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\title{Matematin�s analiz�s konspektai \\
       \Large II~kursas, rudens semestras}
\author{Marius Gedminas \\ pagal prof. V.~Mackevi�iaus paskaitas \\
        VU Matematikos ir Informatikos fakultetas}
%\date{II kursas, rudens semestras (1999--2000~m.)}

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\thispagestyle{empty}

\part{Keli� kintam�j� funkcij� integravimas}
\markboth{Keli� kintam�j� funkcij� integravimas}{}

\section{Papildomos laiptini� funkcij� savyb�s}				% 1

\begin{ap}								% 1.1
\Def{$k$-mate gardele} (\Def{$k$-ma�iu gretasieniu/sta�iakampiu}) vadinama 
aib� $I \subset \R^k$, turinti pavidal�
\[ I = I_1 \times I_2 \times \cdots \times I_k
\]
kur $I_i \subset R$ -- bet kokie intervalai (�skaitant ir vienta�k� $[a, a] 
= \{a\}$).

$k$-mat� gardel� $I$ vadinama \Def{u�dar�ja}, kai visos jos ,,briaunos`` $I_i$
yra u�daros.  Analogi�kai apibr��iama \Def{atviroji}, \Def{apr��toji} ir
t.t. gardel�s.

Gardel�s $I = \bigtimes_{n=1}^k I_n$ \Def{matu} vadinamas skai�ius
\[ m(I) := \prod_{n=1}^k |I_n|
\]
(�iuo atveju laikome, kad $0 \cdot \infty := 0$).
\end{ap}


\begin{ap}\label{ap:1.2}						% 1.2
Aib� $N \subset \R^k$ vadinama \Def{nulinio mato aibe} (�ymime $m(N) = 0$),
jei $\forall \eps > 0 \quad \exists$ $k$-ma�i� gardeli� seka 
$\{I_n: n \in \N\}$, su kuria
\begin{enumerate}
  \item $\displaystyle \bigcup_{n=1}^\infty I_n \supset N$ ir
  \item $\displaystyle \sum_{n=1}^\infty m(I_n) < \eps$
\end{enumerate}

\noindent Vis� toki� aibi� klas� �ym�sime $\mathcal{N}$.
\end{ap}

\begin{pss}\NL								% 1.3
\begin{enumerate}
  \item Nulinio mato aib�s s�voka priklauso nuo dimensijos $k$.  Pvz.,
        ties� $\not\in \mathcal{N}$, kai $k = 1$, bet $\in \mathcal{N}$, kai
        $k > 1$.
  \item \ref{ap:1.2} galima apsiriboti vien u�daromis arba vien atviromis
        gardel�mis; nuo to nulinio mato aib�s s�voka nepasikeis.  

        \begin{ir*} I� ties�, $m(\overline{I_n}) = m(I_n)$, tad galima 
        apsiriboti u�daromis.  �sitikinsime, kad galima apsiriboti ir vien
        tik atviromis gardel�mis.  Remiantis apibr��imu laisvai pasirinkime
        $\eps > 0$.  $\exists$ gardeli� seka $\{I_n: n \in \N\}:$
        $\bigcup_{n=1}^\infty I_n \supset N$ ir $\sum_{n=1}^\infty m(I_n)
        < \eps/2$.  Kiekvienam $I_n$ galime rasti atvir� gardel�
        $\widetilde{I}_n \subset I_n$ ir $m(\widetilde{I}_n) = 2 m(I_n)$.
        Tada $\bigcup_{n=1}^\infty \widetilde{I}_n \supset N$ ir 
        $\sum_{n=1}^\infty m(\widetilde{I}_n) < \eps$.
        \end{ir*}
\end{enumerate}
\end{pss}

\shortpage
\begin{tg}\label{tg:1.4}\NL						% 1.4
\begin{enumerate}
  \item $N \in \mathcal{N}$, $A \subset N$ \tada
        $A \in \mathcal{N}$.
  \item $N_i \in \mathcal{N}$, $i \in \N$ \tada
        $\displaystyle \bigcup_{i=1}^\infty N_i \in \mathcal{N}$.
        \label{tg:1.4.2}
\end{enumerate}
\end{tg}

\begin{pvzz}\NL								% 1.5
\begin{enumerate}
  \item Vienta�k� aib� $\{a\}$ yra nulinio mato aib�. Remiantis \ref{tg:1.4}.%
        \ref{tg:1.4.2}, bet kuri baigtin� arba skaiti aib� taip pat yra 
        nulinio mato aib�.  Pvz., $Q \in \mathcal{N}$ ($k = 1$).
  \item $k$-mat�s gardel�s $\bigtimes_{n=1}^k [a_n, b_n]$ ,,kra�tas`` yra
        nulinio mato aib�.
        \begin{ir*}
        I� tikr�j�, jis sudarytas i� $2k$ nulinio mato gardeli� $\{a_1\} \times
        [a_2, b_2] \times \cdots \times [a_k, b_k]$, $\{b_1\} \times
        [a_2, b_2] \times \cdots \times [a_k, b_k]$, \ldots, $[a_1, b_1] \times
        \cdots \times \{b_k\}$ (,,briaun�``).
        \end{ir*}
  \item �sitikinsime, kad kiekviena glodi kreiv� $\Gamma$ erdv�je $\R^k$ yra
        nulinio mato aib� ($k \ge 2$).
        \begin{ir*}
        Tarkime, $\Gamma = \gamma([a,b])$ ir $\gamma \in C^1([a,b], \R^k)$.

        Imkime bet kok� intervalo $[a,b]$ skaidin� $a = t_0 < t_1 < \ldots
        < t_r = b$ ir pa�ym�kime $\delta := \max |t_{i+1}-t_i|$, 
        $M := \max_{t \in [a,b]} |\gamma'(t)|$.

        Remiantis Lagran�o vidutin�s reik�m�s teorema,
        $|\gamma(t)-\gamma(t_i)| \le M |t - t_i| \le M \delta$, kai 
        $t \in [t_i, t_{i+1}]$.

        Pa�ym�kime $I_i$ $k$-mat� gardel� su centru ta�ke $\gamma(t_i)$ ir
        $2M|t_{i+1}-t_i|$ ilgio kra�tin�mis.

        �vertinkime sumin� gardeli� $I_i$ mat�:
        \begin{align*}
        \sum_{i=0}^{r-1} m(I_i) &= \sum_{i=0}^{r-1} (2M|t_{i+1}-t_i|)^k
        = (2M)^k \sum_{i=0}^{r-1} |t_{i+1}-t_i|^k
        = (2M)^k \sum_{i=0}^{r-1} \underbrace{|t_{i+1}-t_i|^{k-1}}_{\le
          \delta^{k-1}} |t_{i+1}-t_i| \\&
        \le (2M)^k \delta^{k-1} \cdot \underbrace{\sum_{i=0}^{r-1}
        |t_{i+1}-t_i|}_{= b-a} = \underbrace{(b-a)(2M)^k}_{=: C} \delta^{k-1}
        = C\delta^{k-1} \qquad \text{($C$ nepriklauso nuo skaidinio)}
        \end{align*}

        Ai�ku, kad $\bigcup_{i=0}^{r-1} I_i \supset \Gamma$.

        L.p. $\eps > 0$ ir imkime bet kok� $[a,b]$ skaidin� su tokiu $\delta$,
        kad $C\delta^{k-1} < \eps$.  Pagal apibr��im� $\Gamma \in \mathcal{N}$.
        \end{ir*}
\end{enumerate}
\end{pvzz}

\begin{ap}[s�voka ,,beveik visur``]					% 1.6
Sakoma, kad kokia nors aib�s $A \subset \R^k$ ta�k� savyb� galioja \Def{beveik
visur} (b.v.) aib�je $A$, jei ji galioja visuose aib�s $A$ ta�kuose, i�skyrus
galb�t nulinio mato aib�.

\begin{description}
\item[Pvz.:]
\begin{enumerate}
  \item Funkcija $f: A \to \RR := [-\infty, \infty]$ yra baigtin� b.v., jei
        $\{ x \in A : f(x) = \pm\infty \} \in \mathcal{N}$.
  \item Funkcija $f: [a, b] \to \R$ yra b.v. tolydi, jei jos tr�kio ta�k�
        aib� priklauso $\mathcal{N}$.  Pvz., $\forall f \in D[a, b]$ yra
        b.v. tolydi, nes jos tr�kio ta�k� aib� visada baigtin� arba skaiti.
  \item Funkcij� seka $f_n \to f$ b.v., jei $\{x : f_n(x) \not\to f(x)\} \in
        \mathcal{N}$.
  \item Sakoma, kad funkcija $f: A\setminus N \to \R$ yra apibr��ta b.v.
        aib�je $A$, jei $N \in \mathcal{N}$.
\end{enumerate}
\end{description}
\end{ap}

\begin{ap}								% 1.7
Funkcija $\phi: I \to \R$ ($I \subset \R^k$ -- $k$-mat� gardel�) vadinama
\Def{laiptine funkcija} jei $\exists$ baigtin� $k$-ma�i� apr��t� gardeli�
sistema $\{ I_j : j = 1 \ldots m \}$, su kuria $\phi(x) = c_j$, $x \in I_j$,
$j = 1 \ldots m$ ir $\phi(x) = 0$, kai $x \in I \setminus \bigcup_{j=1}^m
I_j$.

Tokios funkcijos \Def{integralu} gardel�je $I$ vadinamas skai�ius
\[
  \int_I \phi := \sum_{j=1}^m c_j m(I_j)
\]

Kiti �ym�jimai:
\[
\int_I \phi(x)\,\dx
\qquad
\int_I f(x_1, \ldots, x_k)\,\dx_1\dx_2\ldots\dx_k
\]

Jei $I = [a_1, b_1] \times \cdots \times [a_k, b_k]$, dar �ymima
\[
  \int_{a_1}^{b_1} \cdots \int_{a_k}^{b_k} 
    f(x_1, \ldots, x_k)\,\dx_1\dx_2\ldots\dx_k
\]

Dvima�iu ir trima�iu atveju dimensija $k$ parodoma integralo �enkl� skai�iumi:
\[
  \iint_I f(x,y)\,\dx\dy
  \qquad
  \iiint_I f(x,y,z)\,\dx\dy\dz
\]

Vis� laiptini� funkcij� klas� gardel�je $I$ �ymima $S(I)$.
\end{ap}

\begin{pss}\NL								% 1.8
\begin{enumerate}
  \item Integralo apibr��imo korekti�kumas �rodomas analogi�kai vienma�iam
        atvejui.
  \item Laiptines funkcijas galima i�reik�ti tiesin�mis gardeli� indikatori�
        kombinacijomis: $\phi = \sum_{j=1}^m c_k \Ind_{I_j}$
        ir atvirk��iai, visos tokio pavidalo funkcijos yra laiptin�s.
\end{enumerate}
\end{pss}

\begin{tg}[laiptini� funkcij� integral� savyb�s]\NL			% 1.9
\begin{enumerate}
  \item (Tiesi�kumas)
        Jei $f, g \in S(I)$, $\alpha, \beta \in \R$, tada
        \[ \int_I (\alpha f + \beta g) = \alpha \int_I f + \beta \int_I g
        \]
  \item (Monotoni�kumas)
        Jei $f, g \in S(I)$, $f \le g$, tai
        \[ \int_I f \le \int_I g
        \]
        Atskiru atveju
        \[ \left| \int_I f \right| \le \int_I |f|
        \]
\end{enumerate}
\begin{ir}
Analogi�ka vienma�iam atvejui.
\end{ir}
\end{tg}

\begin{tr}[pirmoji pagrindin� lema]					% 1.10
Tarkime, kad $\{\phi_n\} \subset S(\R^k)$ yra monotoni�kai ma��janti
neneigiam� laiptini� funkcij� seka, konverguojanti b.v. � $0$, kai $n \to
\infty$.  Tada
\[ \int_{\R^k} \phi_n \to 0, \quad n \to \infty
\]

\noindent
(Trumpai: jei $S(\R^k) \ni \phi_n \downarrow 0$ b.v.\footnote{monotoni�kumas
ir neneigiamumas reikalingi visur, konvergavimo u�tenka ir beveik visur},
tai $\int_{\R^k} \phi_n \downarrow 0$, $n \to \infty$)
\end{tr}

\begin{isv}								% 1.11
Jei $S(\R^k) \ni \phi_n \uparrow \phi \in S(\R^k)$ (arba ,,$\downarrow$``) 
b.v., tai
\[ \int_{\R^k} \phi_n \uparrow \int_{\R^k} \phi
   \qquad \text{(atitinkamai, ,,$\downarrow$``)}
\]
\end{isv}

\begin{tr}[antroji pagrindin� lema]					% 1.12
Jei $\{ \phi_n \} \subset S(\R^k)$ -- monotoni�kai did�janti laiptini� 
funkcij� seka ir $A := \lim \uparrow \int_{\R^k} \phi_n < +\infty$, tai
\[ \lim_{n\to\infty} \phi_n < +\infty \text{ b.v.}
\]
\noindent (t.y., $\{ x \in \R^k : \lim_{n\to\infty} \phi_n(x) = +\infty\} \in
\mathcal{N}$).
\end{tr}

\begin{isv}								% 1.13
Jei $\{ \phi_n \} \subset S(\R^k)$, $\phi_n \ge 0$, $n \in \N$ ir
$\sum_{n=1}^\infty \int_{\R^k} \phi_n < +\infty$, tai
\[ \sum_{n=1}^\infty \phi_n < +\infty \text{ b.v.}
\]
\end{isv}


\section{Lebego integralas}						% 2

Toliau visur $I \subset \R^k$ -- bet kokia $k$-mat� gardel�.

\begin{ap}								% 2.1
Laiptini� funkcij� seka $\{ \phi_n \} \subset S(I)$ vadinama
\Def{fundamentalia} arba \Def{Ko�i} seka, jei
\[ \forall \eps > 0 \quad \exists N \in \N : \quad
   \int_I | \phi_n - \phi_m | < \eps
   \text{, kai $n, m > N$}
\]
\begin{ps*}
Norint atskirti tokias funkcij� Ko�i sekas nuo skai�i� Ko�i sek� kartais
sakoma, kad tokia seka yra Ko�i seka $L^1$ arba vidurkine pasme.
\end{ps*}
\end{ap}

\begin{ap}								% 2.2
Funkcija $f: I \to \RR$ vadinama \Def{integruojama} (Lebego prasme), jei
$\exists$ Ko�i seka $\{\phi_n\}\subset S(I)$, konverguojanti b.v. � $f$.
Tokiu atveju jos \Def{Lebego integralu} gardel�je $I$ vadinamas skai�ius
\[ \int_I f := \lim_{n \to \infty} \int_I \phi_n
\]
\noindent Kiti integralo �ym�jimai yra tokie patys kaip ir laiptini� funkcij�.

