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-rw-r--r--šola/ana1/teor.lyx1282
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diff --git a/šola/ana1/teor.lyx b/šola/ana1/teor.lyx
index af865e2..f716983 100644
--- a/šola/ana1/teor.lyx
+++ b/šola/ana1/teor.lyx
@@ -14424,6 +14424,7 @@ sideways false
status open
\begin_layout Plain Layout
+\align center
\begin_inset Tabular
<lyxtabular version="3" rows="7" columns="3">
<features tabularvalignment="middle">
@@ -15029,5 +15030,1286 @@ a priori
.
\end_layout
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Darbouxove vsote.
+ Imamo torej delitev
+\begin_inset Formula $D=\left\{ \left[t_{j-1},t_{j}\right];j\in\left\{ 1..n\right\} ;t_{0}=1,t_{n}=b\right\} $
+\end_inset
+
+ delitev za
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ in
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+.
+ Imamo tudi množico izbranih točk
+\begin_inset Formula $\xi=\left\{ \xi_{j}\in\left[t_{j-1},t_{j}\right];j\in\left\{ 1..n\right\} \right\} $
+\end_inset
+
+ in
+\begin_inset Formula $R\left(f,D,\xi\right)=\sum_{j=1}^{n}f\left(\xi_{j}\right)\left(t_{j}-t_{j-1}\right)$
+\end_inset
+
+.
+ Ocenimo
+\begin_inset Formula $f\left(\xi_{j}\right)$
+\end_inset
+
+:
+
+\begin_inset Formula $\inf_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)\leq f\left(\xi_{j}\right)\leq\sup_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)$
+\end_inset
+
+.
+ Definirali smo
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+ kot limito Riemannovih vsot s kakršnokoli delitvijo in izbiro
+\begin_inset Formula $\xi$
+\end_inset
+
+,
+ zato lahko pišemo
+\begin_inset Formula $\forall j\in\left\{ 1..n\right\} :\inf_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)=f\left(\xi_{j}\right)=\sup_{x\in\left[t_{j-1},t_{j}\right]}f\left(x\right)$
+\end_inset
+
+.
+ Zato lahko limito Riemannovih vsot obravnavamo neodvisno od
+\begin_inset Formula $\xi$
+\end_inset
+
+:
+\begin_inset Formula
+\[
+s\left(f,D\right)\coloneqq\sum_{j=1}^{n}\left(\inf_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-t_{j-1}\right)\leq R\left(f,D,\xi\right)\leq\sum_{j=1}^{n}\left(\sup_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-f_{j-1}\right)\eqqcolon S\left(f,D\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Definirali smo dva nova pojma,
+ spodnjo Darbouxovo vsoto
+\begin_inset Formula $s\left(f,D\right)$
+\end_inset
+
+ in zgornjo Darbouxovo vsoto
+\begin_inset Formula $S\left(f,D\right)$
+\end_inset
+
+ in velja
+\begin_inset Formula $s\left(f,D\right)\leq R\left(f,D,\xi\right)\leq S\left(f,D\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bosta
+\begin_inset Formula $D$
+\end_inset
+
+ in
+\begin_inset Formula $D'$
+\end_inset
+
+ delitvi za interval
+\begin_inset Formula $J$
+\end_inset
+
+.
+ Pravimo,
+ da je
+\begin_inset Formula $D'$
+\end_inset
+
+ finejša od
+\begin_inset Formula $D$
+\end_inset
+
+,
+ če je ima
+\begin_inset Formula $D'$
+\end_inset
+
+ vse delilne točke,
+ ki jih ima
+\begin_inset Formula $D$
+\end_inset
+
+ in poleg njih še vsaj kakšno.
+ Označimo
+\begin_inset Formula $D\subset D'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Naj bo
+\begin_inset Formula $D\subset D'$
+\end_inset
+
+ (
+\begin_inset Formula $D'$
+\end_inset
+
+ finejša od
+\begin_inset Formula $D$
+\end_inset
+
+).
+ Oglejmo si
+\begin_inset Formula $s\left(f,D\right)$
+\end_inset
+
+ in
+\begin_inset Formula $s\left(f,D'\right)$
+\end_inset
+
+.
+ Tedaj velja
+\begin_inset Formula $s\left(f,D\right)\leq s\left(f,D'\right)$
+\end_inset
+
+,
+ ker je infimum po manjši množici lahko le večji —
+ s finejšo delitvijo smo vsaj neko množico (delitveni interval) razdelili na dva dela.
