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\title[Matematika 1]
{Matematika 1}

\subtitle{7. vaja}

\author[Borut] % 
{B.~Jur\v{c}i\v{c}~Zlobec\inst{1}}

\institute[Univerza v Ljubljani Slovenia] % 
{
  \inst{1}%
    Univerza v Ljubljani,\\Fakulteta za Elektrotehniko\\ 
    1000 Ljubljana, Tr\v{z}a\v{s}ka 25, Slovenija
}


\date[BF 2010] % 
{Matematika FE, Ljubljana, Slovenija \today}

\subject{Matematika}


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\section{Integral funkcije}
\subsection{Nedolo\v{c}eni integral}

\begin{frame}{Pravila za integriranje}
\begin{enumerate}[I]
        \item Linearnost: 
	$\displaystyle \int \left(a\, f(x)+b\, g(x)\right)\,dx=
		a \int f(x)\,dx+b \int g(x)\,dx$.\label{i-lin}
        \item Per partes: $\displaystyle
\int f'(x)g(x)\,dx=f(x)g(x)-\int f(x)g'(x)\,dx$.\label{i-prod}
        \item Substitucija: 
	$\displaystyle \int f(g(x))g'(x)\,dx=F(g(x))$, kjer je 
	$\int f(x)\,dx=F(x)$\label{i-posr}.\\
	$u=g(x)\rightarrow$, $du=g'(x)\,dx\rightarrow$, 
	$\int f(u)\,du=F(u)=F(g(x))$.
	\item Linearna substitucija: 
	$\int f(ax+b)\,dx=\frac{1}{a}F(ax+b)$. \label{i-lin-sub}
\end{enumerate}
\end{frame}

\begin{frame}{Integrali elementarnih funkcij}
\begin{enumerate}
        \item $\int x^n\,dx=
	\frac{x^{n+1}}{n+1}$\label{i-potenca-n},
	$\quad n\in\mathbb{Z}\setminus\{-1\},\quad x\ne 0$.
        \item $\int x^s\,dx=\frac{x^{s+1}}{s+1}$\label{i-potenca-s},
	$\quad s\in\mathbb{R}\setminus\{-1\},\quad x > 0$.
        \item $\int e^x\,dx=e^x$\label{i-exp}.
        \item $\int\frac{dx}{x}=\ln |x|$\label{i-loge}.
        \item $\int\sin x\,dx=-\cos x$\label{i-sin}.
        \item $\int \cos x\,dx=\sin x$\label{i-cos}.
        \item
$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\arctan\frac{x}{a},\quad a>0$
		\label{i-atan}.
        \item $\int\frac{dx}{\sqrt{a^2-x^2}}=
	\arcsin\frac{x}{a},\quad a>0$\label{i-asin}.
	\item $\int\frac{dx}{\sqrt{x^2\pm a^2}}=
	\ln(x+\sqrt{x^2\pm a^2}),\quad a>0$. 
\end{enumerate}
\end{frame}

\begin{frame}{Re\v{s}evanje integralov s pomo\v{c}jo nastavka}
\begin{enumerate}[A]
	\item Integral oblike  
	\formulas{$\displaystyle\int f(x)\,dx=\int\frac{P(x)dx}{(x-a)^n(x^2+px+q)^m}$},\\
	kjer je 
	$m,n\in\mathbb{N}$, $p^2-4q<0$ in $P(x)$ polinom 
	stopnje manj ali enako $2m+n-1$, re"simo s pomo"cjo nastavka:\\
	\formulas{$\displaystyle\int f(x)\,dx=
		\frac{Q_{2m-2+n-2}(x)}{(x-a)^{n-1}(x^2+px+q)^{m-1}}+$\\
		$+A\ln|x-a|+B\ln(x^2+px+q)+C\arctg\frac{2x+p}{\sqrt{4q-p^2}}$}.
	\label{r-rac}
	\item Integral oblike \formulas{$\displaystyle\int
	f(x)\,dx=\int\frac{P_n(x)\,dx}{\sqrt{x^2+px+q}}$}\\
	re"simo s pomo"cjo nastavka:\\
	\formulas{$\displaystyle\int f(x)\,dx=Q_{n-1}(x)\sqrt{x^2+px+q}+
	\lambda\int\frac{dx}{\sqrt{x^2+px+q}}$}.\label{i-rac}
\end{enumerate} 
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=3+\frac{1}{x}+\frac{1}{x^2}$}. 
\begin{itemize}
	\item Uporabimo (\ref{i-lin}) $\rightarrow$ (\ref{i-potenca-n}) 
	$\rightarrow$ (\ref{i-loge}).
	\item $\int\left(3+\frac{1}{x}+\frac{1}{x^2}\right)\,dx=
	\rightarrow$
	\item $\int3\,dx+\int\frac{dx}{x}+\int\frac{dx}{x^2}\,dx=
	\rightarrow$
	\item $\displaystyle\int\left(3+\frac{1}{x}+\frac{1}{x^2}\right)dx=
	3x+\ln |x| - \frac{1}{x} + C$. 
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\frac{3+2x^2}{1+x^2}$}.
\begin{itemize}
        \item Uporabimo (\ref{i-lin}) $\rightarrow$ (\ref{i-potenca-n})
        $\rightarrow$ (\ref{i-atan}).
        \item $\int\frac{3+2x^2}{1+x^2}\,dx=\int\frac{1+2(1+x^2)}{1+x^2}\,dx
        \rightarrow$
        \item $\int 2\,dx + \int\frac{dx}{1+x^2}= \rightarrow$
        \item $\displaystyle\int\frac{3+2x^2}{1+x^2}dx=2x+\arctg x + C$.
\end{itemize}
\end{frame}