Vis� integruojam� gardel�je $I$ funkcij� klas� �ym�sime $L(I)$ arba $L^1(I)$.
\end{ap}

\begin{ps}								% 2.3
Apibr��ime nurodyta riba visada egzistuoja.

\begin{ir*}
I� ties�, remiantis apibr��imu, $\forall \eps > 0$ $\exists N \in \N :$
\( \int_I |\phi_n - \phi_m| < \eps\), kai $n, m > N$, t.y.
\( |\int_I \phi_n - \int_I \phi_m | \le \int_I |\phi_n - \phi_m| < \eps\),
kai $n, m > N$.  Tai rei�kia, kad $\{ \int_I \phi_n \}$ yra skai�i� Ko�i seka
ir tod�l konverguoja.
\end{ir*}
\end{ps}

\begin{tg}[korekti�kumas]						% 2.4
Integralo apibr��imas yra korekti�kas, t.y. jame nurodyta riba
$\lim_{n\to\infty} \int_I \phi_n$ nepriklauso nuo sekos $\{\phi_n\}$
pasirinkimo.
\end{tg}

\begin{tg}[elementarios integralo savyb�s]\NL				% 2.5
\begin{enumerate}
  \item Tiesi�kumas: jei $f, g \in L(I)$, $\alpha, \beta \in \R$, tada
        $\alpha f + \beta g \in L(I)$ ir
        \[ \int_I (\alpha f + \beta g) = \alpha \int_I f + \beta \int_I g
        \]
  \item Jei $f, g \in L(I)$, tai $\max \{f, g\}, \min \{f, g\}, |f| \in L(I)$
  \item Monotoni�kumas: jei $f, g \in L(I)$, $f \le g$ b.v. gardel�je $I$,
        tada
        \[ \int_I f \le \int_I g\]
        \noindent Atskiru atveju
        \[ \left| \int_I f \right| \le \int_I |f| \]
  \item Jei $f \in L(I)$, $g = f$ b.v. gardel�je $I$, tada $g \in L(I)$ ir
        \[ \int_I g = \int_I f \]
\end{enumerate}
\end{tg}

\begin{tr}[Bepo Levi teorema]						% 2.6
Jei $\{ f_n \} \subset L(I)$, $f_n \uparrow f$ ($f_n \downarrow f$) b.v.
gardel�je $I$, $\lim_{n \to \infty} \uparrow \int_I f_n < +\infty$
(atitinkamai $\lim_{n \to \infty} \downarrow \int_I f_n > -\infty$), tai $f
\in L(I)$ ir
\[ \int_I f = \lim_{n \to \infty} \int_I f_n
\]
\begin{ir}
Be �rodymo.
\end{ir}
\end{tr}

\begin{isvv}\NL								% 2.7
\begin{enumerate}
  \item Jei $f_n \in L(I)$, $\sum_{n=1}^\infty \int_I |f_n| < +\infty$, tai
        $\sum_{n=1}^\infty f_n$ konverguoja absoliu�iai b.v., jos suma
        \[f := \sum_{n=1}^\infty f_n \in L(I)\qquad\text{ir}\qquad\int_I f = 
        \sum_{n=1}^\infty \int_I f_n\]
  \item Sakykime, $f \in L(I)$. Tada
        \[ \int_I |f| = 0 \ttt f = 0 \text{ b.v. gardel�je $I$}.
        \]
\end{enumerate}
\end{isvv}

\begin{tr}[Fatu--Lebego\footnotemark{} teorema]\footnotetext{Fatoux--Lebegue}\NL
									% 2.8
\begin{enumerate}
  \item[1a.] $\{ f_n \} \subset L(I)$, $f_n \le g \in L(I)$, $n \in \N$,
            $\limsup_{n\to\infty} \int_I f_n > -\infty$    \tada
            \[ \limsup_{n\to\infty} f_n \in L(I) \quad\textrm{ir}\quad
            \int_I \limsup_{n\to\infty} f_n \ge \limsup_{n\to\infty} \int_I f_n
            \]
  \item[1b.] $\{ f_n \} \subset L(I)$, $f_n \ge g \in L(I)$, $n \in \N$,
            $\liminf_{n\to\infty} \int_I f_n < +\infty$    \tada
            \[ \liminf_{n\to\infty} f_n \in L(I) \quad\textrm{ir}\quad
            \int_I \liminf_{n\to\infty} f_n \le \liminf_{n\to\infty} \int_I f_n
            \]
  \item[2.] (Lebego teorema apie ma�oruot� konvergavim�)
           $\{ f_n \} \subset L(I)$, $|f_n| \le g \in L(I)$, $f_n \to f$ b.v.
           gardel�je $I$ \tada $f \in L(I)$ ir $\int_I f_n \to \int_I f$.
\end{enumerate}
\end{tr}

\begin{tg}								% 2.9
Jei $f: I \to \R$ yra apr��ta ir b.v. tolydi funkcija kompakti�koje gardel�je
$I$, tai $f \in L(I)$. 

\noindent
Atskiru atveju, $C(I) \subset L(I)$ kompakti�koje ardel�je $I$; $D[a, b]
\subset L[a, b]$.
\end{tg}

\begin{ps}								% 2.10
Galima �rodyti, kad funkcijos $f$ apr��tumas ir b.v. tolydumas intervale $[a,
b]$ yra b�tina ir pakankama s�lyga, kad $f \in R[a, b]$ (integruojama Rymano
prasme). Taigi, $R[a, b] \subset L[a,b]$.
\end{ps}

\begin{ap}								% 2.11
Sakoma, kad funkcija $f: \R^k \to \RR$ yra \Def{mati}, jei $\exists$ seka
$\{ \phi_n \} \subset S(\R^k)$, su kuria $\phi_n \to f$ b.v.  Sakoma, kad
funkcija $f: A \to \RR$ ($A \subset \R^k$) yra mati, jei jos t�sinys $\hat{f}
: \R^k \to \R$, apibr��iamas lygybe
\[ \hat{f}(x) := \begin{cases} f(x), & \text{jei $x \in A$}, \\
                               0,    & \text{jei $x \in \R^k \setminus A$}
                 \end{cases}
\]
yra mati funkcija.  Vis� toki� funkcij� aib� �ym�sime $M(A)$, be to
susitarsime �ym�ti $M := M(\R^k)$.
\end{ap}

\medskip
Visi tolimesni teiginiai bus formuluojami funkcij� klasei $M$.

\begin{tg}\NL								% 2.12
\begin{enumerate}
  \item $L := L(\R^k) \subset M$.
  \item $f, g \in M$, $|f| < +\infty$ b.v., $|g| < +\infty$ b.v.  Tada
        $\alpha f + \beta g \in M$ ($\alpha, \beta \in \R$), $fg \in M$,
        $\max \{f, g\} \in M$, $\min \{f, g\} \in M$, $|f| \in M$; jei be to
        $g \ne 0$ b.v., tai $\frac{f}{g} \in M$.
\end{enumerate}
\begin{ps*}
Tais atvejais, kai nurodytas veiksmas neapibr��tas kokiame nors ta�ke (pvz.,
$+\infty - +\infty$, $\frac{0}{0}$ ir t.t.), laikykime, kad rezultatas yra
bet koks skai�ius, pvz., nulis.  Toks funkcijos reik�mi� kaitaliojimas nulinio
mato aib�je nekei�ia nei funkcijos matumo, nei integruojamumo.
\end{ps*}
\end{tg}

\shortpage
\begin{tr}\NL								% 2.13
\begin{enumerate}
\item Tarkime, $f \in M$, $g \in L$, $|f| \le g$ b.v.  Tada $f \in L$.
\item $M \ni f_n \to f$ b.v. \tada $f \in M$.
\item Jei $f: \R^k \to \R$ yra b.v. tolydi, tai $f \in M$.
\end{enumerate}
\end{tr}

\begin{ap}								% 2.14
Jei $f \ge 0$ ir $f \in M \setminus L$, tai laikysime, kad
\[ \int_{\R^k} f = +\infty \]

Jei $f \in M$ ir bent viena i� funkcij� $f^+, f^- \in L$, tai susitarsime, kad
\[ \int_{\R^k} f = \int_{\R^k} f^+ - \int_{\R^k} f^-
\]

\begin{ps*}
Jei $f \in L$, tai pastaroji lygyb� yra jau �inoma savyb�, tod�l �iame
apibr��ime yra nauja tik tai, kad kai kuriom funkcijom apibr��iami begalines
reik�mes �gyjantys integralai.
\end{ps*}

Jei $\int_{\R^k} f$ yra apibr��tas (baigtinis arba begalinis), tai funkcija
$f$ vadinama \Def{kvaziintegruojama}.
\end{ap}

\begin{tg}\NL								% 2.15
\begin{enumerate}
  \item Jei $f, g$ -- kvaziintegruojamos funkcijos, $f \le g$, tai
        \[ \int_{\R^k} f \le \int_{\R^k} g \]
  \item Jei $0 \le f_n \uparrow f$ b.v., $f_n \in M$, tada
        \[ \int_{\R^k} f_n \uparrow \int_{\R^k} f \]
  \item[2$'$.] Jei $0 \le f_n$, $f_n \in M$, tada
        \[ \int_{\R^k} \sum_{n=1}^\infty f_n =
           \sum_{n=1}^\infty \int_{\R^k} f_n
        \]
\end{enumerate}
\end{tg}

\begin{ap}								% 2.16
Sakoma, kad aib� $A \subset \R^k$ yra \Def{mati} (Lebego prasme), jei
$\Ind_A \in M$. Tokios aib�s \Def{matu} (Lebego matu) vadinamas skai�ius
$m(A) := \int_{\R^k} \Ind_A$.

\begin{ps*}
$k$-mat�s gardel�s ir nulinio mato aibi� atveju �is apibr��imas nesiskiria nuo
ankstesnio.
\begin{ir*}
I� tikr�j�, jei $A$ -- $k$-mat� gardel�, tai $\int_{\R^k} \Ind_A = 1 \cdot
m(A) = m(A)$, o jei $A$ -- nulinio mato aib�, tai $\Ind_A = 0$ b.v., tod�l
$\int_{\R^k} \Ind_A = 0$.
\end{ir*}
\end{ps*}
\end{ap}

\begin{tg}[ma�i�j� aibi� ir mat� savyb�s]\NL				% 2.17
\begin{enumerate}
  \item Jei $A$ ir $B$ -- ma�iosios aib�s, tai $A \cap B$, $A \cup B$, $A
        \setminus B$, $A^\complement$ -- taip pat ma�iosios aib�s. Jei, be to,
        $A \cap B = \emptyset$, tai $m(A \cup B) = m(A) + m(B)$ (adityvumas), 
        o jei $A \subset B$, tai $m(A) \le m(B)$ (monotoni�kumas).
  \item Jei $A_n$, $n \in \N$ -- ma�i�j� aibi� seka, tai $\bigcup_{n=1}^\infty
        A_n$ -- taip pat mati aib�.  Jei, be to, $A_i \cap A_j = \emptyset$,
        kai $i \ne j$, tai $m(\bigcup_{n=1}^\infty A_n) = \sum_{n=1}^\infty
        m(A_n)$ ($\sigma$-adityvumas).
  \item Jei $A_n$ ($n \in \N$) -- monotoni�kai did�janti (t.y. $A_n \subset
        A_{n+1}$) ma�i�j� aibi� seka, tai
        \[ m\Big(\bigcup_{n=1}^\infty A_n\Big) = \lim_{n\to\infty} \uparrow 
           m(A_n)
        \]