+ Za zgornjo Darbouxovo vsoto velja obratno,
+ torej
+\begin_inset Formula $S\left(f,D\right)\geq S\left(f,D'\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Za poljubni različni delitvi
+\begin_inset Formula $D_{1},D_{2}$
+\end_inset
+
+ intervala
+\begin_inset Formula $J$
+\end_inset
+
+ velja
+\begin_inset Formula $s\left(f,D_{1}\right)\leq S\left(f,D_{2}\right)$
+\end_inset
+
+ ZDB Katerakoli spodnja Darbouxova vsota je kvečjemu tolikšna kot katerakoli zgornja.
+\end_layout
+
+\begin_layout Proof
+Označimo z
+\begin_inset Formula $D_{1}\cup D_{2}$
+\end_inset
+
+ delitev,
+ ki vsebuje vse delilne točke tako
+\begin_inset Formula $D_{1}$
+\end_inset
+
+ kot tudi
+\begin_inset Formula $D_{2}$
+\end_inset
+
+.
+ Očitno velja,
+ da sta
+\begin_inset Formula $D_{1}\subset D_{1}\cup D_{2}$
+\end_inset
+
+ in
+\begin_inset Formula $D_{2}\subset D_{1}\cup D_{2}$
+\end_inset
+
+.
+ Po prejšnjem izreku veljata leva in desna neenakost,
+ srednja pa iz definicije (očitno).
+\begin_inset Formula
+\[
+s\left(f,D_{1}\right)\leq s\left(f,D_{1}\cup D_{2}\right)\leq S\left(f,D_{1}\cup D_{2}\right)\leq S\left(f,D_{2}\right)
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Definition*
+Naj bo
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ omejena.
+ Označimo
+\begin_inset Formula $s\left(f\right)\coloneqq\sup_{\text{vse možne delitve }D}s\left(f,D\right)$
+\end_inset
+
+ in
+\begin_inset Formula $S\left(f\right)\coloneqq\inf_{\text{vse možne delitve }D}S\left(f,D\right)$
+\end_inset
+
+.
+ Funkcija
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ je Riemannovo integrabilna,
+ če
+\begin_inset Formula $s\left(f\right)=S\left(f\right)$
+\end_inset
+
+ oziroma če
+\begin_inset Formula $\forall\varepsilon>0\exists$
+\end_inset
+
+ delitev
+\begin_inset Formula $D$
+\end_inset
+
+ na
+\begin_inset Formula $J\ni:S\left(f,D\right)-s\left(f,D\right)<\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Note*
+Integrabilnost
+\begin_inset Formula $f$
+\end_inset
+
+ ne pomeni,
+ da
+\begin_inset Formula $\exists D\ni:s\left(f,D\right)=S\left(f,D\right)$
+\end_inset
+
+.
+ Ni namreč nujno,
+ da množica vsebuje svoj supremum.
+ Primer:
+ za
+\begin_inset Formula $f\left(x\right)=x$
+\end_inset
+
+ velja
+\begin_inset Formula $\forall D:S\left(f,D\right)>s\left(f,D\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Vsaka zvezna funkcija je integrabilna na
+\begin_inset Formula $J$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo
+\begin_inset Formula $\varepsilon>0$
+\end_inset
+
+ poljuben.
+ Po definiciji
+\begin_inset Formula $S\left(f,D\right)-s\left(f,D\right)=\sum_{j=1}^{n}\left(\sup_{x\in D_{j}}f\left(x\right)-\inf_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-t_{j-1}\right)$
+\end_inset
+
+.
+ Ker je
+\begin_inset Formula $f$
+\end_inset
+
+ zvezna,
+ je na zaprtem
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ enakomerno zvezna,
+ torej
+\begin_inset Formula $\exists\delta>0\forall x_{1},x_{2}\in J:\left|x_{1}-x_{2}\right|<\delta\Rightarrow\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|<\frac{\varepsilon}{b-a}$
+\end_inset
+
+.
+ Izberimo tako delitev
+\begin_inset Formula $D$
+\end_inset
+
+,
+ da je
+\begin_inset Formula $\forall j\in\left\{ 1..\left|D\right|\right\} :t_{j}-t_{j-1}<\delta$
+\end_inset
+
+.