\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\frac{x^3}{1+x^2}$}.
\begin{itemize}
	\item Preverimo stopnji "stevca in imenovalca. 
	\item Delimo:  $\frac{x^3}{1+x^2}=x-\frac{x}{1+x^2}$
        \item (\ref{i-lin}) $\rightarrow$ (\ref{i-potenca-n}) $\rightarrow$ 
	(\ref{i-posr}) $\rightarrow$ (\ref{i-loge}).
        \item $u=1+x^2, du=2x\,dx\rightarrow$
        \item $\frac{1}{2}x^2-\frac{1}{2}\int\frac{du}{u}=
	\frac{1}{2}x^2-\frac{1}{2}\ln |u|\rightarrow$
        \item $\frac{1}{2}x^2-\frac{1}{2}\ln(1+x^2)\rightarrow$
	\item $\displaystyle\int\frac{x^3}{1+x^2}=
	\frac{1}{2}x^2-\ln\sqrt{1+x^2}+C$
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\frac{3}{2+2x+x^2}+\frac{x}{1+x^2}$}.
\begin{itemize}
        \item Uporabimo 
	(\ref{i-lin}) $\rightarrow$ (\ref{i-lin-sub}) $\rightarrow$
	(\ref{i-posr})
        $\rightarrow$ (\ref{i-atan}) $\rightarrow$ (\ref{i-loge}).
        \item
	$\int\left(\frac{3dx}{2+2x+x^2}+\frac{x}{1+x^2}\right)\,dx=
	3\int\frac{dx}{(x+1)^2+1}+\int\frac{x}{1+x^2}\,dx\rightarrow$
        \item $u=x+1, du=dx,\quad v=1+x^2, dv=2x\,dx\rightarrow$ 
	\item $3\int \frac{du}{1+u^2} + \
	\frac{1}{2}\int\frac{dv}{v}=\rightarrow$
        \item $\displaystyle\int f(x)\,dx=
	3\arctg(x+1)+\frac{1}{2}\ln(1+x^2) + C$.
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\frac{x-1}{x(x+1)}$}.
\begin{itemize}
	\item Razcepimo na parcialne ulomke.\\
	$\frac{x-1}{x(x+1)}=
	\frac{A}{x}+\frac{B}{x+1}=-\frac{1}{x}+\frac{2}{x+1}$.
	\item $\int\frac{x-1}{x(x+1)}\,dx=
	-\int\frac{dx}{x}+\int\frac{2dx}{x+1}$.
	\item $\int\frac{x-1}{x(x+1)}\,dx=-\ln|x|+2\ln|x+1|\rightarrow$
	\item $\displaystyle\int\frac{x-1}{x(x+1)}\,dx=
	\ln\frac{(x+1)^2}{|x|}+C$.
\end{itemize}
\end{frame}


