add ana24
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\documentclass{../../../lecture}
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\usepgfplotslibrary{fillbetween}
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\begin{document}
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\begin{bsp}
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\[
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f(x) = \begin{cases}
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\sin \frac{1}{x} & x \neq 0 \\
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0 & x = 0
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\end{cases}
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.\] auf $I = [0,1]$ ist $f(x)$ R.-integrierbar.
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Auf $I$ hat $f(x)$ eine Unstetigkeit bei $x = 0$.
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Sei $\epsilon > 0$ beliebig. Dann $\exists \delta \in [0,1]$, s.d.
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\[
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\sup_{x \in [0,1]} |f(x)| \cdot \delta < \frac{1}{4} \epsilon
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.\] Auf $[\delta , 1] f(x)$ stetig und R.-integrierbar. Dann
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ex. eine Zerlegung $Z_{\delta } \in \mathcal{Z}(\delta , 1)$, s.d.
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\[
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|\overline{S}_{Z_\delta }(f) - \underline{S}_{Z_\delta }(f)| < \frac{1}{2} \epsilon
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.\] Ergänze $Z_{\delta }$ um das Intervall $[0, \delta ]$
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$\implies Z \in \mathcal{Z}(0,1)$. Und es gilt
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\[
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|\overline{S}_{Z}(f) - \underline{S}_{Z}(f)|
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\le | \overline{S}_{Z_\delta }(f) - \underline{S}_{Z_\delta }(f)|
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+ 2 \sup_{x \in [0, \delta ]} |f(x)| \cdot \delta < \epsilon
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.\]
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\end{bsp}
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\begin{satz}[Linearität]
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Seien $f, g\colon I = [a,b] \to \R$ (beschränkt) R.-integrierbar.
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Dann ist $\alpha f + \beta g$, $\alpha, \beta \in \R$ über
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$I$ R.-integrierbar und es gilt
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\[
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\int_{a}^{b} (\alpha f + \beta g)(x) dx =
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\alpha \int_{a}^{b} f(x) dx + \beta \int_{a}^{b} g(x) dx
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.\]
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\end{satz}
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\begin{proof}
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Es ex. $RS_Z(f)$ und $RS_Z(g)$ s.d.
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\begin{align*}
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\lim_{h \to 0} RS_Z(f) &=
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\int_{a}^{b} f(x) dx \\
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\lim_{h \to 0} RS_Z(g) &=
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\int_{a}^{b} g(x) dx
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.\end{align*}
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o.B.d.A. $Z$ und $\xi_k$ sind gleich für $f$ und $g$. Damit folgt
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\begin{align*}
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RS_Z(\alpha f + \alpha g) &:= RS_Z(\alpha f) + RS_Z(\beta g)
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= \alpha RS_Z(f) + \beta RS_Z(g) \\
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\implies \quad \alpha \int_{a}^{b} f(x) dx + \beta \int_{a}^{b} g(x) dx
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&= \alpha \lim_{h \to 0} RS_Z(f) + \beta \lim_{h \to 0} RS_Z(g) \\
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&= \lim_{h \to 0} (\alpha \cdot RS_Z(f)) +
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\lim_{h \to 0} (\beta \cdot RS_Z(g)) \\
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&= \lim_{h \to 0} RS_Z(\alpha f) + \lim_{h \to 0} RS_Z(\beta g) \\
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&= \lim_{h \to 0} RS_Z(\alpha f + \beta g) \\
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&= \int_{a}^{b} (\alpha f + \beta g)(x)dx \\
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.\end{align*}
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\end{proof}
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\begin{satz}[Monotonie des Riemann-Integrals]
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Seien $f, g\colon I = [a,b] \to \R$ (beschränkte)
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R.-integrierbare Funktionen mit $g(x) \ge f(x)$ $\forall x \in [a,b]$.
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Dann gilt
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\[
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\int_{a}^{b} g(x) dx \ge \int_{a}^{b} f(x) dx
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.\]
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\label{satz:riemann-monoton}
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\end{satz}
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\begin{proof}
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Es gilt für Zerlegung $Z$ und $\xi_k \in I_k$ :
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\begin{align*}
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RS_Z(f) = \sum_{k=1}^{n} f(\xi_k)(x_k - x_{k-1})
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\le \sum_{k=1}^{n} g(\xi_k)(x_k - x_{k-1}) = RS_Z(g)
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.\end{align*} Für $h \to 0$ folgt die Behauptung.