+ Tedaj bo veljalo
+\begin_inset Formula $\sum_{j=1}^{n}\left(\sup_{x\in D_{j}}f\left(x\right)-\inf_{x\in D_{j}}f\left(x\right)\right)\left(t_{j}-t_{j-1}\right)<\sum_{j=1}^{n}\frac{\varepsilon}{b-a}\left(t_{j}-t_{j-1}\right)=\frac{\varepsilon\left(b-a\right)}{b-a}=\varepsilon$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Skratka dokazali smo
+\begin_inset Formula $S\left(f,D\right)-s\left(f,D\right)<\varepsilon$
+\end_inset
+
+ za poljuben
+\begin_inset Formula $\varepsilon$
+\end_inset
+
+,
+ torej je funkcija Riemannovo integrabilna po zgornji definiciji.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $A\subset\mathbb{R}$
+\end_inset
+
+ ima mero
+\begin_inset Formula $0$
+\end_inset
+
+,
+ če
+\begin_inset Formula $\forall\varepsilon>0\exists$
+\end_inset
+
+ družina intervalov
+\begin_inset Formula $I_{j}\ni:A\subset\bigcup I_{j}\wedge\sum\left|I_{j}\right|<\varepsilon$
+\end_inset
+
+.
+ Primer:
+ vse števne in končne množice.
+\end_layout
+
+\begin_layout Theorem*
+Funkcija
+\begin_inset Formula $f$
+\end_inset
+
+ je integrabilna na intervalu
+\begin_inset Formula $J\Leftrightarrow\left\{ x\in J;f\text{ ni zvezna v }x\right\} $
+\end_inset
+
+ ima mero
+\begin_inset Formula $0$
+\end_inset
+
+.
+ ZDB če ima množica točk z definicijskega območja
+\begin_inset Formula $f$
+\end_inset
+
+,
+ v katerih
+\begin_inset Formula $f$
+\end_inset
+
+ ni zvezna,
+ mero
+\begin_inset Formula $0$
+\end_inset
+
+ (recimo če je teh točk končno mnogo),
+ je
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna.
+\end_layout
+
+\begin_layout Fact*
+Označimo z
+\begin_inset Formula $I\left(J\right)$
+\end_inset
+
+ množico vseh integrabilnih funkcij na intervalu
+\begin_inset Formula $J$
+\end_inset
+
+.
+
+\begin_inset Formula $I\left(J\right)$
+\end_inset
+
+ je vektorski prostor za množenje s skalarji iz
+\begin_inset Formula $\mathbb{R}$
+\end_inset
+
+.
+ Naj bodo
+\begin_inset Formula $f,g\in I\left(J\right),\lambda\in\mathbb{R}$
+\end_inset
+
+.
+ Velja aditivnost
+\begin_inset Formula $f\left(x\right)+g\left(x\right)\in J\left(I\right)$
+\end_inset
+
+,
+ kajti
+\begin_inset Formula $\int_{a}^{b}\left(f\left(x\right)+g\left(x\right)\right)dx=\int_{a}^{b}\left(f\left(x\right)\right)dx+\int_{a}^{b}\left(g\left(x\right)\right)dx$
+\end_inset
+
+ in homogenost
+\begin_inset Formula $\int_{a}^{b}\lambda f\left(x\right)dx=\lambda\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Če je
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ in je
+\begin_inset Formula $c\in J$
+\end_inset
+
+,
+ tedaj je
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na
+\begin_inset Formula $\left[a,c\right]$
+\end_inset
+
+ in
+\begin_inset Formula $\left[c,b\right]$
+\end_inset
+
+ in velja
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx=\int_{a}^{c}f\left(x\right)dx+\int_{c}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Če sta
+\begin_inset Formula $f,g$
+\end_inset
+
+ na
+\begin_inset Formula $J$
+\end_inset
+
+ integrabilni funkciji in če je
+\begin_inset Formula $\forall x\in J:f\left(x\right)\leq g\left(x\right)$
+\end_inset
+
+,
+ tedaj
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx\leq\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+ Posledično velja ob isti predpostavki
+\begin_inset Formula $\left|\int_{a}^{b}f\left(x\right)dx\right|\leq\int_{a}^{b}\left|f\left(x\right)\right|dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Če je
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+,
+ definiramo povprečje
+\begin_inset Formula $f$
+\end_inset
+
+ na
+\begin_inset Formula $J$
+\end_inset
+
+ s predpisom
+\begin_inset Formula
+\[
+\left\langle f\right\rangle _{J}\coloneqq\frac{\int_{a}^{b}f\left(x\right)dx}{b-a}\in\mathbb{R}.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Velja
+\begin_inset Formula $\inf_{x\in J}f\left(x\right)\leq\left\langle f\right\rangle _{J}\leq\sup_{x\in J}f\left(x\right)$
+\end_inset
+
+.