\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\frac{x^2+1}{x(1-x^2)}$}.
\begin{itemize}
	\item Razcepimo na parcialne ulomke.\\
	$\frac{x^2+1}{x(1-x^2)}=\frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x}\rightarrow$
	\item $x^2+1=A(1-x)(1+x)+Bx(1+x)+Cx(1-x)\rightarrow$
	\item $-A+B-C=1$, $B+C=0$ in $A=1$. 
	\item $\displaystyle\int\frac{x^2+1}{x(1-x^2)}\,dx=\ln|x|-\ln|1-x^2|$.
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\frac{x-1}{x(x^2+1)}$}.
\begin{itemize}
        \item Razcepimo na parcialne ulomke.\\
        $\frac{x-1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}=
	-\frac{1}{x}+\frac{1+x}{x^2+1}$.
        \item $\int\frac{x-1}{(x^2+1)}\,dx=
        -\int\frac{dx}{x}+\int\frac{dx}{1+x^2}+\int\frac{x\,dx}{1+x^2}$.
        \item $\int\frac{x-1}{x(x+1)}\,dx=
	-\ln|x|+\arctg x+\frac{1}{2}\ln(x^2+1)\rightarrow$
        \item $\displaystyle\int\frac{x-1}{x(x+1)}\,dx=-
        \ln|x|+\arctg x+\ln\sqrt{1+x^2}+C$.
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\frac{x-1}{x^2(x+1)}$}.
\begin{itemize}
        \item Razcepimo na parcialne ulomke.\\
	$\frac{x-1}{x^2(x+1)}=\frac{A}{x^2}+\frac{B}{x}+\frac{C}{x+1}=-\frac{1}{x^2}+\frac{2}{x}-\frac{2}{x+1}$.
        \item $\int\frac{x-1}{x(x+1)}\,dx=
        -\int\frac{dx}{x^2}+\int\frac{2dx}{x}-\int\frac{2dx}{1+x}$.
        \item $\int\frac{x-1}{x(x+1)}\,dx=\frac{1}{x}+2\ln|x|-2\ln|x+1|\rightarrow$
        \item $\displaystyle\int\frac{x-1}{x(x+1)}\,dx=
	\frac{1}{x}+\ln\frac{x^2}{(1+x)^2}+C$.
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\frac{x^2+1}{x^2(1+x)}$}.
\begin{itemize}
        \item Zapi"semo nastavek (\ref{r-rac}).
        \item $\int\frac{x^2+1}{x^2(1+x)}\,dx=
                \frac{A}{x}+B\ln|x|+C\ln|1+x|\rightarrow$
        \item $x^2+1=-A(1+x)+Bx(1+x)+Cx^2\rightarrow$
        \item $B+C=1$, $A+B=0$ in $A=-1$.
	\item $\displaystyle\int\frac{x^2+1}{x^2(1+x)}\,dx=-\frac{1}{x}-\ln|x|+2\ln|1+x|$.
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\frac{x^3-1}{x^2(1+x)}$}.
\begin{itemize}
        \item Stopnji "stevca in imenovalca sta enaki, delimo.\\
	$\frac{x^3-1}{x^2(1+x)}=1-\frac{x^2+1}{x^2(x+1)}$.
        \item Zapi"semo nastavek (\ref{r-rac}).\\
        $\int\frac{x^2+1}{x^2(1+x)}\,dx=
                \frac{A}{x}+B\ln|x|+C\ln|1+x|\rightarrow$
        \item $x^2+1=-A(1+x)+Bx(1+x)+Cx^2\rightarrow$
        \item $B+C=1$, $A+B=0$ in $A=-1$.
        \item $\displaystyle\int\left(1-\frac{x^2+1}{x^2(1+x)}\right)dx=
	x+\frac{1}{x}+\ln|x|-\ln(1+x)^2+C$.
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\frac{x^2-1}{x(2+2x+x^2)}$}.
\begin{itemize}
        \item Preverimo stopnji "stevca in imenovalca.
        \item Zapi"semo nastavek (\ref{r-rac}).\\
        $\int\frac{x^2-1}{x(2+2x+x^2)}\,dx=
                A\ln|x|+B\ln|x^2+x+1|+C\arctg(1+x)\rightarrow$
        \item $x^2-1=\frac{A}{x}+\frac{2Bx+B+C}{x^2+x+1}\rightarrow$
        \item $B+C=1$, $A+B=0$ in $A=-1$.
        \item $\displaystyle\int\frac{x^2-1}{x(x^2+2x+2)}\,dx=
		-\frac{1}{2}\ln|x|
		-\frac{1}{2}\arctg(x+1)+\frac{3}{4}\ln(2+2x+x^2)+C$.
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\frac{x}{\sqrt{x+1}}$}.
\begin{itemize}
        \item (\ref{i-posr}) $\rightarrow$ (\ref{i-potenca-n}).
        \item $\displaystyle \int \frac{x}{\sqrt{x+1}}\,dx
        \rightarrow u=x+1, du=dx\rightarrow$
        \item $\int\frac{u-1}{\sqrt{u}}\,du=
		\int\left(u^\frac{1}{2}-u^{-\frac{1}{2}}\right)du$.
        \item $\int\frac{x}{\sqrt{x+1}}\,dx=
	\frac{2}{3}\sqrt{(x+1)^3}-2\sqrt{x+1}\rightarrow$
	\item $\displaystyle\int\frac{x}{\sqrt{x+1}}\,dx=
	\frac{2}{3}(x-2)\sqrt{1+x}+C$
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=x\sqrt{1+x^2}$}.
\begin{itemize}
	\item (\ref{i-posr}) $\rightarrow$ (\ref{i-potenca-s}).
	\item $\displaystyle \int x\sqrt{1+x^2}
	\rightarrow u=1+x^2, du=2x\,dx\rightarrow$
	\item $\frac{1}{2}\int	u^\frac{1}{2}\,du=
	\frac{1}{3}u^\frac{3}{2}$.  
	\item $\displaystyle\int