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\end{proof}
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\begin{korrolar}
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Sei $f\colon [a,b] \to \R$ (beschr.) R.-integrierbare Funktion,
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$m \le f(x) \le M$. Dann gilt
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\begin{align*}
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m(b - a) \le \int_{a}^{b} f(x) dx \le M(b-a)
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.\end{align*}
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\end{korrolar}
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\begin{proof}
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$g \equiv 1 \implies \int_{a}^{b} 1 dx = (b - a)$ Damit folgt
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\begin{align*}
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m (b-a) = \int_{a}^{b} m \cdot dx
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\stackrel{\text{Satz \ref{satz:riemann-monoton}}}{\le}
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\int_{a}^{b} f(x) dx
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\le \int_{a}^{b} M dx
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= M (b-a)
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.\end{align*}
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\end{proof}
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\begin{korrolar}
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Seien $f, g\colon I \to \R$ zwei beschr. R.-integrierbare
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Funktionen. Dann gilt
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\begin{enumerate}[(a)]
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\item $f_+ := \text{max}\{f, 0\} $ und
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$f_- := \text{min}\{f, 0\} $ sind R.-integrierbar
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\item $|f|$ ist R.-integrierbar und es gilt
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\[
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\left| \int_{a}^{b} f(x) dx \right|
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\le \int_{a}^{b} |f(x)| dx
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.\]
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\item $\forall p \in [1, \infty)$ ist $|f|^{p}$ R.-integrierbar
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\item $f \cdot g$ ist R.-integrierbar.
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\end{enumerate}
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\end{korrolar}
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\begin{proof}
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\begin{enumerate}[(a)]
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\item $Z \in \mathcal{Z}(a,b)$.
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\begin{align*}
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0 &\le \overline{S}_Z(f_+) - \underline{S}_Z(f_+)
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\le \overline{S}_Z(f) - \underline{S}_Z(f) \\
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0 &\le \overline{S}_Z(f_-) - \underline{S}_Z(f_-)
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\le \overline{S}_Z(f) - \underline{S}_Z(f)
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.\end{align*}
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Falls $|\overline{S}_Z(f) - \underline{S}_Z(f)|
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\xrightarrow{h \to 0} 0$
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$\implies$ $|\overline{S}_Z(f_{\pm}) - \underline{S}_Z(f_\pm)
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| \xrightarrow{h \to 0} 0 \implies$ Beh.
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\item $|f| = f_+ - f_- \stackrel{\text{Linearität}}{\implies}
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|f|$ R.-integrierbar.
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$f \le |f|, - f \le |f| \stackrel{\text{Monotonie}}{\implies}
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\int_{a}^{b} f(x) dx \le \int_{a}^{b} |f(x)| dx
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\implies \left| \int_{a}^{b} f(x) dx \right|
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\le \int_{a}^{b} |f(x)| dx $.
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\item Sei $M = \sup_{x \in [a,b]} |f| \stackrel{\text{linear}}{\implies} \frac{|f|}{M}$ integrierbar. $0 \le \frac{|f|}{M} \le 1
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\implies$ z.Zg.: $|f|^{p}$ integr. für $0 \le f \le 1$.
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Sei $0 \le x \le y \le 1$. Aus dem Mittelwertsatz der
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Differentialrechnung folgt
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\begin{align*}
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y^{p} - x^{p} &= p \cdot \xi^{p-1}(y - x) \\
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\implies |y|^{p} - |x|^{p} &= p \cdot |\xi|^{p-1}(|y| - |x|)
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\le p \left( |y| - |x| \right)
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.\end{align*}
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Für $Z \in \mathcal{Z}(0,1)$ gilt
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\begin{align*}
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\underbrace{\overline{S}_Z(|f|^{p}) - \underline{S}_Z(|f|^{p})}_{\xrightarrow{h \to 0} 0}
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\le p \underbrace{(\overline{S}_Z(|f|) - \underline{S}_Z(|f|))}_{ \impliedby \xrightarrow{h \to 0} 0}
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.\end{align*}
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\item $f\cdot g = \frac{1}{4}\left( (f+g)^2 - (f-g)^2 \right) $ und
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c).
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\end{enumerate}
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\end{proof}
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\begin{bem}
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Im Allgemeinen ist
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\[
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\int_{a}^{b} f(x) g(x) dx \neq \left( \int_{a}^{b} f(x) dx \right)
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\left( \int_{a}^{b} g(x) dx \right)
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.\]
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\end{bem}
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\begin{korrolar}[Definitheit des R.-Integrals]
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Sei $f\colon I = [a,b] \to \R$ eine stetige Funktion mit
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$f(x) \ge 0$, $x \in [a,b]$. Dann gilt
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\begin{align*}
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\int_{a}^{b} f(x) dx = 0 \implies f \equiv 0
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.\end{align*}
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\end{korrolar}
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\begin{proof}
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durch Kontraposition.