+ Če je
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ zvezna,
+
+\begin_inset Formula $\exists\xi\in J\ni:f\left(\xi\right)=\left\langle f\right\rangle _{J}$
+\end_inset
+
+ (izrek o vmesni vrednosti).
+\end_layout
+
+\begin_layout Definition*
+Naj bo
+\begin_inset Formula $f:J\to\mathbb{R}$
+\end_inset
+
+ dana funkcija.
+ Nedoločeni integral
+\begin_inset Formula $f$
+\end_inset
+
+ je takšna funkcija
+\begin_inset Formula $F$
+\end_inset
+
+,
+ če obstaja,
+
+\begin_inset Formula $\ni:F'=f\sim\forall x\in J:F'\left(x\right)=f\left(x\right)$
+\end_inset
+
+.
+ Pišemo tudi
+\begin_inset Formula $Pf$
+\end_inset
+
+ ali
+\begin_inset Formula $\mathbb{P}f$
+\end_inset
+
+ in pravimo,
+ da je
+\begin_inset Formula $F=Pf$
+\end_inset
+
+ primitivna funkcija za
+\begin_inset Formula $f$
+\end_inset
+
+.
+ Velja
+\begin_inset Formula $P\left(f+g\right)=Pf+Pg$
+\end_inset
+
+ (aditivnost odvoda) in
+\begin_inset Formula $P\left(\lambda f\right)=\lambda Pf$
+\end_inset
+
+ (homogenost odvoda).
+\end_layout
+
+\begin_layout Definition*
+Nedoločeni integral je na intervalu določen do aditivne konstante natančno.
+ Če je
+\begin_inset Formula $F'_{1}=f=F_{2}'$
+\end_inset
+
+ na intervalu
+\begin_inset Formula $J$
+\end_inset
+
+ oziroma če na
+\begin_inset Formula $J$
+\end_inset
+
+ velja
+\begin_inset Formula $\left(F_{1}-F_{2}\right)'=0$
+\end_inset
+
+,
+ potem
+\begin_inset Formula $F_{1}-F_{2}=c$
+\end_inset
+
+ oziroma
+\begin_inset Formula $F_{1}=F_{2}+c$
+\end_inset
+
+ za neko konstanto
+\begin_inset Formula $c\in\mathbb{R}$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Definition*
+Označimo
+\begin_inset Formula $F\left(x\right)=Pf\left(x\right)=\int f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Theorem*
+Integracija po delih
+\begin_inset Formula $\sim$
+\end_inset
+
+ per partes.
+ Velja
+\begin_inset Formula $\int f\left(x\right)g'\left(x\right)dx=f\left(x\right)g\left(x\right)-\int f'\left(x\right)g\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Izhaja iz odvoda produkza
+\begin_inset Formula $\left(fg\right)'=f'g+fg'$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Claim*
+Naj bo
+\begin_inset Formula $f$
+\end_inset
+
+ integrabilna na
+\begin_inset Formula $J$
+\end_inset
+
+.
+ Definirajmo
+\begin_inset Formula $F\left(x\right)=\int_{a}^{x}f\left(t\right)dt$
+\end_inset
+
+.
+ Velja
+\begin_inset Formula $\left|F\left(x_{1}\right)-F\left(x_{2}\right)\right|=$
+\end_inset
+
+
+\begin_inset Formula
+\[
+=\left|\int_{a}^{x_{1}}f\left(t\right)dt-\int_{a}^{x_{2}}f\left(t\right)dt\right|=\left|\int_{a}^{x_{1}}f\left(t\right)dt+\int_{x_{2}}^{a}f\left(t\right)dt\right|=\left|\int_{x_{2}}^{x_{1}}f\left(t\right)dt\right|=\left|\int_{x_{1}}^{x_{2}}f\left(t\right)dt\right|\leq\int_{x_{1}}^{x_{2}}f\left(t\right)dt
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Osnovni izrek analize/fundamental theorem of calcusus.
+ Naj bo
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ zvezna in
+\begin_inset Formula $F\left(x\right)=\int_{a}^{x}f\left(t\right)dt$
+\end_inset
+
+.