	x\sqrt{1+x^2}=\frac{1}{3}\sqrt{(1+x^2)^3}+C$.
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\sqrt{1-x^2}$}.
\begin{itemize}
        \item (\ref{i-posr}) $\rightarrow$ (\ref{i-potenca-s}).
        \item $\displaystyle \int \sqrt{1-x^2}\,dx
        \rightarrow x=\sin u, dx=\cos u\,du\rightarrow$
        \item $\int \sqrt{1-\sin^2 u}\cos u\,du=
	\int\cos^2 u\,du=\rightarrow$
	\item $u+\frac{1}{2}\sin(2u)=u+\sin u\cos u=
	u + \sin u\sqrt{1-\sin^2u}$.
        \item $\displaystyle \int \sqrt{1-x^2}\,dx=\frac{1}{2}\left(
	\arcsin x+x\sqrt{1-x^2}\right) + C$.
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\sqrt{1-x^2}$}.
\begin{itemize}
	\item Re"sujemo s pomo"cjo nastavka (\ref{i-rac}).
        \item $\displaystyle\int \sqrt{1-x^2}\,dx=\rightarrow$
        \item $\displaystyle\int\frac{1-x^2}{\sqrt{1-x^2}}dx=
		(Ax+B)\sqrt{1-x^2}+C\int\frac{1}{\sqrt{1-x^2}}dx\rightarrow$
        \item $\frac{1-x^2}{\sqrt{1-x^2}}=A\sqrt{1-x^2}
		-\frac{(Ax+B)x}{\sqrt{1-x^2}}+\frac{C}{\sqrt{1-x^2}}\rightarrow$
	\item $1-x^2=-2Ax^2-Bx+C+A\rightarrow$, $A=C=\frac{1}{2}$, $B=0$.
	\item $\displaystyle \int \sqrt{1-x^2}\,dx=\frac{1}{2}\left(
        \arcsin x+x\sqrt{1-x^2}\right) + C$.
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e je}
\formulas{$\displaystyle f(x)=\sqrt{1+x^2}$}.
\begin{itemize}
	\item Re"sujemo s pomo"cjo nastavka (\ref{i-rac}).
        \item $\displaystyle\int\frac{1+x^2}{\sqrt{1+x^2}}dx=
                (Ax+B)\sqrt{1+x^2}+C\int\frac{1}{\sqrt{1+x^2}}dx\rightarrow$
	\item $\frac{1+x^2}{\sqrt{1+x^2}}=A\sqrt{1+x^2}
                +\frac{(Ax+B)x}{\sqrt{1+x^2}}+\frac{C}{\sqrt{1+x^2}}\rightarrow$
        \item $1+x^2=2Ax^2+Bx+C+A\rightarrow$, $A=C=\frac{1}{2}$, $B=0$.
        \item $\displaystyle \int \sqrt{1+x^2}\,dx=\frac{1}{2}\left(
        \ln(x+\sqrt{1+x^2})+x\sqrt{1+x^2}\right) + C$.
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e je}
\formulas{$\displaystyle f(x)=\frac{x+2}{\sqrt{2+2x+x^2}}$}.
\begin{itemize}
	\item Re"sujemo s pomo"cjo nastavka (\ref{i-rac}).
        \item $\displaystyle\int\frac{x+2}{\sqrt{2+2x+x^2}}dx=
                A\sqrt{2+2x+x^2}+B\int\frac{dx}{\sqrt{2+2x+x^2}}\rightarrow$
	\item $\frac{x+2}{\sqrt{1+x^2}}=
                +\frac{A(x+1)}{\sqrt{2+2x+x^2}}+
		\frac{B}{\sqrt{2+2x+x^2}}\rightarrow$
        \item $x+2=A(x+1)+B\rightarrow$, $A=1$, $B=1$.
        \item $\displaystyle \int \sqrt{x+2}\,dx=
        \sqrt{2+2x+x^2}+\ln(x+1+\sqrt{x^2+2x+2}) + C$.
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e je}
\formulas{$\displaystyle f(x)=\frac{x+2}{\sqrt{1+2x-x^2}}$}.
\begin{itemize}
	\item Re"sujemo s pomo"cjo nastavka (\ref{i-rac}).
	\item $\displaystyle\int\frac{x+2}{\sqrt{1+2x-x^2}}dx=
		A\sqrt{1+2x-x^2}+B\int\frac{dx}{\sqrt{1+2x-x^2}}\rightarrow$
	\item $\frac{x+2}{\sqrt{1+2x-x^2}}=
	\frac{A(2-2x)}{2\sqrt{1+2x-x^2}}+\frac{B}{\sqrt{1+2x-x^2}}\rightarrow$
	\item $x+2=A-Ax+B$, $A=-1$ in $B=3$. 
	\item $\int\frac{1}{\sqrt{1+2x-x^2}}dx=
		\int\frac{dx}{\sqrt{2-(x-1)^2}}=
	\rightarrow$(\ref{i-lin-sub})$
		\rightarrow$(\ref{i-asin})$=\arcsin(x-1)$.
	\item $\displaystyle\int\frac{x+2}{\sqrt{2x-x^2}}dx=
		-\sqrt{1+2x-x^2}-3\arcsin\frac{x-1}{\sqrt{2}}+C$.
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\frac{1}{x\ln x}$}.
\begin{itemize}
        \item (\ref{i-posr}) $\rightarrow$ (\ref{i-loge}).
        \item $\displaystyle \int \frac{dx}{x\ln x}
        \rightarrow u=\ln x, du=\frac{dx}{x}\rightarrow$
        \item $\int \frac{du}{u}=\ln u\rightarrow$
	\item $\displaystyle \int \frac{dx}{x\ln x}=\ln|\ln x|+C$.
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\frac{e^x}{1+e^{2x}}$}.
\begin{itemize}
	\item (\ref{i-posr}) $\rightarrow$ (\ref{i-atan}).
 	\item $\displaystyle\int\frac{e^x}{1+e^{2x}}\,dx\rightarrow$
	\item $u=e^x, du=e^x\,dx\rightarrow$
	\item $\int\frac{du}{1+u^2}=\arctg u\rightarrow$
	\item $\displaystyle\int\frac{e^x}{1+e^{2x}}\,dx =\arctg e^x+C$.
\end{itemize}
\end{frame}
	