        Jei $A_n$ -- monotoni�kai ma��janti ma�i�j� aibi� seka ir $\exists
        n_0 \in \N:$ $m(A_{n_0}) < +\infty$, tai
        \[ m\Big(\bigcap_{n=1}^\infty A_n\Big) = \lim_{n\to\infty}
           \downarrow m(A_n)
        \]
\end{enumerate}
\begin{pss*}\NL
\begin{enumerate}
  \item Did�jan�ios aibi� sekos atveju priimta ra�yti $A_n \uparrow A :=
        \bigcup_{n=1}^\infty A_n$, ma��jan�ios -- $A_n \downarrow A :=
        \bigcap_{n=1}^\infty A_n$.  Tod�l tre�i� savyb� galima u�ra�yti taip:
        $A_n \uparrow A$ \tada $m(A_n) \uparrow m(A)$; $A_n \downarrow A$ ir
        $m(A_{n_0}) < +\infty$ \tada $m(A_n) \downarrow m(A)$.
  \item S�lyga $m(A_{n_0}) < +\infty$ su bent vienu $n_0$ -- esmin�.
        Pvz., \(A_n := [n, +\infty) \downarrow \emptyset\), bet
        $m(A_n) = +\infty \not\downarrow m(\emptyset) = 0$.
\end{enumerate}
\end{pss*}
\end{tg}

\begin{tg}								% 2.18
Visos atviros ir u�daros aib�s yra ma�ios.
\end{tg}

\begin{ap}								% 2.19
Funkcijos $f: \R^k \to \RR$ \Def{integralu} ma�iojoje aib�je $A$ vadinamas
skai�ius
\[ \int_A f := \int_{\R^k} f \cdot \Ind_A
\]
(su s�lyga, kad $f \cdot \Ind_A \in M$). Jei $\int_A f \in \R$ (t.y. $\ne
\pm\infty$), tai ra�ysime, kad $f \in L(A)$ ir sakysime, kad funkcija $f$ yra
\Def{integruojama} (Lebego prasme) aib�je $A$.
\end{ap}

\begin{tg}\NL								% 2.20
\begin{enumerate}
  \item (Adityvumas) $f \in L(A_1) \cap L(A_2)$, $A_1 \cap A_2 = \emptyset$
        \tada $f \in L(A_1 \cup A_2)$ ir
        \[ \int_{A_1 \cup A_2} f = \int_{A_1} f + \int_{A_2} f
        \]
  \item[1$'$.] ($\sigma$-adityvumas)
        Jei $f \in L(A)$, $A = \bigcup_{n=1}^\infty A_n$, $A_n \in M$, $n \in
        \N$, $A_i \cap A_j = \emptyset$, kai $i \ne j$, tada
        \[ \int_A f = \sum_{n=1}^\infty \int_{A_n} f
        \]
  \item Tarkime, kad $A_n \uparrow A$ ($A_n \in M$, $n \in \N$). Tada
        $\int_{A_n} f \to \int_A f$, jei i�pildyta bent viena i� �i� trij�
        s�lyg�:
        \begin{enumerate}
          \item $f \in L(A)$
          \item $\displaystyle \lim_{n\to\infty} \int_{A_n} |f| < +\infty$
          \item $M \ni f \ge 0$
        \end{enumerate}
\end{enumerate}
\end{tg}


\section[Fubinio ir kintam�j� keitimo teoremos]{Fubinio (integralo suvedimo � 
kartotinius integralus) ir kintam�j� keitimo teoremos}

\begin{tr}[Fubinio teorema]\label{tr:3.1}				% 3.1
Jei \(f \in L(I_1 \times I_2)\), \(I_1, I_2 \subset \R\) -- intervalai, tai
\[
  \int_{I_1 \times I_2} f = \int_{I_1} \Big( \int_{I_2} f(x,y)\,\dy \Big)\,\dx
                          = \int_{I_2} \Big( \int_{I_1} f(x,y)\,\dx \Big)\,\dy
\]

\begin{ps*}
Pirmoji lygyb� suprantama taip: integralas \(F_1(x) := \int_{I_2}
f(x,y)\,\dy\) egzistuoja su b.v. $x \in I_1$.  Tuose ta�kuose $x$, kuriuose
$F_1(x)$ neapibr��tas, susitarsime, laikyti, kad $F_1(x) := 0$.  Tuomet
$\int_{I_1} F_1(x)\,\dx = \int_{I_1 \times I_2} f$.

Analogi�kai galima komentuoti ir antr�j� lygyb�.
\end{ps*}
\begin{kir}
\begin{description}
  \item[1 �ingsnis:] $f = \Ind_{J_1 \times J_2}$, $J_1 \subset I_1$,
                     $J_2 \subset I_2$ -- intervalai. Tuomet
    \[
    \int_{I_1 \times I_2} f = 1 \cdot m_2(J_1 \times J_2) = m(J_1)
    \cdot m(J_2)
    \]
    ($m_2$ -- matas plok�tumoje, $m$ -- matas ties�je).

    \begin{align*}
    \int_{I_1} \left( \int_{I_2} f(x,y)\,\dy\right)\,\dx &=
    \int_{I_1} \left( \int_{I_2} \Ind_{J_1\times J_2}(x,y)\,\dy
               \right)\,\dx
    = \int_{I_1} \left( \int_{I_2} \Ind_{J_1}(x) \Ind_{J_2}(y)\,\dy
                 \right)\,\dx \\&
    = \int_{I_1} \left( \Ind_{J_1}(x) \int_{I_2} \Ind_{J_2}(y)\,\dy
                 \right)\,\dx
    = \int_{I_1} \Ind_{J_1}(x) \cdot 1 \cdot m(J_2) \,\dx \\&
    = m(J_2) \cdot 1 \cdot m(J_1)
    = m(J_1) \cdot m(J_2)
    \end{align*}
  \item[2 �ingsnis:] $f \in S(I_1 \times I_2)$

    $f$ galima i�reik�ti suma
    \[ f = \sum_{i=1}^m c_i \Ind_{J_i}
      \qquad \text{$J_i$ -- sta�iakampiai (dvimat�s gardel�s)}
    \]

    Remiantis pirmuoju �ingsniu ir integral� tiesi�kumu gauname
    \begin{align*}
    \int_{I_1 \times I_2} f &= 
    \int_{I_1 \times I_2} \sum_{i=1}^m c_i \Ind_{J_i} = % tiesi�kumas
    \sum_{i=1}^m c_i \int_{I_1 \times I_2} \Ind_{J_i} =
    \sum_{i=1}^m c_i \int_{I_1} \left( \int_{I_2} \Ind_{J_i}(x,y)\,\dy
                                \right)\,\dx \\&=
    \int_{I_1} \left( \int_{I_2} \sum_{i=1}^m c_i \Ind_{J_i}(x,y)\,\dy 
               \right)\,\dx =
    \int_{I_1} \left( \int_{I_2} f(x,y) \,\dy \right)\,\dx
    \end{align*}
  \item[3 �ingsnis:] $f \in L(I_1 \times I_2)$.

    Imkime laiptini� funkcij� Ko�i sek� $S(I_1\times I_2) \ni f_n \to
    f$ b.v.  Tada kiekvienam $n$ turime (remdamiesi antruoju �ingsniu)
    \[ \int_{I_1 \times I_2} f_n =
       \int_{I_1} \left( \int_{I_2} f_n(x,y) \,\dy \right)\,\dx
    \]
    Per�je prie ribos, kai $n \to \infty$, gauname
    \[ \int_{I_1 \times I_2} f =
       \int_{I_1} \left( \int_{I_2} f(x,y) \,\dy \right)\,\dx
    \]

    �rodymo techninis sunkumas -- grie�tai pagr�sti per�jima prie
    ribos de�in�je lygyb�s pus�je.  �ia tai nebus padaryta, tad
    �rodymas nepilnas.
\end{description}
\enlargethispage*{10cm}
\end{kir}
\pagebreak
\end{tr}

\begin{pss}\NL								% 3.2
\begin{enumerate}
  \item Teorema bus teisinga ir tada, kai $I_1 \subset \R^k$, $I_2 \subset
        \R^m$ -- gardel�s ($k$-mat� ir $m$-mat�).
  \item Tarkime, $D = \{ (x, y) \in \R^2 : a \le x \le b, y_1(x) \le y \le
        y_2(x) \} \subset \R^2$, $y_1, y_2 \in M([a, b])$, $y_1 \le y_2$.

        Jei $f \in L(D)$, tai
        \begin{align*}
        \int_D f &= \iint_{\R^2} f \Ind_D
                  = \int_{-\infty}^{+\infty} \left(
                    \int_{-\infty}^{+\infty} f(x,y) \Ind_D(x,y) \,\dy
                                             \right)\,\dx
              \\ &= \int_{-\infty}^{+\infty} \left(
                    \int_{-\infty}^{+\infty} f(x,y) \Ind_{[a,b]}(x) 
                                                    \Ind_{[y_1(x),y_2(x)]}(y)
                                       \,\dy \right)\,\dx
                  = \int_{-\infty}^{+\infty} \left( \Ind_{[a,b]}(x)
                    \int_{y_1(x)}^{y_2(x)} f(x,y)\,\dy \right) \,\dx
              \\ &= \int_a^b \Big( \int_{y_1(x)}^{y_2(x)} f(x,y)\,\dy
                             \Big)\,\dx
        \end{align*}

        Analogi�kai, jei $V = \{ (x, y, z) \in \R^3: (x, y) \in D,
        z_1(x, y) \le z \le z_2(x, y) \} \subset \R^3$ ir $f \in L(V)$, tai
        \[ \int_V f = \iint_D \Big( \int_{z_1(x, y)}^{z_2(x, y)} f(x, y, z) 
           \,\dz\Big)\,\dx\dy
        \]
\end{enumerate}
\end{pss}

\begin{tr}[Tonelio teorema]						% 3.3
Jei $f \in M(I_1 \times I_2)$, $I_1, I_2$ -- intervalai, $f \ge 0$, tai
\begin{equation}
   \int_{I_1 \times I_2} f = \int_{I_1} \Big( \int_{I_2} f(x,y)\,\dy \Big)
   \,\dx = \int_{I_2} \Big( \int_{I_1} f(x,y)\,\dx \Big) \,\dy
   \tag*{$(\le +\infty)$}
\end{equation}

�ia galioja ta pati pastaba kaip ir Fubinio teoremai (\ref{tr:3.1}).  �i
teorema tinka visoms kvaziintegruojamoms funkcijoms.
\end{tr}

\begin{isv}								% 3.4
Jei $f \in M(I_1 \times I_2)$ ir $\int_{I_1} \big( \int_{I_2} |f(x,y)|\,\dy
\big)\,\dx < +\infty$, tai $f \in L(I_1 \times I_2)$ ir tuo pa�iu funkcijai
$f$ teisinga Fubinio teorema (\ref{tr:3.1}).
\end{isv}

\begin{pvzz}\NL								% 3.5
\begin{enumerate}
  \item Tarkime, $f \in L(I_1)$, $g \in L(I_2)$. Tada
        \[ \int_{I_1 \times I_2} f \circledast g = \int_{I_1} f \cdot 
           \int_{I_2} g
        \]
        ($f \circledast g (x,y) := f(x) \cdot g(y)$ -- \Def{tenzorin�
        sandauga}).
        \begin{ir*}
        \[
        \iint_{I_1 \times I_2} f(x)g(y)\,\dx\dy 
         = \int_{I_1} \left( f(x) \int_{I_2} g(y)\,\dy \right)\,\dx
         = \int_{I_1} f(x)\,\dx \cdot \int_{I_2} g(y) \,\dy
        \]
        \end{ir*}

        Pvz:
        \[
        \iint_{\R^2} \frac{\dx\dy}{(1+x^2)(1+y^2)} 
        = \left(\int_{-\infty}^{+\infty} \frac{\dx}{1+x^2}\right)^2 = \pi^2
        \]
  \item Tarkime, $\displaystyle f(x, y) := \frac{x^2 - y^2}{(x^2 + y^2)^2}$
        \begin{align*}
          \int_0^1 \dx \int_0^1 f(x,y)\,\dy &= \frac{\pi}{4} \\
          \int_0^1 \dy \int_0^1 f(x,y)\,\dx &= -\frac{\pi}{4}
        \end{align*}
        Gauname skirtingus rezultatus, nes $f \not\in L([0,1]^2)$.
\end{enumerate}
\end{pvzz}

\medskip
Toliau $D$ ir $\widetilde{D}$ -- bet kokios atviros aib�s erdv�je $\R^k$.

\begin{ap}								% 3.6
Diferencijuojamos funkcijos $f: D \to \R^k$ \Def{Jakobianu} ta�ke $x \in D$
vadinamas skai�ius
\[
  J_f(x) := \frac{D(f_1, f_2, \ldots, f_k)}{D(x_1, x_2, \ldots, x_k)} :=
            \det f'(x) = \begin{vmatrix}
              \frac{\partial{f_1}}{\partial{x_1}} &
                \frac{\partial{f_1}}{\partial{x_2}} &
                  \ldots &
                    \frac{\partial{f_1}}{\partial{x_k}} \\
              \frac{\partial{f_2}}{\partial{x_1}} &
                \frac{\partial{f_2}}{\partial{x_2}} &
                  \ldots &
                    \frac{\partial{f_2}}{\partial{x_k}} \\
              \vdots &
                \vdots &
                  \ddots &
                    \vdots \\
              \frac{\partial{f_k}}{\partial{x_1}} &
                \frac{\partial{f_k}}{\partial{x_2}} &
                  \ldots &
                    \frac{\partial{f_k}}{\partial{x_k}} \end{vmatrix}
\]
\end{ap}

\begin{tr}[kintam�j� keitimo teorema]					% 3.7
Jei $g \in C^1(D, \widetilde{D})$ yra bijekcija, $\forall x \in D$ $J_g(x) 
\ne 0$ ir $f \in L(\widetilde{D})$, tai
\[ \int_{\widetilde{D}} f(y)\,\dy = \int_D f\big(g(x)\big) \, |J_g(x)| \,\dx
\]
arba
\[ \int_{\widetilde{D}} f(y_1, \ldots, y_k)\,\dy_1\dots\dy_k = 
   \int_D f\big(g_1(x_1, \ldots, x_k), \ldots, g_k(x_1, \ldots, x_k)\big) \, 
   \left|\frac{D(g_1, \ldots, g_k)}{D(x_1, \ldots, x_k)}\right| 
   \,\dx_1 \dots \dx_k
\]
(�ia $g = (g_1, \ldots, g_k)$).

\begin{ir}
Be �rodymo.
\end{ir}
\end{tr}

\begin{pvzz}\NL								% 3.8
\begin{enumerate}
  \item (Polinis kintam�j� keitimas)