+ Tedaj je
+\begin_inset Formula $F$
+\end_inset
+
+ odvedljiva na
+\begin_inset Formula $J$
+\end_inset
+
+ in velja
+\begin_inset Formula $F'\left(x\right)=f\left(x\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+\begin_inset Formula
+\[
+F\left(x+h\right)-F\left(x\right)=\int_{x}^{x+h}f\left(t\right)dt\quad\quad\quad\quad/:h
+\]
+
+\end_inset
+
+
+\begin_inset Formula
+\[
+\frac{F\left(x+h\right)-F\left(x\right)}{h}=\frac{\int_{x}^{x+h}f\left(t\right)dt}{h}=\left\langle f\right\rangle _{\left[x,x+h\right]}
+\]
+
+\end_inset
+
+
+\begin_inset Formula
+\[
+F'\left(x\right)=\lim_{h\to0}\left\langle f\right\rangle _{x,x+h}=f\left(x\right).
+\]
+
+\end_inset
+
+
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+glej ANA1P FMF 2024-01-15.pdf/str.
+ 5 za dokaz,
+ ki ga ne razumem,
+ zakaj je
+\begin_inset Formula $\lim_{h\to0}\left\langle f\right\rangle _{\left[x,x+h\right]}-f\left(x\right)=0$
+\end_inset
+
+...
+ ampak sej to je nekak očitno
+\end_layout
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Corollary*
+Naj bo
+\begin_inset Formula $f:\left[a,b\right]\to\mathbb{R}$
+\end_inset
+
+ zvezna in
+\begin_inset Formula $G=Pf$
+\end_inset
+
+ (
+\begin_inset Formula $G'=f$
+\end_inset
+
+).
+ Tedaj je
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx=G\left(b\right)-G\left(a\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Proof
+Naj bo
+\begin_inset Formula $F\left(x\right)=\int_{a}^{x}f\left(t\right)dt$
+\end_inset
+
+.
+ Ker je
+\begin_inset Formula $F'=f=G'$
+\end_inset
+
+,
+ je
+\begin_inset Formula $\left(F-G\right)'=0\Rightarrow F-G=c\in\mathbb{R}$
+\end_inset
+
+,
+ torej
+\begin_inset Formula $G\left(x\right)=F\left(x\right)+c$
+\end_inset
+
+,
+ sledi
+\begin_inset Formula $G\left(a\right)=F\left(a\right)=0$
+\end_inset
+
+ po definiciji
+\begin_inset Formula $F$
+\end_inset
+
+,
+ torej je
+\begin_inset Formula $G\left(a\right)=c$
+\end_inset
+
+.
+ Sledi
+\begin_inset Formula $F\left(x\right)=G\left(x\right)-G\left(a\right)$
+\end_inset
+
+ in
+\begin_inset Formula $F\left(b\right)=G\left(b\right)-G\left(a\right)$
+\end_inset
+
+ in zato
+\begin_inset Formula $F\left(b\right)=\int_{a}^{b}f\left(t\right)dt$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Iskanje primitivne funkcije
+\end_layout
+
+\begin_layout Itemize
+Uganemo jo
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(x^{n}\right)=\frac{x^{n+1}}{n+1}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(e^{x}\right)=e^{x}$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(\sin x\right)=-\cos x$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Itemize
+\begin_inset Formula $P\left(\ln x\right)=x\left(\ln x-1\right)$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Theorem*
+Substitucija/uvedba nove spremenljivke
+\begin_inset Foot
+status open
+
+\begin_layout Plain Layout
+ne razumem.
+ mogoče bom v naslednjem življenju.
+\end_layout
+
+\end_inset
+
+.
+ Naj bo
+\begin_inset Formula $F\left(x\right)$
+\end_inset
+
+ nedoločeni integral funkcije
+\begin_inset Formula $f\left(x\right)$
+\end_inset
+
+ ter
+\begin_inset Formula $\phi\left(x\right)$
+\end_inset
+
+ odvedljiva funkcija.