\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\tg x$}.
\begin{itemize}
\item Upo"stevamo, da je $\displaystyle\tg x =\frac{\sin x}{\cos x}\rightarrow$
\item (\ref{i-posr}) $\rightarrow$ (\ref{i-loge}).
\item $u=\cos x, dx=-\sin x\,dx\rightarrow$ 
\item $\displaystyle\int\frac{\sin x}{\cos x}=
	\int\frac{du}{u}=-\ln |u|\rightarrow$
\item $\displaystyle\int\tg x\,dx=-\ln |cos x|+C$
\end{itemize}
\end{frame}
\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\cos^2x$}.
\begin{itemize}
	\item Upo"stevamo, da je
	$\cos^2x=\frac{1}{2}\left(1+\cos(2x)\right)$ 
	\item (\ref{i-lin}) $\rightarrow$ (\ref{i-potenca-n}) 
		$\rightarrow$ (\ref{i-lin-sub}) $\rightarrow$ (\ref{i-cos}).
	\item $\int \cos^2x\,dx = \int dx + \int\cos(2x)\,dx=\rightarrow$
	\item $\displaystyle\int \cos^2x\,dx=
		\frac{x}{2}+\frac{1}{4}\sin(2x)+C$. 
\end{itemize}
\end{frame}


\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$. \v{c}e  je}
\formulas{$\displaystyle f(x)=xe^{-x}$}
\begin{itemize}
        \item (\ref{i-prod}) $\rightarrow$ (\ref{i-potenca-n})
	$\rightarrow$ (\ref{i-lin-sub}) $\rightarrow$ (\ref{i-exp}).
        \item $\displaystyle \int xe^{-x}\,dx
        \rightarrow u=x, dv=e^{-x}\,dx,\quad du=dx, v=-e^{-x}\rightarrow$
        \item $-xe^{-x}+\int e^{-x}\,dx=-xe^{-x}-e^{-x}$
        \item $\displaystyle \int xe^{-x}\,dx=-e^{-x}(x+1)+C$
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=\frac{\sin x}{1+\cos x}$}.
\begin{itemize}
	\item $\displaystyle\int\frac{\sin x}{1+\cos x}\,dx\rightarrow$
	\item $u=\cos x, du=-\sin x\,dx\rightarrow$
	\item $-\int\frac{du}{1+u}=-\ln|1+u|=-\ln|1+\cos x|\rightarrow$
	\item $\displaystyle\int\frac{\sin x}{1+\cos x}\,dx=
		-\ln|1+\cos x|+C$ 
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=e^{-x}\cos(2x)$}.
\begin{itemize}
	\item Dvakrat uporabimo (\ref{i-prod}).
	\item $\int e^{-x}\cos(2x)\,dx\rightarrow$
	\item $u=e^{-x},dv=\cos(2x)\,dx,\quad du=-e^{-x}\,dx, 
	v=\frac{1}{2}\sin(2x)$
	\item $\int e^{-x}\cos(2x)\,dx=
	\frac{1}{2}e^{-x}\sin(2x)+\frac{1}{2}\int e^{-x}\sin(2x)\,dx$
	\item $\frac{1}{2}\int e^{-x}\sin(2x)\,dx\rightarrow$
	\item $u=e^{-x}, dv=\sin(2x)\,dx,\quad 
	du=-e^{-x}, v=-\frac{1}{2}\cos(2x)$
	\item $\frac{1}{2}\int e^{-x}\sin(2x)\,dx=-
	\frac{1}{4}\left(e^{-x}\cos(2x)+\int e^{-x}\cos(2x)\right)$
	\item $\displaystyle\int e^{-x}\cos(2x)\,dx=
	\frac{1}{5}e^{-x}\left(2\sin(2x)-\cos(2x)\right)+C$
\end{itemize}
\end{frame}

\begin{frame}{Dolo\v{c}i $\int f(x)\,dx$, \v{c}e  je}
\formulas{$\displaystyle f(x)=x^2\ln x$}.
\begin{itemize}
	\item Uporabimo (\ref{i-prod}).
	\item $\displaystyle\int x^2\ln x\,dx\rightarrow$
	\item $u=\ln x, dv=x^2\,dx,\quad du=\frac{dx}{x},
	v=\frac{x^3}{3}\rightarrow$
	\item $\int x^2\ln x\,dx=
		\frac{1}{3}\left(x^3\ln x-\int x^2\,dx\right)$.
	\item $\displaystyle\int x^2\ln x\,dx=
                \frac{x^3}{3}\ln x-\frac{x^3}{9}+C$
\end{itemize}
\end{frame}

\subsection{Dolo\v{c}eni integral}
\subsubsection{Dolo\v{c}eni integrali}

\begin{frame}{Izra\v{c}unaj dolo\v{c}eni integral.}
\formulas{$\displaystyle S=\int_{0}^4e^{\sqrt{x}}\,dx$}.
\begin{itemize}
	\item Uvedemo $t=\sqrt{x}\rightarrow t^2=x\rightarrow 2t\,dt=dx$.
	\item Ker je $x=0\rightarrow t=0$ in $x=4\rightarrow t=2$,
		$S=2\int_0^2te^tdt$.
	\item Integriramo \emph{per partes}: 
	$u=t,\quad dv=e^tdt\rightarrow du=dt,\quad v=e^t$.
	\item $S=\left.2te^t\right|_0^2-2\int_0^2e^tdt=4e^2-2e^2-2=
	2(e^2-1)$.
	\item $\displaystyle\int_0^4 e^{\sqrt{x}}dx=2(e^2-1)$. 
\end{itemize}
\end{frame}

\begin{frame}{Izra\v{c}unaj dolo\v{c}eni integral.}
\formulas{$\displaystyle S=\int_{0}^2xe^{x^2}\,dx$}.
\begin{itemize}
        \item Uvedemo $t=x^2\rightarrow dt=2xdx
	\rightarrow dx=\frac{dt}{2}$.
        \item Ker je $x=0\rightarrow t=0$ in $x=2\rightarrow t=4$,
                $S=\frac{1}{2}\int_0^4e^tdt$.
        \item  $\displaystyle S=\frac{1}{2}\int_0^4e^tdt=\frac{1}{2}(e^4-1)$.
\end{itemize}
\end{frame}

\begin{frame}{Izra\v{c}unaj dolo\v{c}eni integral.}
\formulas{$\displaystyle S=\int_{0}^{\pi^2} \sin\sqrt{x}\,dx$}.
\begin{itemize}
        \item Uvedemo $t=\sqrt{x},\rightarrow t^2=x,\rightarrow 2tdt=dx$.
        \item Ker je $x=0\rightarrow t=0$ in $x=\pi^2\rightarrow t=\pi$, velja
                $S=2\int_0^\pi t\sin t\,dt$.
        \item Integriramo \emph{per partes}: $u=t,\quad dv=\sin t\,dt,\rightarrow du=dt,\quad v=-\cos t$.
	\item $S=\left.-2t\cos t\right|_0^\pi + 2\int_0^\pi\cos t \,dt=\pi $
	$\displaystyle S=2\int_0^\pi t \sin t\,dt=2\pi$.
\end{itemize}
\end{frame}

