k} - \tfrac{1}{4} \cdot 2^4 - \Big( e^{k \cdot 0} - \tfrac{1}{4} \cdot 0^4 \Big) & = e - 5 \\[12pt] e^{2k} - \tfrac{1}{4} \cdot 16 - 1 & = e - 5 \\[12pt] e^{2k} - 4 - 1 & = e - 5 \\[12pt] e^{2k} - 5 & = e^1 - 5 \quad | +5 \\[12pt] e^{2k} & = e^1 \qquad | \hspace{4pt} ln \\[12pt] 2k & = 1 \qquad | :2 \\[12pt] k & = \tfrac{1}{2} \end{align} \)

\(\\\)