+ Potem velja
+\begin_inset Formula
+\[
+F\left(\phi\left(t\right)\right)=\int f\left(\phi\left(t\right)\right)\phi'\left(t\right)dx
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Proof
+Formula je posledica odvoda kompozituma:
+\begin_inset Formula
+\[
+\left(F\left(\phi\left(t\right)\right)\right)'=F'\left(\phi\left(t\right)\right)\phi'\left(t\right)=f\left(\phi\left(t\right)\right)\phi'\left(t\right)
+\]
+
+\end_inset
+
+integrirajmo levo in desno stran:
+\begin_inset Formula
+\[
+\int\left(F\left(\phi\left(t\right)\right)\right)'dt=\int f\left(\phi\left(t\right)\right)\phi'\left(t\right)dt.
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+\begin_inset Separator plain
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subsection
+Izlimitirani integrali
+\end_layout
+
+\begin_layout Standard
+Doslej smo računali določene integrale omejene funkcije na omejenem intervalu,
+ torej
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+ Kaj pa neomejen interval,
+ torej
+\begin_inset Formula $\lim_{b\to\infty}\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+?
+\end_layout
+
+\begin_layout Definition*
+Naj bo
+\begin_inset Formula $f:[a,\infty)\to\mathbb{R}$
+\end_inset
+
+ in naj bo
+\begin_inset Formula $\forall m>a:f$
+\end_inset
+
+ integrabilna na
+\begin_inset Formula $\left[a,-m\right]$
+\end_inset
+
+.
+ Če
+\begin_inset Formula $\exists\lim_{m\to\infty}\int_{a}^{m}f\left(x\right)dx$
+\end_inset
+
+,
+ pracimo,
+ da integral
+\begin_inset Formula $\int_{a}^{\infty}f\left(x\right)dx$
+\end_inset
+
+ konvergira,
+ sicer pa divergira.
+ Označimo
+\begin_inset Formula $\int_{a}^{\infty}f\left(x\right)dx\coloneqq\lim_{m\to\infty}\int_{a}^{m}f\left(x\right)dx$
+\end_inset
+
+.
+ Podobno definiramo
+\begin_inset Formula $\int_{-\infty}^{a}f\left(x\right)dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Pomemben primer.
+
+\begin_inset Formula $\int_{1}^{\infty}x^{\alpha}dx=?$
+\end_inset
+
+.
+
+\begin_inset Formula $\int_{1}^{M}x^{\alpha}dx=\frac{M^{\alpha+1}}{\alpha+1}-\frac{1}{\alpha+1}=\frac{M^{\alpha+1}-1}{\alpha+1}$
+\end_inset
+
+.
+ Torej
+\begin_inset Formula $\exists\lim_{M\to\infty}\int_{1}^{M}x^{\alpha}dx\Leftrightarrow\alpha\not=-1$
+\end_inset
+
+.
+ Poglejmo,
+ kaj se zgodi v
+\begin_inset Formula $\alpha=-1$
+\end_inset
+
+:
+
+\begin_inset Formula $\int_{1}^{\infty}x^{-1}dx=\ln M-\ln1=\ln M$
+\end_inset
+
+.
+ Toda
+\begin_inset Formula $\lim_{n\to\infty}\ln M=\infty$
+\end_inset
+
+,
+ torej
+\begin_inset Formula $\int_{1}^{\infty}x^{-1}dx$
+\end_inset
+
+ divergira.
+\end_layout
+
+\begin_layout Definition*
+\begin_inset Formula $\int_{a}^{\infty}f\left(x\right)dx$
+\end_inset
+
+ je absolutno konvergenten,
+ če je
+\begin_inset Formula $\int_{a}^{\infty}\left|f\left(x\right)\right|dx<\infty$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Fact*
+Velja
+\begin_inset Formula $\int_{a}^{\infty}\left|f\left(x\right)\right|dx<0\Rightarrow\int_{a}^{\infty}f\left(x\right)dx<\infty$
+\end_inset
+
+.
+ Velja
+\begin_inset Formula $\left|\int_{a}^{\infty}f\left(x\right)dx\right|\leq\int_{a}^{\infty}\left|f\left(x\right)\right|dx$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Standard
+Ali je predpostavka,
+ da je
+\begin_inset Formula $f$
+\end_inset
+
+ omejena,
+ sploh potrebna?
+\end_layout
+
+\begin_layout Definition*
+Naj bo
+\begin_inset Formula $f:[a,b)\to\mathbb{R}\ni:\forall c<b:f$
+\end_inset
+
+ integrabilna na
+\begin_inset Formula $\left[a,c\right]$
+\end_inset
+
+.
+ V točki
+\begin_inset Formula $b$
+\end_inset
+
+ je
+\begin_inset Formula $f$
+\end_inset
+
+ lahko neomejena.