\begin{frame}{Izra\v{c}unaj dolo\v{c}eni integral.}
\formulas{$\displaystyle S=\int_{0}^{\frac{\pi}{4}} \frac{dx}{\cos^2x}$}.
\begin{itemize}
	\item Uvedemo $t=\tg x$.
	\item $dt = \frac{dx}{\cos^2x}$. Nove meje $x=0\rightarrow t=0$,  
	$x=\frac{\pi}{4}\rightarrow t=1$. 
	\item $\displaystyle\int_0^1dt=\left.t\right|_0^1=1$. 
\end{itemize}
\end{frame}

\begin{frame}{Izra\v{c}unaj dolo\v{c}eni integral.}
\formulas{$\displaystyle\int_4^9\frac{dx}{\sqrt{x}-1}$}.
\begin{itemize}	
	\item Uvedemo $x=t^2,\rightarrow dx=2t\,dt,\quad x=4\rightarrow
	t=2,\quad x=9\rightarrow t=3$. 
	\item $2\int_2^3\frac{t\,dt}{t-1}=
	\left.t+\ln|t-1|\right|_2^3=2+\ln4$. 
	\item $\displaystyle\int_4^9\frac{dx}{\sqrt{x}-1}=2+\ln4$.
\end{itemize}
\end{frame}

\begin{frame}{Izra\v{c}unaj dolo\v{c}eni integral.}
\formulas{$\displaystyle\int_{-1}^2\frac{dx}{9-x^2}$}.
\begin{itemize}
	\item Razcepimo na parcialne ulomke: 
	$\frac{1}{9-x^2}=\frac{1}{6}\left(\frac{1}{x-3}+\frac{1}{x+3}\right)$
	\item $\int_{-1}^2\frac{dx}{9-x^2}=
	\int_{-1}^2\frac{1}{6}\left(\frac{1}{x-3}+
		\frac{1}{x+3}\right)
		dx=\left.\frac{1}{6}\ln|x^2-3|\right|_{-1}^2$. 
	\item $\displaystyle\int_{-1}^2\frac{dx}{9-x^2}=\frac{\ln 10}{6}$.
\end{itemize}
\end{frame}


\subsubsection{Posplo\v{s}eni integrali}
\begin{frame}{Doloka\v{z}i, da je  $\int_0^\infty x^ne^{-x}dx=n\,!$.} 
	\begin{itemize}
		\item Uporabimo (\ref{i-posr}) 
		$\rightarrow u=x^n,dv=e^{-x},\quad du=nx^{n-1}, v=-e^{-x}$
		\item Pi"simo $\Pi(n)=\int_0^\infty x^ne^{-x}dx$.
		\item $\int_0^\infty x^ne^{-x}dx=
		-\left.x^ne^{-x}\right|_0^\infty + 
		n\int_0^\infty x^{n-1}e^{-x}dx$
		\item Uporabimo l'H\^ospitalovo 
		pravilo pri ra"cunanju limite
		$\lim_{x\to\infty}x^ne^{-x}=
		\lim_{x\to\infty}\frac{x^n}{e^x}=0$,
		\item in dobimo zvezo $\Pi(n)=n\Pi(n-1)$.
		\item Ker je 
		$\Pi(0)=\int_0^\infty e^{-x}=
		\left.-e^{-x}\right|_0^\infty=1$,
		\item lahko z matemati"cno indukcijo poka"zemo, da je
			$\Pi(n)=n\,!$
	\end{itemize}
\end{frame}

\begin{frame}{Izra\v{c}unaj (posplo\v{s}eni) dolo\v{c}eni integral.}
\formulas{$\displaystyle\int_1^\infty\frac{dx}{x^2}$}.
\begin{itemize}
	\item $\displaystyle\int_1^\infty\frac{dx}{x}=
		\left.-\frac{1}{x}\right|_1^\infty=1$.
	\item $\displaystyle\int_0^\infty\frac{dx}{x^2}=1$.
\end{itemize}
\end{frame}


\begin{frame}{Izra\v{c}unaj (posplo\v{s}eni) dolo\v{c}eni integral.}
\formulas{$\displaystyle\int_0^4\frac{dx}{\sqrt{4-x}}$}.
\begin{itemize}
	\item Uvedemo $t^2=4-x\rightarrow 2t\,dt=-dx$. 
	\item Meje $x=0\rightarrow
	t=2,\quad x=4\rightarrow t=0$. 
	\item $\displaystyle-2\int_2^0\frac{t\,dt}{t}=\left.-2t\right|_2^0=4$
	\item $\displaystyle\int_0^4\frac{dx}{\sqrt{4-x}}=4$
\end{itemize}
\end{frame}
	
\begin{frame}{Izra\v{c}unaj (posplo\v{s}eni) dolo\v{c}eni integral.}
\formulas{$\displaystyle\int_{-2}^1\frac{|x|}{x}dx$}.
\begin{itemize}
	\item $\displaystyle\int_{-2}^1\frac{|x|}{x}\,dx=
		\int_{-2}^0(-1)dx+\int_0^1dx=
		\left.-x\right|_{-2}^0+\left.x\right|_0^1=-1$
	\item $\displaystyle\int_{-2}^1\frac{|x|}{x}dx=-1$
\end{itemize}
\end{frame}