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Sei $f \not\equiv 0$, d.h. $\exists x_0 \in [a,b]$ mit $f(x_0) > 0$.
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$\stackrel{f\text{ stetig}}{\implies} \exists I_{\epsilon}
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:= [x_0, x_0 + \epsilon]$ oder $I_{\epsilon} := [x_0 - \epsilon, x_0]$, s.d.
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$f(x) \ge \delta > 0$ $\forall x \in I_{\epsilon}$.
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Sei $Z \in \mathcal{Z}(a,b)$ mit $h$ klein genug, s.d.
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für ein $k$ $I_k \subset I_{\epsilon}$. Dann gilt
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\begin{align*}
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0 &< \delta (x_{k} - x_{k-1}) \le
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\inf_{x \in I_k} f(x) (x_k - x_{k-1})
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\le \underline{S}_Z(f) \le \int_{a}^{b} f(x) dx
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.\end{align*}
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\end{proof}
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\begin{definition}
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Sei $a \le b$ Dann ist
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\begin{align*}
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\int_{b}^{a} f(x) dx &:= - \int_{a}^{b} f(x) dx \\
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\int_{a}^{a} f(x) dx &:= 0
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.\end{align*}
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\end{definition}
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\begin{satz}[1. Mittelwertsatz]
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Sei $f\colon I = [a,b] \to \R$ stetig, $g \colon I \to \R$
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R.-integrierbar. $g$ habe in $I$ keinen Vorzeichenwechsel.
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Dann $\exists \xi \in [a,b]$ s.d. gilt
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\[
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\int_{a}^{b} f(x) g(x) dx = f(\xi) \int_{a}^{b} g(x) dx
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.\]
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\end{satz}
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\begin{proof}
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Sei $g \ge 0$ (o.B.d.A.). $f$ stetig
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$\implies$ $\exists m = \min_{x \in I} f(x)$, $M = \max_{x \in I} f(x)$.
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Dann folgt
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\begin{align*}
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m \int_{a}^{b} g(x) dx \le \int_{a}^{b} f(x) g(x) dx
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\le M \int_{a}^{b} g(x) dx
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.\end{align*}
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Betrachte $\varphi(t) := (m (1-t) + M \cdot t) \int_{a}^{b} g(x) dx $,
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$t \in [0,1]$. Nach ZWS $\exists \theta \in [0,1]$, s.d.
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\begin{align*}
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\varphi(\theta) &= y = (m(1-\theta) + M\cdot \theta) \int_{a}^{b} g(x) dx \\
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\varphi(0) &\le y \le \varphi(1) \\
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m \int_{a}^{b} g(x) dx &\le y \le M \int_{a}^{b} g(x) dx \\
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\implies \int_{a}^{b} f(x) g(x) dx &= \mu \int_{a}^{b} g(x) dx
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.\end{align*} Nach dem ZWS für $f$ $\exists \xi \in [a,b]$, s.d.
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$f(\xi) = \mu \in [m, M]$.
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\end{proof}
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\begin{korrolar}
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Sei $f\colon I \to \R$ stetig.
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\begin{enumerate}
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\item $\exists \xi \in I$, s.d. $\int_{a}^{b} f(x) dx =
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f(\xi)(b-a)$
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\item Sei $f\colon [a,b] \to \R$ stetig mit $m \le f(x) \le M$.
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$x \in I$. Sei $g\colon I \to \R$ R.-integrierbar mit $g \ge 0$.
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Dann gilt
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\[
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m \int_{a}^{b} g(x) dx \le \int_{a}^{b} f(x) g(x) dx
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\le M \int_{a}^{b} g(x) dx
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.\]
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\end{enumerate}
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\end{korrolar}
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\begin{bem}
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Voraussetzungen sind unverzichtbar!
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Stetigkeit: $f(x) = \begin{cases}
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0 & 0 \le x < 1 \\
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1 & 1 \le x \le 2
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\end{cases}$ unstetig.
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\begin{align*}
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f(\xi)(b-a) = f(\xi) \cdot 2 = \begin{cases}
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0 & 0 \le \xi < 1 \\
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2 & 1 \le \xi \le 2
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\end{cases} \neq 1 = \int_{0}^{2} f(x) dx
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.\end{align*}
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Positivität: $f(x) = x$, $g(x) = \begin{cases}
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-1 & 0 \le x < 1 \\
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1 & 1 \le x \le 2
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\end{cases}$.