        Imkime \(\widetilde{D} = \R^2 \ni (x, y) \longleftrightarrow (r, \phi)
        \in D = [0; +\infty) \times [0; 2\pi]\)
        \[ \begin{cases} x = r \cos \phi \\ y = r \sin \phi \end{cases}
        \]

        Kadangi n�ra vienareik�mi�kumo, galime imti \(\widetilde{D}' :=
        \widetilde{D} \setminus [0; +\infty) \times \{0\}\) ir \(D' := (0;
        +\infty) \times (0; 2\pi)\). Integralas nuo to nepasikeis.

        \[ \iint_{\R^2} f(x, y)\,\dx\dy = \int_0^{2\pi} \dphi
           \int_0^{\infty} f(r\cos\phi, r\sin\phi) \, \left|\frac{D(x, y)}%
           {D(r, \phi)}\right|\,\dr
        \]

        Jakobianas
        \[ \frac{D(x, y)}{D(r, \phi)} =
        \begin{vmatrix}
             \frac{\partial{x}}{\partial{r}} &
               \frac{\partial{x}}{\partial{\phi}} \\
             \frac{\partial{y}}{\partial{r}} &
               \frac{\partial{y}}{\partial{\phi}}
           \end{vmatrix} =
           \begin{vmatrix}
             \cos\phi & -r\sin\phi \\
             \sin\phi & -r\cos\phi \\
           \end{vmatrix} =
           r (\cos^2 \phi + \sin^2 \phi) = r
        \]

        Pvz.:
        \[ \iint_{x^2+y^2 \le R^2} f(x, y) \,\dx\dy =
           \int_0^{2\pi} \dphi \int_0^R f(r\cos\phi, r\sin\phi)\,r\,\dr
        \]

        Kitas pvz.:
        \[ I = \iint_{\R^2} e^{-(x^2+y^2)}\,dx\dy
             = \int_0^{2\pi} \dphi \int_0^{+\infty} e^{-r^2}\,
               \underbrace{r\dr}_{= \frac{1}{2}\dr^2}
             = 2\pi \cdot \left(-\frac{e^{-r^2}}{2}\right)\Bigg|_0^{+\infty}
             = \pi
        \]
        Beje,
        \[ I = \left(\int_{-\infty}^{+\infty} e^{-x^2}\,\dx\right)^2 = \pi
           \tada
           \int_{-\infty}^{+\infty} e^{-x^2}\,\dx = \sqrt{\pi}
        \]
  \item (Sferin�s koordinat�s)
        \[ \begin{cases} x = r \sin \theta \cos \phi \\
                         y = r \sin \theta \sin \phi \\
                         z = r \cos \theta \end{cases}
        \]
        Jei $(x, y) \ne (0, 0)$, tai $(x, y, z)$ atitinka vienint�lis trejetas
        $(r, \phi, \theta)$, kai $r \in (0; +\infty)$, \(\phi \in [0; 2\pi)\),
        $\theta \in (0; \pi)$ (kartais imama \(\theta \in (-\frac{\pi}{2};
        \frac{\pi}{2})\)).

        Jakobianas
        \[\begin{split}
           \frac{D(x, y, z)}{D(r, \phi, \theta)} &=
           \begin{vmatrix}
             \frac{\partial{x}}{\partial{r}} &
               \frac{\partial{x}}{\partial{\phi}} &
                 \frac{\partial{x}}{\partial{\theta}} \\
             \frac{\partial{y}}{\partial{r}} &
               \frac{\partial{y}}{\partial{\phi}} &
                 \frac{\partial{y}}{\partial{\theta}} \\
             \frac{\partial{z}}{\partial{r}} &
               \frac{\partial{z}}{\partial{\phi}} &
                 \frac{\partial{z}}{\partial{\theta}}
           \end{vmatrix} =
           \begin{vmatrix}
             \cos\phi\sin\theta & -r\sin\phi\sin\theta & r\cos\phi\cos\theta \\
             \sin\phi\sin\theta &  r\cos\phi\sin\theta & r\sin\phi\cos\theta \\
             \cos\theta & 0 & -r\sin\theta
           \end{vmatrix} = \\ &=
           \cos\theta \cdot (-r^2)\sin\theta\cos\theta (\sin^2\phi + \cos^2\phi)
           - r\sin\theta \cdot r\sin^2\theta (\cos^2\phi + \sin^2\phi) = \\ &=
           -r^2 \sin\theta (\cos^2\theta + \sin^2 \theta) =
           -r^2 \sin\theta
        \end{split}\]

        Imkime, pvz., $B_R := \{ (x, y, z): x^2+y^2+z^2 < R^2 \}$.
        �i� srit� atitiks sritis \([0; R) \times [0; 2\pi) \times [0; \pi)\)
        polin�se koordinat�se.

        Nor�dami u�tikrinti teoremos s�lygas imkime srit� $D := (0; R) \times 
        (0; 2\pi) \times (0; \pi)$. J� atitiks $B_R \supset \widetilde{D} =
        \{(x, y, z): x^2+y^2+z^2 < R^2 \text{ ir } (y \ne 0 \text{ arba }
        x < 0) \}$.
        \begin{align*}
        \iiint_{B_R} f(x,y,z)\,\dx\dy\dz &= \iiint_{\widetilde{D}} f(x,y,z)
        \,\dx\dy\dz = \\ &=
        \int_0^{2\pi} \dphi \int_0^\pi \dtheta \int_0^R f(r\cos\phi\sin\theta,
        r\sin\phi\sin\theta, r\cos\theta) \, r^2\sin\theta\,\dr
        \end{align*}

        Imkime, pvz., $f(x, y, z) = (x^2+y^2+z^2)^\alpha$.
        \begin{align*}
        \iiint_{B_R} f &= \int_0^{2\pi}\dphi \int_0^\pi \dtheta \int_0^R
        r^{2\alpha} \, r^2\sin\theta \, \dr =
        2\pi \underbrace{\int_0^\pi \sin\theta \, \dtheta}_{=2} \cdot
        \int_0^R r^{2\alpha+2} \,\dr = 4\pi\frac{R^{2\alpha+3}}{2\alpha+3}
        \end{align*}
  \item[2$'$.] (Cilindrin�s koordinat�s)

        Kai $\widetilde{D} = \{ (x, y, z) : x^2 + y^2 < R^2, 0 \le z \le H\}$,
        patogu naudotis cilindrin�mis koordinat�mis
        \[ \begin{cases} x = r \cos \phi \\
                         y = r \sin \phi \\
                         z = z \end{cases}
        \]

        Jakobianas toks pats kaip ir polinio koordina�i� keitinio:
        \[ \frac{D(x, y, z)}{D(r, \phi, z)} = r
        \]
  \item Paskai�iuoti plot� fig�ros, apibr��tos kreiv�mis $y^2 = 2x$,
        $y^2 = x$, $y = x^2$, $2y = x^2$.

        \begin{eqnarray*}
        \begin{cases} x \le y^2 \le 2x \\ y \le x^2 \le 2y \end{cases}
        \ttt
        \begin{cases} 1 \le \frac{y^2}{x} \le 2 \\ 
                      1 \le \frac{x^2}{y} \le 2 \end{cases}
        \end{eqnarray*}

        Pa�ym�kime $u := \frac{y^2}{x}$, $v := \frac{x^2}{y}$
        \begin{eqnarray*}
        \begin{cases} x^3 = uv^2 \\ y^3 = u^2v \end{cases}
        \ttt
        \begin{cases} x = u^{1/3}v^{2/3} \\ y = u^{2/3}v^{1/3} \end{cases}
        \end{eqnarray*}

        \[ \frac{D(x, y)}{D(u, v)} =
           \begin{vmatrix}
             \frac{1}{3} u^{-2/3}v^{2/3} & \frac{2}{3} u^{1/3}v^{-1/3} \\
             \frac{2}{3} u^{-1/3}v^{1/3} & \frac{1}{3} u^{2/3}v^{-2/3}
           \end{vmatrix} = \frac{1}{9} - \frac{4}{9} = - \frac{1}{3}
        \]

        Plotas
        \[ S(D) = \iint_D \dx\dy = \int_1^2 \int_1^2 \frac{1}{3} \,\du\dv
                = \frac{1}{3}
        \]
\end{enumerate}
\end{pvzz}


\section{Daugialypi� integral� taikymas}				% 4

\begin{pvzz}[plot� ir t�ri� skai�iavimas]\NL				% 4.1
\begin{enumerate}
  \item Aib�s (,,fig�ros``) $A \subset \R^2$ \Def{plotu} vadinamas jos Lebego
        matas $m(A)$ (su s�lyga, kad $A$ -- mati aib�).

        Imkime $A = \{(x, y) : a \le x \le b, y_1(x) \le y \le y_2(x) \}$,
        $y_1 \le y_2$ -- ma�ios funkcijos (t.y., $A$ -- kreivin� trapecija).
        \begin{align*}
          m(A) &= \iint_A\dx\dy 
                = \iint_{R^2} \Ind_A(x,y)\,\dx\du 
                = \iint_{R^2} \Ind_{[a;b]}(x)\Ind_{[y_1(x);y_2(x)]}(y)\,\dx\dy
            \\ &= \int_{-\infty}^{+\infty} \Big(\int_{-\infty}^{+\infty} 
                       \Ind_{[a;b]}(x)\Ind_{[y_1(x);y_2(x)]}(y)\,\dy\Big)\,\dx
            \\ &= \int_a^b \Big(\int_{y_1(x)}^{y_2(x)} \dy\Big)\,\dx
                = \int_a^b \big(y_2(x)-y_1(x)\big)\,\dx
        \end{align*}

  \item Analogi�kai aib�s (,,k�no``) $A \subset \R^3$ \Def{t�riu} vadinamas jo
        Lebego matas $m(A)$.

        Jei $A = \{(x,y,z): (x,y) \in D, z_1(x, y) \le z \le z_2(x, y) \}$,
        $D \subset \R^2$ ($A$ -- kreivinis cilindras), tai
        \[ m(A) = \iint_D \big(z_2(x,y) - z_1(x,y)\big)\,\dx\dy
        \]
        (su s�lyga, kad $D$ -- mati aib� ir $z_1 \le z_2$ -- ma�ios funkcijos
        aib�je $D$).

  \item Aib�s $A \subset \R^3$ \Def{pj�viu} ta�ke $x \in \R$ vadinama aib�
        $A_x := \{(y,z): (x,y,z) \in A\}$.

        Tada t�ris yra lygus
        \begin{align*}
        m(A) &= \iiint_{\R^3} \Ind_A(x,y,z)\,\dx\dy\dz
              = \iiint_{\R^3} \Ind_{A_x}(y,z)\,\dx\dy\dz
          \\ &= \int_{-\infty}^{+\infty} \Big( \iint_{\R^2} \Ind_{A_x} (y,z)
                \,\dy\dz \Big)\,\dx
              = \int_{-\infty}^{+\infty} S(A_x)\,\dx
        \end{align*}
        �ia $S(A_x) = m_2(A_x)$ -- aib�s $A_x$ plotas.
\end{enumerate}
\end{pvzz}

\begin{pvz}[pavir�iaus ploto skai�iavimai]				% 4.2
Imkime pavir�i� $\Phi = \{ (x,y,z) : (x,y)\in D, z = f(x,y) \}$,
$f: D\to\R$ -- ,,gera`` funkcija.
Kam lygus pavir�iaus $\Phi$ plotas?

Padalinkime $D$ � daug ma�� sriteli�:
\[ D = \bigcup_{i=1}^m D_i \qquad D_i \cap D_j = \emptyset,\text{ kai $i\ne j$}
\]

Kiekvien� $D_i$ atitinka pavir�iaus dalis $\Phi_i = \{(x,y,z): (x,y) \in D_i,
z = f(x,y) \}$.  Kiekvienam $\Phi_i$ imkime ta�k� $(x_i, y_i, z_i) \in
\Phi_i$.  I�veskime per j� lie�iam�j� plok�tum� ir pa�ym�kime
$\widetilde{\Phi}_i$ tos plok�tumos dal�, esan�i� vir� srities $D_i$.

Intuityviai ai�ku, kad
\[ S(\Phi_i) \approx S(\widetilde{\Phi}_i) = \frac{S(D_i)}{|\cos \alpha_i|}
\]
kur $\alpha_i$ -- kampas tarp $\widetilde{\Phi}_i$ ir plok�tumos $Oxy$.

I�veskime $\widetilde{\Phi}_i$ lygt�:
\[ f(x,y) = f(x_i, y_i) + \frac{\partial{f}}{\partial{x}}(x_i,y_i)(x-x_i)
                        + \frac{\partial{f}}{\partial{y}}(x_i,y_i)(y-y_i)
                        + o(\sqrt{(x-x_i)^2+(y-y_i)^2})
\]
kai $x \to x_i$, $y \to y_i$.

Atmet� ,,ma��`` nar� $o(\sqrt{(x-x_i)^2+(y-y_i)^2})$, gauname plok�tum�
\[ z = \underbrace{f(x_i, y_i)}_{= z_i} + 
        \frac{\partial{f}}{\partial{x}}(x_i,y_i)(x-x_i) +
        \frac{\partial{f}}{\partial{y}}(x_i,y_i)(y-y_i)
\]

�i plok�tuma ir vadinama \Def{pavir�iaus lie�iam�ja plok�tuma} ta�ke $(x_i,
y_i, z_i)$.