\begin{frame}{Izra\v{c}unaj (posplo\v{s}eni) dolo\v{c}eni integral.}
\formulas{$\displaystyle\int_{0}^\infty e^{-\sqrt{x}} dx$}.
\begin{itemize}
	\item Uvedemo $t^2=x\rightarrow 2t\,dt=dx$.
	\item Meje $x=0\rightarrow t=0,\quad x\to\infty\rightarrow
	t\to\infty$. 
	\item $\int_{0}^\infty e^{-\sqrt{x}}dx=
	2\int_{0}^\infty te^{-t}\,dt$
	\item Integriramo \emph{per partes}:  
	$du=e^{-t}dx, v=t\rightarrow u=-e^{-t}, dv=dt$. 
	\item $2\int_{0}^\infty te^{-t}\,dt=
	\left.-2te^{-t}\right|_0^\infty + 2\int_0^\infty e^{-t} dt$
	\item Ker je $\lim_{t\to\infty}te^{-t}=0$, je 
	$2\int_0^\infty te^{-t}dt=2$.
	\item $\displaystyle\int_{0}^\infty e^{-\sqrt{x}} dx=2$.
\end{itemize}
\end{frame}

\begin{frame}{Izra\v{c}unaj (posplo\v{s}eni) dolo\v{c}eni integral.}
\formulas{$\displaystyle\int_{0}^\infty 
	\frac{e^{-\sqrt{x}} dx}{\sqrt{x}}$}.
\begin{itemize}
	\item Uvedemo $t^2=x\rightarrow 2t\,dt=dx$.
        \item Meje $x=0\rightarrow t=0,\quad x\to\infty\rightarrow
	t\to\infty$.
	 \item $\displaystyle\int_{0}^\infty
	\frac{e^{-\sqrt{x}}dx}{\sqrt{x}}=
        2\int_{0}^\infty e^{-t}\,dt=\left.-2e^{-t}\right|_0^\infty=2$.
	\item $\displaystyle\int_{0}^\infty 
        \frac{e^{-\sqrt{x}} dx}{\sqrt{x}}=2$.
\end{itemize}
\end{frame}

\begin{frame}{Izra\v{c}unaj (posplo\v{s}eni) dolo\v{c}eni integral.}
\formulas{$\displaystyle\int_{0}^1\sqrt{x}\ln\frac{1}{x}\,dx$}.
\begin{itemize}
	\item $\int_{0}^1\sqrt{x}\ln\frac{1}{x}\,dx=-
	\int_{0}^1\sqrt{x}\ln x\,dx $
	\item Integriramo \emph{per partes}. 
	$u=\ln x, dv=\sqrt{x}\,dx\rightarrow du=\frac{dx}{x},
	v=\frac{2}{3}\sqrt{x^3}$
	\item $\displaystyle\int_{0}^1\sqrt{x}\ln x\,dx=
	\left.\frac{2}{3}\sqrt{x^3}\ln x\right|_0^1-
	\frac{2}{3}\int_0^1\sqrt{x}\,dx$
	\item $\displaystyle\int_{0}^1\sqrt{x}\ln\frac{1}{x}\,dx=\frac{4}{9}$ 
\end{itemize}
\end{frame}


\end{document}
\begin{frame}{Dolo\v{c}i plo\v{s}\v{c}ino lika, ki jo omejujeta grafa\\ 
$f(x)=x$ in $g(x)=2-x^2$} 
\begin{itemize}
	\item Poi"s"cemo prese"ci"s"ci grafov, 
	$x=2-x^2$, $x_1=-2$ in $x_2=1$.
	\item $\displaystyle\int_{-2}^1\left(2-x^2\right)dx-
	\int_{-2}^1x\,dx\rightarrow$.
	\item $\displaystyle\left.
2x-\frac{1}{3}x^3-\frac{1}{2}x^2\right|_{-2}^1=\frac{9}{2}$
\end{itemize}
\end{frame}

\begin{frame}{Slika k prej\v{s}nji nalogi.}
	\includegraphics[scale=1]{func-52.pdf}
\end{frame}


\begin{frame}{Dolo\v{c}i plo\v{s}\v{c}ino med osjo $x$ in grafom
funkcije $f(x)=x^2e^{-x}$ med dvema ekstremoma.}
\begin{itemize}
	\item Odvod $f'(x)=e^{-x}(2x-x^2)$.
	\item Ekstrema $f'(x)=0$, $x_1=0$ in $x_2=2$. 
	\item Plo"s"cina $\int_0^2x^2e^{-x}\,dx$.
	\item Dvakrat uporabimo (\ref{i-prod}) in dobimo $\rightarrow$
	\item $\displaystyle\int_0^2x^2e^{-x}\,dx=
	\left.e^{-x}(-2-2x-x^2)\right|_0^2=2-\frac{10}{e^2}$	
\end{itemize}  
\end{frame}

\begin{frame}{Slika k prej\v{s}nji nalogi.}
        \includegraphics[scale=1]{func-54.pdf}
\end{frame}

\begin{frame}{Dolo\v{c}i plo"s"cino zanke, ki jo omejuje krivulja\\ 
$y^2=x^2(1+x)$}
\begin{itemize}
        \item Zanko sestavljata grafa $y=\pm\sqrt{x^2(1+x)}$,
		$-1\le x \le 0$.  
	\item Plo"s"cina zanke je enaka $-2\int_{-1}^0x\sqrt{1+x}$.
        \item Uporabimo (\ref{i-posr}) 
	$\rightarrow u=1+x, du=dx,x=u-1,\quad x=-1, u=0, \quad x=0, u=1$ 
        \item $\displaystyle -2\int_0^1(u-1)\sqrt{u}=\left.
		-\frac{4}{5}u^\frac{5}{2}+
		\frac{4}{3}u^\frac{3}{2}\right|_0^1=\frac{8}{15}$
	\item Plo"s"cina zanke je $\displaystyle P=\frac{8}{15}$
\end{itemize}
\end{frame}
\begin{frame}{Slika k prej\v{s}nji nalogi.}
        \includegraphics[scale=1]{func-53.pdf}
\end{frame}