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\begin{align*}
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\int_{0}^{1} f(x) g(x) dx = \int_{0}^{1} (-x) dx
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+ \int_{1}^{2} x dx
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= - \left( \frac{1}{2} - 0 \right) + \left( \frac{4}{2} - \frac{1}{2} \right) = 1
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.\end{align*} aber
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\begin{align*}
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\xi \cdot \int_{0}^{2} g(x) dx = \xi \cdot 0 = 0 \quad \forall \xi \in [0,2]
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.\end{align*}
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\end{bem}
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\subsection{Hauptsatz der Differential- und Integralrechnung}
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\begin{satz}[Hauptsatz der Differential- und Integralrechnung]
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Es sei $a < b$, $f \colon [a,b] \to \R $ stetig.
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\begin{enumerate}[(a)]
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\item Die Funktion $F_0 \colon [a,b] \to \R$ definiert durch
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$F_0(x) := \int_{a}^{x} f(t) dt $ ist
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stetig differenzierbar auf $[a,b]$ und
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\[
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F_0'(x) = f(x) \quad x \in [a,b] \quad (*)
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.\]
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Jede Funktion $F \in C^{1}([a,b], \R)$ welche $(*)$ erfüllt,
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heißt Stammfunktion von $f$.
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\item Jede Stammfunktion $F$ von $f$ hat die Form
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\[
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F(x) = C + \int_{a}^{x} f(t) dt = C + F_0(x)
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.\]
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\item Ist $F$ eine Stammfunktion von $f$, dann gilt
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\[
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\int_{a}^{b} f(t) dt = F(b) - F(a) \; \text{d.h. insb.}
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\; \int_{a}^{b} F'(t) dt = F(b) - F(a)
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\quad \forall F \in C^{1}([a,b], \R)
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.\]
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\end{enumerate}
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\end{satz}
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\begin{proof}
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\begin{enumerate}[(a)]
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\item
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Für $h \neq 0$, $x \in [a,b]$
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\begin{align*}
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\frac{F_0(x+h) - F_0(x)}{h}
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= \frac{1}{h} \left( \int_{a}^{x+h} f(t) dt
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- \int_{a}^{x} f(t) dt \right)
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= \frac{1}{h} \int_{x}^{x+h} f(t) dt
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\quad \stackrel{\text{1. MWS}}{=}
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\quad \frac{1}{h} f(\xi_h) \cdot h
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\xrightarrow[h \to 0]{f\text{ stetig}} f(x) \\
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\implies F_0'(x) = \lim_{h \to 0} \frac{F_0(x+h) - F_0(x)}{h} = f(x)
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.\end{align*}
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\item Sei $F$ eine Stammfunktion ovn $f$, dann gilt
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$(F - F_0)' = f - f = 0
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\stackrel{\text{MWS Diff.}}{\implies} F - F_0 \equiv
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\text{konstant} = C \implies F = F_0 + C$ für ein $C \in \R$.
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\item $F(b) - F(a) \stackrel{\text{(b)}}{=} F_0(b) - F_0(a)
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= \int_{a}^{b} f(t) dt $
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\end{enumerate}
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\end{proof}
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\begin{bem}
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\begin{enumerate}
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\item $F(b) - F(a) =: F(x) \Big|_a^b$
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$\implies \int_{a}^{b} F'(t) dt = F(x) \Big|_a^b $.
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Man bezeichnet eine Stammfunktion auch als unbestimmtes Integral
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\[
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F(x) = \int f(x) dx
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.\] (math. nicht korrekte Bezeichnung)
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\item Integration und Differentiation sind inverse zu einander
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\begin{align*}
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\frac{d}{dx} \int_{a}^{x} f(t) dt = f(x) \\
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F(x) = F(a) + \int_{a}^{x} F'(t) dt
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.\end{align*}
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\end{enumerate}
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\end{bem}
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\end{document}
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Binary file not shown.
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\documentclass{../../../lecture}
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\begin{document}
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\begin{align*}
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\left[\begin{matrix}1 & 1 & -1 & -0.5\\0 & -2 & 6 & 6\\0 & 0 & -6 & -15\\0 & 0 & 0 & 10\end{matrix}\right]
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\left[\begin{matrix}1 & 0 & 0 & 0\\0 & 0.333333333333333 & 2.22044604925031 \cdot 10^{-16} & 4.44089209850063 \cdot 10^{-16}\\0 & 0 & 0.200000000000001 & 3.33066907387547 \cdot 10^{-15}\\0 & 0 & 1.33226762955019 \cdot 10^{-15} & 0.0357142857142889\end{matrix}\right]
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.\end{align*}
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sympy integrate((-1+6*x+6*x^2)*(-0.5 +6*x + -15*x^2 + 10*x^3), (x, 0, 1)) sympy
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\end{document}
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