Primename kampo tarp plok�tum� $ax+by+cz+d=0$ ir $\tilde{a}x+\tilde{b}y+
\tilde{c}z+\tilde{d}=0$ formul�:
\[ \cos\alpha = \frac{|a\tilde{a}+b\tilde{b}+c\tilde{c}|}
                {\sqrt{a^2+b^2+c^2}\sqrt{\tilde{a}^2+\tilde{b}^2+\tilde{c}^2}}
\]


\newcommand{\BjauriSaknis}[1][_i]
           {\sqrt{1 + \left(\frac{\partial{f}}{\partial{x}}(x#1,y#1)\right)^2
                    + \left(\frac{\partial{f}}{\partial{y}}(x#1,y#1)\right)^2}}

Taigi, kampas $\alpha_i$ yra
\[ \cos\alpha_i = \frac{1}{\BjauriSaknis}
\]

�stat� j� � $S(\widetilde{\Phi}_i)$ i�rai�k� gauname
\[ S(\widetilde{\Phi}_i) = S(D_i) \cdot \BjauriSaknis
\]

Tada
\[
S(\Phi) = \sum_i S(\Phi_i) \approx \sum_i S(\widetilde{\Phi}_i) 
        = \sum_i \BjauriSaknis S(D_i)
\]

Pa�ym�kime
\( \phi(x, y) := \BjauriSaknis \), $(x, y) \in D_i$,
t.y.
\[ \phi := \sum_i \BjauriSaknis \Ind_{D_i}
\]

Tuomet
\begin{align*}
S(\Phi) &\approx \sum_i \BjauriSaknis S(D_i) \\
        &=        \sum_i \BjauriSaknis \iint_D \Ind_{D_i} \\
        &= \iint_D \sum_i \BjauriSaknis \Ind_{D_i} \\
        &= \iint_D \phi
\end{align*}

Funkcija $\phi$ priklauso nuo srities $D$ padalinimo � sriteles $D_i$ ir ta�k�
$(x_i, y_i) \in D_i$ pasirinkimo.  $\phi$ art�ja prie $\psi$:
\[ \psi(x,y) := \BjauriSaknis[]
\]
su s�lyga, kad $\frac{\partial f}{\partial x}$ ir $\frac{\partial f}{\partial
y}$ -- tolyd�ios funkcijos.

Taigi,
\[ S(\Phi) = \iint_D \BjauriSaknis[]\,\dx\dy
\]
arba sutrumpintai
\[ S(\Phi) = \iint_D \sqrt{1+(z'_x)^2+(z'_y)^2}\,\dx\dy
\]

Pvz., imkime $z := \sqrt{R^2-x^2-y^2}$, $(x,y) \in D := \{(x,y): x^2+y^2 \le
R^2\}$
\begin{align*}
S_R &= 2 \iint_D \sqrt{1+\left(\frac{-2x}{2\sqrt{R^2-x^2-y^2}}\right)^2
                    +\left(\frac{-2y}{2\sqrt{R^2-x^2-y^2}}\right)^2}\,\dx\dy \\
    &= 2 \iint_D \sqrt{1 + \frac{x^2+y^2}{R^2-x^2-y^2}}\,\dx\dy 
     = 2 \iint_D \sqrt{\frac{R^2}{R^2-x^2-y^2}}\,\dx\dy \\
    &= 2R \iint_D \frac{\dx\dy}{\sqrt{R^2-x^2-y^2}} 
     = 2R \int_0^{2\pi}\dphi \int_0^R \frac{r\,\dr}{\sqrt{R^2-r^2}} \\
    &= 4\pi R \int_0^R \frac{r\,\dr}{\sqrt{R^2-r^2}}
     = 4\pi R \Bigg(\Big(-\frac{1}{2}\Big)\Big(2\sqrt{R^2-r^2}\Big)\Bigg)
              \Bigg|_0^R
     = 4\pi R^2
\end{align*}
\end{pvz}

\begin{pvz}[k�no mas�s skai�iavimas]					% 4.3
Tarkime, kad ,,k�nas`` $A \subset \R^3$ kiekvienam savo ta�ke $(x,y,z)$ turi
\Def{tank�} $\rho(x,y,z) \ge 0$ ir galima $A$ padalinti � atskirus gabalus, 
kuriuose tankis $\rho$ yra pastovus:
\[
  \rho(x,y,z) = \rho_i \qquad x,y,z \in A_i
                       \qquad A = \bigcup_i A_i
                       \qquad A_i \cap A_j = \emptyset,\text{ kai $i \ne j$}
\]

Tada k�no $A$ \Def{mas�} lygi
\begin{align*}
  M(A) &= \sum_i \rho_i \underbrace{V(A_i)}_{\text{t�ris}}
        = \sum_i \rho_i \iiint_{\R^3} \Ind_{A_i}
        = \iiint_{\R^3} \left(\sum_i \rho_i \Ind_{A_i} \right)
    \\ &= \iiint_A \rho(x,y,z)\,\dx\dy\dz
\end{align*}

�ia lygybe apibr��ime \Def{mas�} bet kokiam k�nui:
\[ M(A) = \iiint_A \rho(x,y,z)\,\dx\dy\dz
   \qquad \rho  \ge 0 \text{ -- bet kokia mati funkcija}
\]
\end{pvz}

\begin{pvz}[materialaus k�no statinis momentas ir mas�s centras]	% 4.4
Materialaus ta�ko $a = (x,y,z)$ su mase $m$ \Def{statinis momentas} $Oxy$
plok�tumos at�vilgiu -- tai skai�ius $z \cdot m$ (atstumo iki plok�tumos ir
mas�s sandauga).

Jei turime keli� materiali� ta�k� sistem�, tai jos statinis momentas
apibr��iamas kaip atskirt� ta�k� statini� moment� suma.

Turime material� k�n� $A \subset \R^3$, kurio tankis ta�ke $(x,y,z)$ yra
$\rho(x,y,z) \ge 0$.

Tarkime, kad $A = \bigcup_{i=1}^m A_i$, $A_i \cap A_j = \emptyset$, kai
$i \ne j$, ir $\rho(x,y,z) = \rho_i$, kai $(x,y,z) \in A_i$.

Tada k�no $A$ statinis momentas
\[ M_{xy}(A) \approx \sum_{i=1}^m z_i \underbrace{\rho_i V(A_i)}_{= M(A_i)}
\qquad\text{(�ia $(x_i, y_i, z_i)$ -- bet koks ta�kas i� $A_i$)}
\]

T�sdami galime u�ra�yti
\begin{align*}
M_{xy}(A) &\approx \sum_i z_i \rho_i \iiint_{\R^3} \Ind_{A_i}
          = \iiint_{\R^3} \left( \sum_i \rho_i z_i \Ind_{A_i} \right)
          \approx \iiint_{\R^3} z \underbrace{\sum_i \rho_i \Ind_{A_i}}_{= 
                                                                       \rho}
      \\ &= \iiint_A z \cdot \rho(x,y,z)\,\dx\dy\dz
\end{align*}

Nat�ralu �i� formul� laikyti statinio momento apibr��imu bet kokiam k�nui.
Taip pat galime apibr��ti statinius momentus ir kitoms koordina�i�
plok�tumoms:
\begin{align*}
  M_{xy}(A) &= \iiint_A z \cdot \rho(x,y,z)\,\dx\dy\dz \\
  M_{xz}(A) &= \iiint_A y \cdot \rho(x,y,z)\,\dx\dy\dz \\
  M_{yz}(A) &= \iiint_A x \cdot \rho(x,y,z)\,\dx\dy\dz
\end{align*}

K�no $A$ \Def{mas�s centru} vadinamas ta�kas $(x_0, y_0, z_0)$, kuriame
sukoncentravus vis� k�no mas�, statiniai momentai nepasikeist�.%
\footnote{Galima �rodyti, kad jei statiniai momentai i�liks pastov�s
$Oxy$, $Oxz$ ir $Oyz$ plok�tum� at�vilgiu, tai jie i�liks pastov�s ir vis� kit�
plok�tum� at�vilgiu.}

Mas�s centro koordinat�s yra tokios:
\[
  x_0 = \frac{M_{yz}(A)}{M(A)} \qquad
  y_0 = \frac{M_{xz}(A)}{M(A)} \qquad
  z_0 = \frac{M_{xy}(A)}{M(A)}
\]
\end{pvz}

\begin{pvz}								% 4.5
Apskai�iuosime pusrutulio su spinduliu $R$ ir tankiu $q = 1$ mas�s centr�.

$B^+_R = \{ (x,y,z) : x^2+y^2+z^2 < R^2 \text{ ir } z > 0\}$

\begin{align*}
M_{xy}(B^+_R) &= \iiint_{B^+_R} z\,\dx\dy\dz
               = \int_0^{2\pi}\dphi \int_0^{\pi/2}\dtheta \int_0^R r\cos\theta
                 \cdot r^2\sin\theta \,\dr
           \\ &= 2\pi \int_0^{\pi/2} \sin\theta
                      \underbrace{\cos\theta\,\dtheta}_{= \ud\sin\theta}
                 \cdot \int_0^R r^3\,\dr 
               = 2\pi \cdot \frac{1}{2} \cdot \frac{R^4}{4}
               = \frac{\pi R^4}{4}
\end{align*}
\[ M(B^+_R) = \frac{1}{2} M(B_R) = \frac{2\pi R^3}{3}
\]
\[ z_0 = \frac{\pi R^4}{4}\cdot\frac{3}{2\pi R^3} = \frac{3R}{8}
\]

Taigi, $B^+_R$ mas�s centras yra ta�ke $(0, 0, \frac{3R}{8})$.
\end{pvz}

% I dalies pabaiga.

% FIXME: maybe force odd page?
\part{IPP, kreiviniai bei Oilerio integralai}
\markboth{IPP, kreiviniai bei Oilerio integralai}{}

\section{Integralai, priklausantys nuo parametro}

Da�nai tenka nagrin�ti tokio pavidalo funkcijas:
\[ J(x) := \int_Y f(x, y) \,\dy.
\]
Jos vadinamos integralais, priklausan�iais nuo parametro (sutrumpintai IPP).

Toliau laikysime, kad $X \subset \R^k$, $Y \subset \R^n$ -- ma�ios aib�s, $f:
X \times Y \to \R$ tokia, kad $\forall x \in X \quad f(x, \cdot) \in L(Y)$
(�ia ir kitur, $f(x, \cdot)$ �ymi funkcij� $g_x(y) : Y \to \R$, $\forall y \in
Y \quad g_x(y) := f(x, y)$).

\begin{ap}								% 5.1
Sakysime, kad IPP $J(x) := \int_Y f(x,y)\,\dy$ yra \Def{normaliai
integruojamas} aib�je $A \subset X$ (arba ,,$x \in A$ at�vilgiu``, arba
,,pagal $x \in A$``), jei $\exists g \in L(Y):$
\[ |f(x,y)| \le g(y), \qquad y \in Y, \quad x \in A
\]
(kartais dar sakoma ,,integralas normaliai konverguoja``).
\end{ap}

\begin{tg}								% 5.2
Tarkime, kad funkcija $f: X \times Y \to \R$ tenkina �ias s�lygas:
\begin{enumerate}
\item $\forall y \in Y \quad \exists f_0(y) := \lim_{x \to x^0} f(x,y)$.
\item $\exists$ pradurtoji ta�ko $x^0$ aplinka $\dot{U}$ tokia, kad IPP $J$
yra normaliai integruojamas aib�je $\dot{U} \cap X$ (t.y. $\exists g \in L(Y)
\quad |f(x,y)| \le g(y) \quad y \in Y, \quad x \in \dot{U} \cap X$).
\end{enumerate}
Tada $\exists J_0 := \lim_{x \to x^0} J(x) := \int_Y f_0(y) \,\dy$
(t.y. \[\lim_{x \to x^0} \int_Y f(x,y)\,\dy = \int_Y \lim_{x\to x^0}
f(x,y)\,\dy ).\]
Atskiru atveju, kai vietoje pirmos s�lygos i�plildyta tokia:
\begin{enumerate}
\item[1$'$.] $\forall y \in Y$ funkcija $f(\cdot, y)$ yra tolydi ta�ke
             $x^0$ (t.y. $f_0(y) = f(x^0,y)$),
\end{enumerate}
$J$ yra tolydus ta�ke $x^0$.
\end{tg}

\begin{pvzz}\NL								% 5.3
\begin{enumerate}
\item Funkcijos \(f \in L[0; +\infty)\) \Def{Laplaso transformacija} vadinama
      funkcija \[ F(p) := \int_0^{+\infty} e^{-px}f(x)\,\dx, \qquad p \ge 0 \]

      Du faktai:
      \begin{enumerate}
        \item pointegralin� funkcija $h(p, x) = e^{-px} f(x)$ yra tolydi $p$
              at�vilgiu;
        \item $|h(p,x)| \le |f(x)|$, $p \ge 0$, $x \ge 0$.
      \end{enumerate}

      Kadangi \(|f| \in L[0; +\infty)\), tai IPP $F(p)$ normaliai
      integruojamas visame intervale \([0; +\infty)\).  Remdamiesi praeita
      teorema gauname, kad Laplaso transformacija visada yra toldydi funkcija.
\item Tarkime, $f \in L(\R)$.  Kompleksin� funkcija \[\hat{f}(t) :=
      \int_{-\infty}^{+\infty} e^{itx} f(x) \,\dx, \qquad t \in \R,\] vadinama
      funkcijos $f$ \Def{Furje transformacija}.