\begin{frame}{Koliko je plo"s"cina pod krivuljo Marie Gaetane Agnesi.}
\formulas{$\displaystyle f(x)=\frac{1}{1+x^2}$}.
	\begin{itemize}
		\item
		$\displaystyle P=\int_{-\infty}^\infty\frac{dx}{1+x^2}$
		\item $\int_{-\infty}^\infty\frac{dx}{1+x^2}=
		\left.\arctg x\right|_{-\infty}^\infty$.
		\item $\lim_{x\to-\infty}\arctg x=-\frac{\pi}{2}$, 
			$\lim_{x\to\infty}\arctg x=\frac{\pi}{2}$.
		\item Od tod sledi, da je
		$P=\frac{\pi}{2}+\frac{\pi}{2}=\pi$.
	\end{itemize}
\end{frame}

\begin{frame}{Koliko je plo"s"cina elipse.}
\formulas{$\displaystyle\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$}.
	\begin{itemize}
		\item Parametra $a>0$ in $b>0$ sta polosi elipsi.  
		\item Izrazimo $y=\pm\frac{b}{a}\sqrt{a^2-x^2}$ 
		\item Plo"s"cina je
		$P=2\frac{b}{a}\int_{-a}^a\sqrt{a^2-x^2}dx$.
		\item Vpeljemo $x=a\sin t, dx=a\cos t\,dt,
		-\frac{\pi}{2}\le t \le \frac{\pi}{2}\rightarrow$
		\item $P=2 a\,b\,\int_{-\frac{\pi}{2}}^\frac{\pi}{2}
		\cos^2t\,dt=2 a\,b\,
		\frac{1}{2}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}
		\left(1+\cos(2t)\right)dt
		\rightarrow$
		\item Plo"s"cina je 
		$P=a\,b\,\left.\left(t+\frac{1}{2}\sin(2t)
		\right)\right|_{-\frac{\pi}{2}}^\frac{\pi}{2}=\pi\,a\,b$
	\end{itemize}
\end{frame}


\end{document}
\subsection{Krivulje v ravnini}
\begin{frame}{Parametri\v{c}na oblika}
	\begin{enumerate}
		\item Parametri"cna oblika $x=x(t)$ in $y=y(t)$.
		\item Smerni koeficient tangente
		$\displaystyle\frac{dy}{dx}=\frac{\dot{y}(t)}{\dot{x}(t)}$.
		\item Plo"s"cina zanke 
		$\displaystyle S=
		\frac{1}{2}\int_{t_1}^{t_2}\left(x(t)\dot{y}(t)-
		y(t)\dot{x}(t)\right)dt$.
		\item Dol"zina loka 
		$\displaystyle
		s=\int_{t_1}^{t_2}\sqrt{\dot{x}(t)^2+\dot{y}(t)^2}dt$.
	\end{enumerate}
\end{frame}

\begin{frame}{Eksplicitna oblika}
	\begin{itemize}
		\item Kot poseben primer parametri\v{c}ne oblike 
		$x=x$, $y=y(x)$.
		\item Smerni koeficient tangente 
		$\displaystyle\frac{dy}{dx}=y'(x)$.
		\item Plo"s"cina zanke, ki jo dolo"ca krivulja in os $x$
		$\displaystyle S=\int_{x_1}^{x_2}y(x)\,dx$.
		\item Dol"zina loka 
		$\displaystyle s=\int_{x_1}^{x_2}\sqrt{1+y'(x)^2}dx$.
	\end{itemize}
\end{frame}

\begin{frame}{Polarna oblika}
        \begin{itemize}
		\item $r=r(\varphi)$.
                \item Kot poseben primer parametri\v{c}ne oblike\\ 
                $x=r(\varphi)\cos\varphi$, $y=r(\varphi)\sin\varphi$.
                \item Smerni koeficient tangente $\frac{dy}{dx}$.
                \item Plo"s"cina obmo"cja, ki jo dolo"ca krivulja in
		kota $\varphi_1$, $\varphi_2$ je  
                $\displaystyle S=\frac{1}{2}\int_{\varphi_1}^{\varphi_2}
		r(\varphi)^2\,d\varphi$.
                \item Dol"zina loka med kotoma 
		$\varphi_1$ in $\varphi_2$ je\\ 
		$\displaystyle s=\int_{\varphi_1}^{\varphi_2}
		\sqrt{r(\varphi)^2+r'(\varphi)^2}d\varphi$.
        \end{itemize}
\end{frame}

\begin{frame}{Graf sr\v{c}nice $(x^2+y^2-x)^2=(x^2+y^2)$}
Vpeljemo polarne koordinate: 
$r^2=x^2+y^2$,\\ $x=r\,\cos\varphi$ in dobimo
\formulas{$r=1+\cos\varphi$}.
\formulas{\bf\tt 
PolarPlot[1+Cos[t],\{t,0,2Pi\},PlotStyle->Thick]
}\\
\begin{center}
	\includegraphics[scale=0.5]{func-57.pdf}
\end{center}
\end{frame}

\begin{frame}{Graf lemniskate $(x^2+y^2)^2=(x^2-y^2)$}
Vpeljemo polarne koordinate in dobimo
\formulas{$r=\sqrt{\cos(2\varphi)}$},
$-\frac{\pi}{4}\le\varphi\le\frac{\pi}{4}$,
\formulas{\bf\tt
PolarPlot[Sqrt[Cos[2t]],\{t,-Pi/4,Pi/4\},PlotStyle->Thick]
}\\
\end{frame}