+ Če
+\begin_inset Formula $\exists$
+\end_inset
+
+ končna limita
+\begin_inset Formula $\lim_{c\to b}\int_{a}^{c}f\left(x\right)dx$
+\end_inset
+
+,
+ je integral
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+ konvergenten,
+ sicer je divergenten.
+ Podobno definiramo,
+ če je funkcija definirana na intervalu
+\begin_inset Formula $(a,b]$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+\begin_inset Formula $\int_{0}^{1}x^{\alpha}dx$
+\end_inset
+
+.
+ Za
+\begin_inset Formula $\alpha<0$
+\end_inset
+
+ ima graf
+\begin_inset Formula $x^{\alpha}$
+\end_inset
+
+ v
+\begin_inset Formula $x=0$
+\end_inset
+
+ pol.
+ Računajmo
+\begin_inset Formula
+\[
+\lim_{\varepsilon\to0}\int_{\varepsilon}^{1}x^{\alpha}dx=\lim_{\varepsilon\to0}\frac{x^{\alpha+1}}{\alpha+1}\vert_{\varepsilon}^{1}=\lim_{\varepsilon\to0}\left(\frac{1}{\alpha+1}-\frac{\varepsilon^{\alpha+1}}{\alpha+1}\right)=\lim_{\varepsilon\to0}\frac{1-\varepsilon^{\alpha+1}}{\alpha+1}=\lim_{\varepsilon\to0}\frac{1-\cancelto{0}{e^{\left(\alpha+1\right)\ln\varepsilon}}}{\alpha+1}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Example*
+Pridobimo pogoj
+\begin_inset Formula $\alpha\not=-1$
+\end_inset
+
+ (imenovalec) in
+\begin_inset Formula $\alpha+1>0$
+\end_inset
+
+ (da bo
+\begin_inset Formula $\left(\alpha+1\right)\ln\varepsilon\to-\infty$
+\end_inset
+
+),
+ torej skupaj s predpostavko
+\begin_inset Formula $\alpha\in\left(-1,0\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Example*
+Torej
+\begin_inset Formula $\int_{0}^{1}x^{\alpha}dx=\frac{1}{\alpha+1}$
+\end_inset
+
+ za
+\begin_inset Formula $\alpha\in\left(-1,0\right)$
+\end_inset
+
+.
+\end_layout
+
+\begin_layout Subsection
+Uporaba integrala
+\end_layout
+
+\begin_layout Itemize
+Ploščine:
+
+\begin_inset Formula $f\geq0$
+\end_inset
+
+ na
+\begin_inset Formula $J=\left[a,b\right]$
+\end_inset
+
+ in je
+\begin_inset Formula $f\in I\left(J\right)$
+\end_inset
+
+,
+ je ploščina lika med
+\begin_inset Formula $x$
+\end_inset
+
+ osjo in grafom
+\begin_inset Formula $f$
+\end_inset
+
+ definirana kot
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx$
+\end_inset
+
+.
+ Če
+\begin_inset Formula $f$
+\end_inset
+
+ ni pozitivna,
+ pa je
+\begin_inset Formula $\int_{a}^{b}f\left(x\right)dx=pl\left(L_{1}\right)-pl\left(L_{2}\right)$
+\end_inset
+
+,
+ kjer je
+\begin_inset Formula $L_{1}$
+\end_inset
+
+ lik nad
+\begin_inset Formula $x$
+\end_inset
+
+ osjo in
+\begin_inset Formula $L_{2}$
+\end_inset
+
+ lik pod
+\begin_inset Formula $x$
+\end_inset
+
+ osjo.
+\end_layout
+
+\begin_layout Example*
+Ploščina kroga:
+ Enačba krožnice je
+\begin_inset Formula $x^{2}+y^{2}=r^{2}$
+\end_inset
+
+ za
+\begin_inset Formula $r>0$
+\end_inset
+
+.
+
+\begin_inset Formula $y=\sqrt{r^{2}-x^{2}}$
+\end_inset
+
+.
+ Ploščina kroga z radijem
+\begin_inset Formula $r$
+\end_inset
+
+ je torej
+\begin_inset Formula $2\int_{-r}^{r}\sqrt{r^{2}-x^{2}}dx=\cdots=\pi r^{2}$
+\end_inset
+
+.
+\end_layout
+
\end_body
\end_document
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