      (Primename, kad \(e^{ix} := \cos x + i \sin x\); \(\int(f(x) + i
      g(x))\,\dx := \int f(x)\,\dx + i \int g(x)\,\dx\), \(f, g: \R \to \R\))

      Faktai:
      \begin{enumerate}
        \item IPP tolydus $t$ at�vilgiu;
        \item $|e^{itx} f(x)| = |f(x)|$.
      \end{enumerate}

      Taigi Furje transformacija visada yra tolydi.
\item Tarkime, kad $f \in C(X \times Y)$ ir $X, Y$ -- kompaktai.  Tada
      IPP $J(x) := \int_Y f(x,y) \,\dy$, $x \in X$ -- tolydi funkcija.
\item $F(x) := \int_0^{+\infty} \underbrace{xe^{-xy}}_{=: f(x,y)}\,\dy$, $x
      \ge 0$, $f(x,y)$ tolydi pagal $x$.

      \begin{align*}
      F(0) &= 0 \\
      F(x) &= -e^{-xy} \Big|_{y=0}^{+\infty} = 1 \quad x \ge 0
      \end{align*}

      Taigi, $F(x) \not\in C$ (n�ra tolydus ta�ke $x = 0$).
\end{enumerate}
\end{pvzz}

\begin{tr}								% 5.4
Tarkime, kad funkcija $f: X \times Y \to \R$ tenkina �ias s�lygas:
\begin{enumerate}
\item $\forall y \in Y \quad \exists \frac{\partial f(x^0,y)}{\partial x_i}$
      (pagal ka�koki� vien� koordinat� $x_i$).
\item $\exists$ ta�ko $x^0$ aplinka $U$, kurioje funkcij� sistema
      $\{ \frac{\partial f(x,y)}{\partial x_i}, x \in U \}$ yra normaliai
      integruojama (t.y. $\exists g \in L(Y):$
      \[ \left|\frac{\partial f(x,y)}{\partial x_i}\right| \le g(y), \qquad
      y \in Y, \quad x \in U \text{\/)}.\]
\end{enumerate}
Tada \[\exists \frac{\partial J(x^0)}{\partial x_i} = \frac{\partial}{\partial
x_i} \Big(\int_Y f(x^0, y)\,\dy \Big) = \int_Y \frac{\partial
f(x^0,y)}{\partial x_i} \,\dy.\]
\end{tr}

\begin{pvzz}\NL								% 5.5
\begin{enumerate}
\item $\displaystyle J(a) := \int_0^\infty \frac{\dx}{a+x^2} =
\frac{\pi}{2\sqrt{a}}$, $a > 0$.

Diferencijuodami gausime kair�je pus�je auk�tesnius laipsnius: $\int_0^\infty
\frac{\dx}{(a+x^2)^n}$.  Taip skai�iuoti papras�iau, nei taikyti anks�iau �inom�
rekurentin� formul�.

Formaliai diferencijuokime $a$ at�vilgiu:
\[ J'(a) = - \int_0^\infty \frac{\dx}{(a+x^2)^2} = \frac{\pi}{2} \cdot
\left(-\frac{1}{2}\right) a^{-3/2} = - \frac{\pi}{2 \cdot 2} \cdot
\frac{1}{a^{3/2}}, \qquad a > 0.
\]

Kad �sitikintume diferencijavimo po integralo �enklu teis�tumu, turime
patikrinti, ar gautasis integralas normaliai konverguoja $\forall$ ta�ko $a_0
> 0$ aplinkoje (pakankamai ma�oje).

T�sdami diferencijavim� matome, kad
\[ \int_0^\infty \frac{\dx}{(a+x^2)^{n+1}} = \frac{(2n-1)!! \pi}{2\cdot (2n)!!
a^{n+1/2}}.
\]

(Primename, kad $n!! := \begin{cases} n \cdot (n-2)!! & n > 2 \\ n & n \le 2
\end{cases}$.)

Diferencijavimo teis�tumu �sitikiname lygiai taip pat.

\item $\displaystyle \int_0^1 x^{a-1}\,\dx = \frac{1}{a}$, $a > 0$.

Diferencijuojame pagal $a$:
\[ \int_0^1 x^{a-1} \ln x \,\dx = - \frac{1}{a^2}, \qquad a > 0.
\]

Teis�tumu �sitikaniname pana�iai.

T�sdami toliau gauname, kad
\[ \int_0^1 x^{a-1} \ln^n x \,\dx = (-1)^n \frac{n!}{a^{n+1}}, \qquad a > 0.
\]

% "geras u�davinukas egzaminui" -- V. Mackevi�ius

\item $\displaystyle J(x) = \int_{a(x)}^{b(x)} f(x,y)\,\dy$, $J'(x) = ?$

Atlikime kintam�j� keitim�: $y = a(x)(1-z) + b(x)z$, $z \in [0,1]$.

\begin{align*}
J(x) &= \int_0^1 \underbrace{f\big(x, (1-z)a(x) + zb(x)\big) \cdot
\big(b(x)-a(x)\big)}_{=: \tilde{f}(x,z)} \,\dz
      = \int_0^1 \tilde{f}(x,z)\,\dz
\end{align*}

Jei $a(x)$, $b(x)$ yra ,,geros`` (tiksliau, tolyd�iai diferencijuojamos)
funkcijos, tai
\begin{align*}
J'(x) &= \int_0^1 \frac{\partial \tilde{f}(x,z)}{\partial x} \,\dz
       = \dots % FIXME
       = \int_{a(x)}^{b(x)} \frac{\partial f(x,y)}{\partial x}\,\dy +
         f\big(x, b(x)\big) b'(x) - f\big(x, a(x)\big) a'(x).
\end{align*}

\item $\displaystyle F(p) = \int_0^{+\infty} e^{-px} x f(x) \,\dx$, $p \ge 0$,
\(l \in L[0; +\infty)\) (Laplaso transformacija).  Kada egzistuoja $F'(p)$?

Formaliai diferencijuokime:
\[ F'(p) = -\int_0^{+\infty} e^{-px} x f(x) \,\dx, \qquad p \ge 0.
\]

Ai�ku, kad
\[ |-e^{-px} x f(x) | \le x |f(x)|, \qquad x \ge 0, \quad p \ge 0.
\]

Reikia, kad funkcija $g(x) := x | f(x) |$ b�t� integruojama, t.y.
$\int_0^{+\infty} x|f(x)| \,\dx < +\infty$.

Apjungdami �i� s�lyga su \(f \in L[0; +\infty)\) gauname bendr� s�lyg�:
\[ \int_0^{+\infty} (1+x)|f(x)|\,\dx < +\infty.
\]

Analogi�kai �sitikiname, kad $F^{(k)}(p)$ egzistuoja tada ir tik tada, kai
\[ \int_0^{+\infty} (1+x^k)|f(x)|\,\dx < +\infty.
\]

\end{enumerate}
\end{pvzz}

\begin{ap}[netiesioginiai IPP]						% 5.6
Tarkime, \(f \in C(I \times [a,b))\), $I \subset \R$ -- intervalas.  Tada
\[ J(x) = \int_a^b f(x,y)\,\dy := \lim_{c\uparrow b} \int_a^c f(x,y)\,\dy.
\]

\Def{Netiesioginio integralo} $J$ tolydumui pakanka �ios s�lygos:
\[ |f(x,y)| \le g(y) \text{ kiekvieno ta�ko $x^0 \in I$ aplinkoje}, g \in
L[a,b).
\]

�i s�lyga reikalauja, kad $\int_a^b |f(x,y)| \,\dy < +\infty$, vadinasi
reliatyviai konverguojantiems integralams su parametru reikia ka�ko kito.
\end{ap}

\begin{ap}								% 5.7
Sakysime, kad netiesioginis IPP (NIPP)
\[ J(x) = \int_a^b f(x,y)\,\dy := \lim_{c\uparrow b} \int_a^c f(x,y)\,\dy,
\qquad x \in I,
\]
konverguoja \Def{tolygiai} pagal $x \in I$, jei
\[ \sup_{x \in I} \Big| \int_c^b f(x,y) \,\dy \Big| \to 0, \quad c \uparrow b.
\]
\end{ap}

\begin{tg}								% 5.8
\label{5.8}
Jei \(f \in C(I\times[a,b))\), $I \subset \R$ -- bet koks intervalas, o NIPP
$J(x) = \int_a^b f(x,y) \,\dy$ konverguoja tolygiai ta�ko $x^0 \in I$
aplinkoje, tai $J$ tolydus ta�ke $x^0$.
\end{tg}

\begin{tg}[NIPP tolygaus konvergavimo Abelio-Dirichle po�ymis]		% 5.9
Tarkime, kad \(f \in C(I \times [a,b))\), ir \(g \in C^1[a,b)\) tenkina �ias
s�lygas:
\begin{enumerate}
\item $F(x, t) := \int_a^t f(x,y)\,\dy$, \(t \in [a,b)\) -- tolygiai apr��ta
      funkcija ($|F(x,t)| \le M < \infty$, $x \in I$, \(t \in [a,b)\)).
\item $g(x) \downarrow 0$, kai $x \uparrow b$
\end{enumerate}
Tada NIPP $J(x) := \int_a^b f(x,y) \cdot g(y)\,\dy$, $x \in I$, konverguoja
tolygiai ir $|J(x)| \le Mg(a)$, $x \in I$ ($J$ tolydus remiantis \ref{5.8}
teiginiu).
\end{tg}

\begin{pvz}[!]								% 5.10
\[ J(x) = \int_0^{+\infty} e^{-xy} \frac{\sin y}{y}\,\dy, \qquad x \ge 0.
\]

Pa�ym�kime $f(x,y) := e^{-xy}\sin y$, $g(y) := \frac{1}{y}$. Jei $J_1(x) :=
\int_1^{+\infty} f(x,y)g(y)\,\dy$ konverguos tolygiai, tai tolygiai konverguos
ir $J(x) = \int_0^1 f(x,y)g(y) \,\dy + J_1(x)$ (nes $|f(x,y)g(y)| \le M$, kai
$x \in [0,1]$).

\begin{align*}
|F(x,t)| &= \Big|\int_1^t f(x,y)\,\dy\Big|
          = \Big|\int_1^t e^{-xy}\sin y \,\dy\Big|
          = \left|\frac{e^{-xy}(x \sin y - \cos y)}{1+x^2} \Big|_{y=1}^t \right|
       \\&= \left|\frac{e^{-xt}(x \sin t - \cos t)}{1+x^2} -
                  \frac{e^{-x} (x \sin 1 - \cos 1)}{1+x^2} \right|
          \le \frac{x|\sin t| + |\cos t|}{1+x^2} + \frac{x \sin 1 + \cos
              1}{1+x^2}
       \\&\le \frac{2(1+x)}{1+x^2} \le 4, \qquad x \ge 0, \quad t \ge 1.
\end{align*}

$g(y) \downarrow 0$, kai $y \to \infty$.  Pritaik� Abelio-Dirichle po�ym�
gauname, kad $J_1(x)$ -- tolydi funkcija, tad tolydus ir $J(x)$, $x \ge 0$.

Atskiru atveju, kai $x = 0$, gauname, kad
\[ J(0) = \int_0^{+\infty} \frac{\sin y}{y}\,\dy
        = \lim_{x \downarrow 0} J(x) 
        = \lim_{x \downarrow 0} \int_0^{+\infty} e^{-xy} \frac{\sin
          y}{y}\,\dy.
\]

Diferencijuokime $J(x)$, $x > 0$:
\[ J'(x) = \int_0^{+\infty} -e^{-xy}\sin y \,\dy.
\]

Diferencijavimas yra teis�tas, nes $|e^{-xy}\,\sin y| \le x^{-xy} \le
e^{-\frac{x_0}{2}y}$, $y \ge 0$, $x \in U_{\frac{x_0}{2}}(x_0) =
(\frac{x_0}{2}, \frac{3x_0}{2})$.

Tai rodo, kad pointegrin� funkcija yra normaliai integruojama kiekvieno ta�ko
$x_0 > 0$ aplinkoje, kadangi de�in�je esanti funkcija integruojama.

Apskai�iuokime $J'(x)$:
\[ J'(x) = -\frac{e^{-xy}(x \sin y - \cos y)}{1+x^2} \bigg|_{y=0}^{+\infty}
         = - \frac{1}{1+x^2}, \qquad x > 0.
\]

I� �ia
\[ J(x) = - \int \frac{1}{1+x^2} \,\dx = -\arctan x + C.
\]

Pereikime prie ribos, kai $x \uparrow +\infty$:
\[ 0 = -\frac{\pi}{2} + C.
\]

Taigi, $C = \frac{\pi}{2}$ ir $J(x) = \frac{\pi}{2} - \arctan x$.
Per�j� prie ribos, kai $x \downarrow 0$, gauname
\[ \int_0^{+\infty} \frac{\sin y}{y} \,\dy = \frac{\pi}{2}.
\]

Pa�m� $y = ax$ turime
\[ \int_0^{+\infty} \frac{\sin ax}{x} \,\dx = \frac{\pi}{2} \sgn a, \qquad a
\in \R.
\]
\end{pvz}


\section{Oilerio integralai}

\begin{ap}								% 6.1
\Def{Gama funkcija} arba \Def{antro tipo Oilerio integralu} vadinama funkcija
\[ \Gamma(x) := \int_0^{+\infty} e^{-t} t^{x-1} \,\dt, \qquad x > 0.
\]
\end{ap}

\begin{tg}								% 6.2
Funkcija $\Gamma$ pasi�ymi �iomis savyb�mis:
\begin{enumerate}
\item $\Gamma \in C^\infty(0, +\infty)$.
\item $\forall x > 0 \quad \Gamma(x+1) = x \Gamma(x)$.
\item $\forall n \in \N \quad \Gamma(n+1) = n!$ ir
      $\Gamma(n+\frac{1}{2}) = \frac{(2n-1)!!}{2^n} \sqrt{\pi}$.
\item $\displaystyle \lim_{x \downarrow 0} \Gamma(x) = \lim_{x \to +\infty}
      \Gamma(x) = +\infty$.
\item $\Gamma$ i�kila �emyn ir turi minimum� intervalo $(1;2)$ ta�ke $x_{min}
      \approx 1.4616$ ($\Gamma(x_{min}) \approx 0.8856$).
\end{enumerate}
\end{tg}

\begin{ap}								% 6.3
\Def{Pirmo tipo Oilerio IPP} arba \Def{Beta funkcija} apibr��iama lygybe
\[ \Beta(x, y) := \int_0^1 t^{x-1}(1-t)^{y-1} \,\dt, \qquad x > 0, \quad y > 0.
\]
\end{ap}

\shortpage
\begin{tgg}\NL								% 6.4
\begin{enumerate}
\item $\displaystyle \Beta(x, y) = \int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}}
      \,\dt, \qquad x > 0, \quad y > 0$.
\item $\Beta(x, y) = \Beta(y, x)$, $x > 0$, $y > 0$.
\item $\displaystyle \Beta(x+1, y) = \frac{x}{x+y} \Beta(x, y)$; \\
      $\displaystyle \Beta(x, y+1) = \frac{y}{x+y} \Beta(x, y)$.
\end{enumerate}
\end{tgg}

\begin{tr}\NL								% 6.5
\begin{enumerate}
\item $\displaystyle \Beta(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$,
      $x > 0$, $y > 0$.
\item $\displaystyle \Gamma(x)\Gamma(1-x) = \frac{\pi}{\sin \pi x}$,
      $0 < x < 1$.
\end{enumerate}
\begin{ir}
Be �rodymo.
\end{ir}
\end{tr}

\begin{pvzz}\NL								% 6.6
\begin{enumerate}
\item $\displaystyle \int_0^1 \sqrt{x-x^2}\,\dx = \frac{\pi}{8}$.
\item $\displaystyle \int_0^{+\infty} \frac{\sqrt[4]{x}}{(1+x)^2}\,\dx =
      \frac{\pi\sqrt{2}}{4}$.
\item $\displaystyle \int_0^{\pi/2} \sin^{p-1} x \cos^{q-1} x\,\dx =
      \frac{1}{2} \Beta(p/2, q/2)$.
\item $\displaystyle \int_0^{+\infty} x^m e^{-x^n}\,\dx = \frac{1}{n}
      \Gamma\left(\frac{m+1}{n}\right)$.
\end{enumerate}
\end{pvzz}


\section{Kreiviniai integralai}

\begin{ap}[su prisiminimu]						% 7.1
\label{7.1}
\Def{Kreive} erdv�je $\R^k$ vadinama bet kokia funkcija $\gamma \in C([a,b],
\R^k)$.  Kreiv� $\gamma$ vadinama \Def{glod�ia}, jei ji yra tolyd�iai
diferencijuojama ir $\gamma'(t) \ne 0$ (kai kada pastaroji s�lyga � apibr��im�
ne�traukiama).  Kreiv� $\gamma$ vadinama \Def{gabalais} (\Def{dalimis})
\Def{glod�ia}, jei $\gamma'(t)$ egzistuoja ir yra apr��ta i�skyrus galb�t
baigtin�\footnote{Galima b�t� kalb�ti apie skai�i� aib�, bet taikymuose tai
retai pasitaiko.} skai�i� ta�k�.

Funkcija $\tau : [a,b] \to [\alpha, \beta]$ vadinama \Def{diffeomorfizmu}, jei
ji yra tolyd�iai diferencijuojama bijekcija, kurios atvirk�tin� $\tau^{-1}$
taip pat yra tolyd�iai diferencijuojama.

\shortpage
Sakoma, kad kreiv�s $\gamma_1$ ir $\gamma_2$ yra \Def{ekvivalen�ios}, jei
egzistuoja grie�tai did�jantis diffeomorfizmas $\tau$ toks, kad $\gamma_1 =
\gamma_2 \circ \tau$ (t.y. $\gamma_1(t) = \gamma_2(\tau(t))$ su visais $t$).
Tokiu atveju ra�oma $\gamma_1 \sim \gamma_2$ ir sakoma, kad kreiv� $\gamma_1$
yra \Def{gaunama} i� kreiv�s $\gamma_2$ parametro keitimu $\tau$.

Glod�ios kreiv�s $\gamma : [a,b] \to \R^k$ \Def{liestine} ta�ke $t_0 \in
[a,b]$ vadinama ties�
\[ x(t) = \gamma(t_0) + \gamma'(t_0) (t-t_0), \qquad t \in \R.
\]

Vektorius $\gamma'(t_0)$ yra \Def{lie�iamasis vektorius} kreivei $\gamma$
ta�ke $t_0$ (geometri�kai -- ta�ke $\gamma(t_0)$).

Vektorius $T(t_0) := \frac{\gamma'(t_0)}{|\gamma'(t_0)|}$ vadinamas kreiv�s
$\gamma$ \Def{vienetiniu lie�iamuoju vektoriumi}.
\end{ap}

\begin{tg}[kreivi� ekvivalentumo savyb�s]\NL				% 7.2
\begin{enumerate}
\item \label{7.2.1} $\gamma \sim \gamma$.
\item \label{7.2.2} $\gamma_1 \sim \gamma_2$ \tada $\gamma_2 \sim \gamma_1$.
\item \label{7.2.3} $\gamma_1 \sim \gamma_2$ ir $\gamma_2 \sim \gamma_3$ \tada
      $\gamma_1 \sim \gamma_3$.
\item Vienetinis lie�iamasis vektorius nesikei�ia kei�iant parametr�.
\end{enumerate}
\begin{ps*} \ref{7.2.1}--\ref{7.2.3} savyb�s leid�ia visas kreives erdv�je
$\R^k$ suskirstyti � nesikertan�ias klases.  Kiekvien� toki� klas� galima
interpretuoti kaip vien� geometrin� kreiv� su pasirinkta ,,ap�jimo kryptimi``.
\end{ps*}
\end{tg}

\begin{ap}								% 7.3
\Def{Orientuota} kreive (erdv�je $\R^k$) vadinama bet kokia klas� ekvivalen�i�
kreivi� erdv�je $\R^k$.

Kreiv� $\overline{\gamma}(t) := \gamma(a+b-t)$, $t \in [a,b]$, vadinama
kreive, gauta i� kreiv�s $\gamma : [a, b] \to \R^k$ \Def{orientacijos}
(,,ap�jimo krypties``) pakeitimu.
\end{ap}

\begin{pss}\NL								% 7.4
\begin{enumerate}
\item Kreive erdv�je $\R^k$ \Def{geometrine prasme} vadainama aib� $\Gamma \in
      \R^k$, kuriai egzistuoja kreiv� (\ref{7.1} apibr��imo prasme) $\gamma :
      [a, b] \to \R^k$, su kuria vaizdas $\gamma([a,b]) = \Gamma$.  Tokiu
      atveju funkcija $\gamma$ vadinama kreiv�s $\Gamma$ parametrizacija.

      $\Gamma$ vadinama \Def{glod�ia} (\Def{gabalais glod�ia}), jei tokia yra
      $\gamma$.

\item Orientuota kreiv� kartais vadinama tiesiog ,,kreive``, o kreiv� pagal
      m�s� supratim� vadinama ,,keliu``.

\item Jei $\gamma$ -- glodi kreiv� ($\gamma \in C^1([a,b], \R^k)$), kreiv�s
      $\gamma$ ilgis yra
      \[ l(\gamma) = \int_a^b |\gamma'(t)|\,\dt,
      \]
      be to, $l(\gamma_1) = l(\gamma_2)$, jei $\gamma_1 \sim \gamma_2$ arba
      $\gamma_1 \sim \overline{\gamma_2}$.

      Geometrin�s kreiv�s ilgis $l(\Gamma) := l(\gamma)$, kur $\gamma$ -- bet
      kokia kreiv�s $\Gamma$ parametrizacija.

\item Vietoje kreivi� $\gamma : [a, b] \to \R^k$ galime nagrin�ti ir kreives
      \(\gamma : [0, +\infty) \to \R\) arba $\gamma : (-\infty, +\infty) \to
      \R^k$ (su tam tikru atsargumu).
\end{enumerate}
\end{pss}

\begin{ap}								% 7.5
Tarkime, $\gamma : [\alpha, \beta] \to D \subset \R^k$ -- glodi kreiv�, $f: D
\to \R$.  Funkcijos $f$ \Def{pirmo tipo kreiviniu integralu} kreive $\gamma$
vadinamas skai�ius
\[ \int_\gamma f \,\dl := \int_\alpha^\beta f\big(\gamma(t)\big)\,\dl(t)
   = \int_\alpha^\beta f\big(\gamma(t)\big)|\gamma'(t)|\,\dl
\]
(su s�lyga, kad de�in�je esantis integralas egzistuoja).
\end{ap}

\begin{tgg}[pirmo tipo kreivini� integral� savyb�s]\NL			% 7.6
\begin{enumerate}
\item Jei $\gamma_1 \sim \gamma_2$ arba $\gamma_1 \sim \overline{\gamma_2}$,
tai \[ \int_{\gamma_1} f\,\dl = \int_{\gamma_2} f\,\dl \]
(�i savyb� leid�ia apibr��ti pirmo tipo kreivin� integral� geometrinei kreivei
$\Gamma$:
\[ \int_\Gamma f\,\dl := \int_{\gamma} f\,\dl,
\text{ kur $\gamma$ -- bet kokia $\Gamma$ parametrizacija).}
\]
\item (Tiesi�kumas)
\[ \int_\gamma (a f + b g)\,\dl = a \int_\gamma f\,\dl + b \int_\gamma g\,\dl
\]
(jei egzistuoja de�in�je esantys integralai).
\item (Adityvumas)
Jei $\gamma = \gamma_1 \cup \gamma_2$, tai
\[ \int_\gamma f\,\dl = \int_{\gamma_1} f\,\dl + \int_{\gamma_2} f\,\dl.
\]
\item
\[ \Big|\int_\gamma f\,\dl \Big| \le \int_\gamma |f|\,\dl.
\]
\end{enumerate}
\begin{ps*}
$\int_\gamma f\,\dl$ galima interpretuoti kaip kreiv�s $\Gamma$ mas�, kai
jos tankis ta�ke $x \in \Gamma$ yra lygus $f(x)$.
\end{ps*}
\end{tgg}

\begin{pvz}								% 7.7
$\Gamma = \{(x,y): \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\}$, $f = f(x,y)$,
$(x,y) \in \Gamma$.

Parametrizacija:
\[ \begin{cases} x = a\cos t \\ y = b\sin t \end{cases} \qquad 
   t \in [0, 2\pi].
\]

Tada
\[ \int_\Gamma f\,\dl = \int_0^{2\pi} f(a\cos t, b\sin t) \underbrace{\sqrt{
   a^2 \sin^2 t + b^2 \cos^2 t}}_{= |\gamma'(t)| = c\sqrt{1 - k \sin^2t} }.
\]

\( E(k) := \int \sqrt{1 - k \sin^2 t} \) -- \Def{elipsinis integralas}.
Bjaurus daiktas.
\end{pvz}

\begin{ap}								% 7.8
Tarkime, $\gamma: [\alpha, \beta] \to D \subset \R^k$ -- glodi kreiv�, $f : D
\to \R^k$.  Funkcijos $f$ \Def{antro tipo kreiviniu integralu} kreive $\gamma$
vadinamas skai�ius 
\begin{align*}
   \int_\gamma \langle f, \dx \rangle = &
   \int_\gamma f_1 \dx_1 + f_2 \dx_1 + \dots + f_k \dx_k := \\ &
   \int_\alpha^\beta \Big(f_1\big(\gamma(t)\big)\gamma'_1(t) +
                          f_2\big(\gamma(t)\big)\gamma'_2(t) + \dots +
                          f_k\big(\gamma(t)\big)\gamma'_k(t) \Big)\,\dt =
   \int_\alpha^\beta \langle f\big(\gamma(t)\big), \gamma'(t) \rangle \,\dt
\end{align*}
(su s�lyga, kad de�in�je esantis integralas egzistuoja).
\end{ap}

\begin{pss}[ir komentarai]\NL						% 7.9
\begin{enumerate}
\item Kai $k=2$ �prasta �ym�ti $f = (P, Q)$, $x = (x, y)$:
\[ \int_\gamma P(x,y)\,\dx + Q(x,y)\,\dy. \]

Kai $k = 3$, �ymima $f = (P, Q, R)$, $x = (x, y, z)$:
\[ \int_\gamma P(x,y,z)\,\dx + Q(x,y,z)\,\dy + R(x,y,z)\,\dz. \]

\item $\int_\gamma \langle f, \dx \rangle$ galima interpretuoti kaip darb�,
      kur� atlieka ,,j�g� laukas`` f, ,,stumiantis`` dalel� kreive $\gamma$.

\item Pirmo ir antro tipo kreivinius integralus sieja �i lygyb�:
\[ \int_\gamma \langle f, \dx \rangle = \int_\gamma \langle f, T \rangle\,\dl.
\]
\begin{ir*}
I� tikr�j�,
\begin{align*}
\int_\gamma \langle f, T \rangle \,\dx = \int_\alpha^\beta \Big\langle f\big(
\gamma(t) \big), \frac{\gamma'(t)}{|\gamma'(t)|} \Big\rangle |\gamma'(t)|\,\dt
= \int_\alpha^\beta \big\langle f\big( \gamma(t) \big), \gamma'(t)
\big\rangle\,\dt = \int_\gamma \langle f, \dx \rangle
\end{align*}
\end{ir*}
\end{enumerate}
\end{pss}

\begin{tgg}\NL								% 7.10
\begin{enumerate}
\item
\begin{align*}
\gamma_1 \sim \gamma_2 &\tada \int_{\gamma_1} \langle f, \dx \rangle =
                              \int_{\gamma_2} \langle f, \dx \rangle; \\
\gamma_1 \sim \overline{\gamma_2} &\tada \int_{\gamma_1} \langle f, \dx \rangle
                                      = -\int_{\gamma_2} \langle f, \dx \rangle.
\end{align*}

�ios savyb�s leid�ia aibr��ti antro tipo kreivin� integral� geometrine kreive
$\Gamma \subset \R^k$ su pasirinkta ,,ap�jimo`` kryptimi:
\[ \int_\Gamma \langle f, \dx \rangle := \int_\gamma \langle f, \dx \rangle,
\]
kur $\gamma$ -- $\Gamma$ parametrizacija su ,,atitinkamai`` pasirinkta
kryptimi.  Pakeitus ,,ap�jimo krypt�`` prie�inga, integralo reik�m� keis
�enkl�.

\item (Tiesi�kumas)
\[ \int_\gamma \langle af+bg, \dx \rangle = a \int_\gamma \langle f, \dx
   \rangle + b \int_\gamma \langle g, \dx \rangle.
\]

\item (Adityvumas)
Jei $\gamma = \gamma_1 \cup \gamma_2$, tai
\[ \int_\gamma \langle f, \dx \rangle = \int_{\gamma_1} \langle f, \dx
   \rangle + \int_{\gamma_2} \langle f, \dx \rangle.
\]

\item
\[ \Big| \int_\gamma \langle f, \dx \rangle \Big| \le \int_\gamma |f|\,\dl.
\]

\end{enumerate}
\end{tgg}

\begin{pvz}								% 7.11
$\Gamma = \{(x,y): \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\}$, $f(x,y) =
\big(P(x,y), Q(x,y)\big)$, $(x,y) \in D \supset \Gamma$.

Parametrizacija:
\[ \begin{cases} x = \gamma_1(t) = a\cos t \\ y = \gamma_2(t) = b\sin t 
   \end{cases} \qquad t \in [0, 2\pi].
\]

\[ \int_\Gamma P(x,y)\,\dx + Q(x,y)\,\dy =
   \int_0^{2\pi} \big(
     P(a\cos t, b\cos t) \cdot \underbrace{(-a\sin t)}_{= \gamma_1'(t)} +
     Q(a\cos t, b\cos t) \cdot \underbrace{( b\cos t)}_{= \gamma_2'(t)} +
                  \big)\,\dt.
\]
\end{pvz}

\begin{ap}								% 7.12
Aib� $D \subset \R^k$ vadinama \Def{lanki�kai jungia}, jei $\forall x, y \in
D$ egzistuoja gabalais glodi kreiv� $\gamma : [\alpha, \beta] \to D$ su galais
$\gamma(\alpha) = x$, $\gamma(\beta) = y$.

\Def{Sritimi} erdv�je $\R^k$ vadinsime bet koki� atvir�j� lanki�kai jungi�
aib� $D \subset \R^k$.

Toliau visur $D \subset \R^k$ -- sritis.

Funkcijos $f : D \to \R$ \Def{gradientu} vadinama funkcija $\grad f : D \to
\R^k$:
\[ \grad f(x) := \left( \frac{\partial f}{\partial x_1}, \dots,
                        \frac{\partial f}{\partial x_k} \right),
              \qquad x \in D
\]
(kitas �ym�jimas: $\grad f(x) = f'(x)$, jei vartojamas ne eilutinis, o
stulpelinis vektorius).

Funkcija $f: D \to \R^k$ vadinama \Def{potencialo lauku}, jei egzistuoja
funkcija $U: D \to \R$, su kuria $\grad U = f$ srityje $D$.  Tokiu atveju $U$
vadinama ,,j�g� lauko`` $f$ \Def{potencialu}.
\end{ap}

\begin{tr}[kreiviniai integralai ir integravimo kreiv�]			% 7.13
Tarkime, kad $f \in C(D, \R^k)$.  Tada �ie trys teiginiai yra ekvivalent�s:
\begin{enumerate}
\item $f$ -- potencialo laukas.
\item $\forall x^0, x^1 \in D$ kreivinis integralas $\int_\gamma \langle f,
      \dx \rangle$ �gyja t� pa�i� reik�m� su visomis gabalais glod�iomis
      kreiv�mis $\gamma : [\alpha, \beta] \to D$ su galais $\gamma(\alpha)
      = x^0$, $\gamma(\beta) = x^1$.
\item $\int_\gamma \langle f, \dx \rangle = 0$ su visomis u�daromis gabalais
      glod�iomis kreiv�mis $\gamma : [\alpha, \beta] \to D$ ($\gamma(\alpha) =
      \gamma(\beta)$).
\end{enumerate}
\end{tr}

\begin{tr}[b�tina ir pakankama s�lyga, kad funkcija b�t� potencialo laukas]
\label{7.14}								% 7.14
Tarkime, $f \in C^1(D, \R^k)$.
\begin{enumerate}
\item Jei $f$ -- potencialo laukas, tai
\begin{equation*}
   \frac{\partial f_i}{\partial x_j} =
   \frac{\partial f_j}{\partial x_i},
   \qquad i, j = 1 \dots k, \text{ srityje $D$.}
   \tag*{($*$)}\label{eq:7.14.1}
\end{equation*}
\item Jei $D$ -- atviroji gardel�, ir joje patenkintos \ref{eq:7.14.1}
lygyb�s, tai $f$ -- potencialo laukas ir funkcija $U(x)$ yra jos potencialas:
\begin{align*}
  U(x) = U(x^0) &+ \int_{x^0_1}^{x_1} f_1(y, x^0_2, x^0_3, \dots, x^0_k)\,\dy \\
                &+ \int_{x^0_2}^{x_2} f_2(x_1, y, x^0_3, \dots, x^0_k)\,\dy \\
                &+ \ldots \\
                &+ \int_{x^0_k}^{x_k} f_k(x_1, \dots, x_{k-1}, y)\,\dy,
\qquad x \in D
\end{align*}
(ta�kas $x^0 \in D$ ir $U(x^0)$ reik�m� pasirenkami laisvai).
\end{enumerate}
\end{tr}

\begin{pss}\NL								% 7.15
\begin{enumerate}
\item Kai $k = 2$, $f = (P, Q)$, tada \ref{eq:7.14.1} atrodo taip:
\[ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}.
\]

Kai $k = 3$, $f = (P, Q, R)$, tada \ref{eq:7.14.1} atrodo taip:
\[ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}, \qquad
   \frac{\partial P}{\partial z} = \frac{\partial R}{\partial x}, \qquad
   \frac{\partial Q}{\partial z} = \frac{\partial R}{\partial y}.
\]

\item \ref{7.14} treorema yra teisinga ir bendresn�ms sritims nei gardel�s.
Galima �rodyti, kad ji galioja vadinamosioms \Def{vienajung�ms sritims}
(sritims ,,be skyli�``; ,,beveik grie�tas`` apibr��imas b�t� toks: vienajung�
sritis -- tokia, kurios kra�tas yra lanki�kai jungi aib�).

\item Jei $f$ -- potencialo laukas, tai turi prasm� integralas
\[ \int_A^B \langle f, \dx \rangle = U(B) - U(A),
   \qquad A, B \in D, \quad \grad U = f
\]
(Niutono-Leibnico formul� daugiama�iu atveju).

\item Jei $\grad U = f$, tai $\ud U = f_1(x)\,\dx_1 + \dots + f_k(x)\,\dx_k$
-- \Def{pilnasis diferencialas}.  \ref{7.14} teorema nusako s�lyg�, kada
$f_1(x)\,\dx_1 + \dots + f_k(x)\,\dx_k$ yra pilnasis diferencialas.

\end{enumerate}
\end{pss}

\shortpage
\begin{pvzz}\NL								% 7.16
\begin{enumerate}
\item $\displaystyle \oint_{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}
(x^2+y^2)(x\,\dx+y\,\dy) = 0$ ($\oint$ rei�kia, kad integruojama u�dara kreive
prie� laikrod�io rodykl�).  Galima tai su�inoti neskai�iuojant integralo,
u�tenka patikrinti \ref{7.14} s�lygas:
\[ \frac{\partial P}{\partial y} = 2xy = \frac{\partial Q}{\partial x}
\]
($P(x,y) = x(x^2+y^2)$, $Q(x,y) = y(x^2+y^2)$).

\item $\displaystyle I = \oint_{x^2+y^2=1} \frac{x\,\dy - y\,\dx}{x^2+y^2}$,
$(x,y) \ne (0,0)$.

Parametrizacija:
\[ \begin{cases} x = \cos \phi \\ y = \sin \phi \end{cases} \qquad \phi \in
   [0, 2\pi).
\]

\[ I = \int_0^{2\pi} \frac{\cos^2 \phi + \sin^2 \phi}{\cos^2 \phi + \sin^2
   \phi}\,\ud\phi = 2\pi \ne 0.
\]

U�tenka vieno ,,blogo`` ta�ko $(0, 0)$, kad funkcija neb�t� potencialo lauku,
nors $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$.

\item 
$\displaystyle J = \int_{(1,1)}^{(2,2)} (4x^3y^3-3y^2+5)\,\dx +
                                        (3x^4y^2-6xy-4)\,\dy$.

Reikia patikrinti, ar $P\,\dx + Q\,\dy$ yra pilnas diferencialas.

$\frac{\partial P}{\partial y} = 12x^3y^2 - 6y = \frac{\partial Q}{\partial
x}$.  Yra.  Vadinasi $P\,\dx + Q\,\dy = \ud U$.

\begin{align*}
U(x,y) &= \underbrace{U(0,0)}_{= 0 \text{ (laisvas pasirinkimas)}} +
          \int_0^x P(u, 0)\,\du +
          \int_0^y Q(x, u)\,\du
        = \int_0^x 5\,\du +
          \int_0^y (3x^4u^2 - 6xu - 4)\,\du
     \\&= 5x + x^4y^3 - 3xy^2 - 4y.
\end{align*}

Patikrinimas: $\frac{\partial U}{\partial x} = P$, $\frac{\partial U}{\partial
y} = Q$.

Vadinasi $J = U(2,2) - U(1,1)$.

Antras b�das: $\frac{\partial U}{\partial x} = P$, tada $U(x,y) = \int
P(x,y)\,\dx = \int (4x^3y^3-3y^2+5)\,\dx = x^4y^3-3y^2x+5x+C(y)$,  kur $C(y)$
-- konstanta, priklausanti nuo $y$.

$\frac{\partial U}{\partial y} = Q$ \tada $3x^4y^2-6yx+C'(y)= 3x^4y^2-6xy-4$
\tada $C'(y) = -4$ \tada $C(y) = \int (-4)\,\dy = -4y + C$.

Taigi, $U(x,y) = x^4y^3-3y^2x+5x - 4y+C$, o $J = U(2,2)-U(1,1)$.
\end{enumerate}
\end{pvzz}

\begin{ap}								% 7.17
\Def{Horizontali�ja kreivine trapecija} vadinsime aib� $D \subset \R^2$,
kurios pavidalas toks:
\[ D = \{(x,y): a \le x \le b, y_1(x) \le y \le y_2(x)\}, \qquad
   y_1\le y_2 \text{-- tolyd�iai diferencijuojamos funkcijos.}
\]

\Def{Vertikali�ja kreivine trapecija} vadinsime aib� $D \subset \R^2$, kurios
pavidalas toks:
\[ D = \{(x,y): a \le y \le b, x_1(x) \le x \le x_2(x)\}, \qquad
   x_1\le x_2 \text{-- tolyd�iai diferencijuojamos funkcijos.}
\]

Aib� $D \subset \R^2$ susitarsime vadinti \Def{,,gera``}, jei
\begin{enumerate}
\item j� galima padalinti baigtiniu skai�iumi horizontali� ir vertikali�
tiesi� tiek � horizontalias, tiek � vertikalias kreivines trapecijas
\item j� riboja viena gabalais glodi kreiv� $\Gamma = \partial D$ ($\partial
D$ -- aib�s $D$ kra�tas).  Laikysime, kad ji apeinama teigiama kryptimi --
prie� laikrod�io rodykl�.
\end{enumerate}
\end{ap}

\begin{tr}[Gryno formul�]						% 7.18
Jei $P, Q: D \to \R$ -- tolyd�iai diferencijuojamos funkcijos ,,geroje``
aib�je $D$, tai
\[
   \int_{\partial D} P\,\dx + Q\,\dy = \iint_D \left(\frac{\partial
   Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,\dx\dy.
\]
\end{tr}

\begin{pvzz}\NL								% 7.19
\begin{enumerate}
\item Parink� funkcijas $P$ ir $Q$ taip, kad
\[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \equiv 1,
\]
Gryno formul�s de�in�je pus�je gauname srities plot� $S(D)$.

Toki� funkcij� pavyzd�iai:
\begin{enumerate}
\item $P = -y$, $Q = 0$;
\item $P = 0$, $Q = x$;
\item $P = -\frac{y}{2}$, $Q = \frac{x}{2}$.
\end{enumerate}

I� �ia
\[ S(D) = -\int_{\partial D} y\,\dx =
           \int_{\partial D} x\,\dy =
          \frac{1}{2} \int_{\partial D} x\,\dy - y\,\dx.
\]

\item Elips�s plotas: $D = \{ (x,y) : \frac{x^2}{a^2} + \frac{x^2}{b^2} \le 1
\}$.

$\partial D$ parametrizacija:
\[ \begin{cases} x = a \cos t\\ y = b \sin t \end{cases}
   \qquad t \in [0, 2\pi).
\]

\begin{align*}
S(D) &= - \int_0^{2\pi} b\sin t a(-\sin t)\,\dt = ab\int_0^{2\pi} \sin^2
t\,\dt = ab\int_0^{2\pi} \frac{1-\cos 2t}{2}\,\dt = ab\pi.
\end{align*}

\end{enumerate}
\end{pvzz}

\end{document}

