%% This document created by Scientific Notebook (R) Version 3.5 %% Starting shell: article \documentclass{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Version=5.00.0.2570} %TCIDATA{} %TCIDATA{Created=Wednesday, February 10, 1999 13:29:48} %TCIDATA{LastRevised=Monday, October 08, 2012 11:59:36} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=On line bluem.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{Counters=arabic,1} %TCIDATA{AllPages= %H=36 %F=36,\PARA{038
\QTR{small}{Matematick\U{e1} anal\U{fd}za I online - Krivkov\U{e9} integr\U{e1}ly - Cvi\U{10d}enia\dotfill \thepage }} %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \author{A. U. Thor} \title{Lab Report} \date{The Date } \maketitle \begin{abstract} A Laboratory report created with Scientific Notebook \end{abstract} \section{Krivkov\'{e} integr\'{a}ly} \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{maiindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{maiindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{Ma6.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{Ma6.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{O6.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{O6.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{Glos.tex}}% %BeginExpansion \msihyperref{Index}{}{}{Glos.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \section{Cvi\v{c}enia} \begin{enumerate} \item Vypo\v{c}\'{\i}tajte krivkov\'{e} integr\'{a}ly \item $\int_{C}\frac{1}{x-y}ds,\,$kde $C\ $je \'{u}se\v{c}ka od bodu $\left[ 0,-2\right] $ po bod $\left[ 4,0\right] .% \CustomNote{Answer}{$\sqrt{5}\ln 2.$}$ \item $\int_{C}xds,\,$kde $C\ $je \v{c}as\v{t} paraboly $y=x^{2}$\ medzi bodmi $\left[ 2,4\right] $ a $\left[ 1,1\right] .% \CustomNote{Answer}{$\frac{17\sqrt{17}-5\sqrt{5}}{12}.$}$ \item $\int_{C}x^{2}ds,\,$kde $C\ $je \v{c}as\v{t} grafu $y=\ln x,\,$kde $% 1\leq x\leq 2.% \CustomNote{Answer}{$\frac{5\sqrt{5}-2\sqrt{2}}{3}.$}$ \item $\int_{C}\sqrt{x^{2}+y^{2}}ds,\,$kde $C\ $je kru\v{z}nica $% x^{2}+y^{2}=x.% \CustomNote{Answer}{$2.$}$ \item $\int_{C}x^{2}ds,\,$kde $C\ $je kru\v{z}nica $x^{2}+y^{2}+z^{2}=a^{2},% \,x+y+z=0.% \CustomNote{Answer}{$\frac{2\pi a^{3}}{3}.$}$ \item $\int_{C}\left( x^{2}+y^{2}\right) dx+\left( x^{2}-y^{2}\right) dy,\,$% kde $C\ $je \v{c}as\v{t} grafu funkcie $y=1-\left\vert 1-x\right\vert ,\,0\leq x\leq 2,\,$\ so za\v{c}iato\v{c}n\'{y}m bodom $\left[ 0,0\right] .% \CustomNote{Answer}{$\frac{4}{3}.$}$ \item $\int_{C}\left( x^{2}-2xy\right) dx+\left( y^{2}-2xy\right) dy,\,$kde $% C\ $je krivka $y=x^{2},\,$ z bodu $\left[ -1,1\right] $ po bod $\left[ 1,1% \right] .% \CustomNote{Answer}{$-\frac{14}{15}.$}$ \item $\int_{C}ydx+xdy,\,$kde $C\ $je \v{c}as\v{t} kru\v{z}nice $x=a\cos t,y=a\sin t,\,t\in \left\langle 0,\frac{\pi }{2}\right\rangle ,\,$\ kde $% \left[ a,0\right] $ je za\v{c}iato\v{c}n\'{y} bod.% \CustomNote{Answer}{$0.$} \item $\int_{C}xdx+ydy+\left( x+y-1\right) dz,\,$kde $C\ $je \'{u}se\v{c}ka so za\v{c}iato\v{c}n\'{y}m bodom $\left[ 1,1,1\right] $ a koncov\'{y}m bodom $\left[ 2,3,4\right] .$% \CustomNote{Answer}{$13.$} Zistite, \v{c}i s\'{u} nasleduj\'{u}ce integr\'{a}ly z\'{a}visl\'{e} od integra\v{c}nej cesty: \item $\int_{C}\left( 2x+3y\right) dx+\left( 3x-4y\right) dy.$ \item $\int_{C}\left( x^{4}+4xy^{3}\right) dx+\left( 6x^{2}y^{2}-5y^{4}\right) dy.$ N\'{a}jdite nasleduj\'{u}ce integr\'{a}ly po krivke $C,$\ ktor\'{a} sp\'{a}% ja za\v{c}iato\v{c}n\'{y} bod $A$ s koncov\'{y}m bodom $B,$ ak \item $\int_{C}2xydx+x^{2}dy,\,A=\left[ 0,0\right] ,\,B=\left[ 2,1\right] .% \CustomNote{Answer}{$4.$}$ \item $\int_{C}\left( x^{4}+4xy^{3}\right) dx+\left( 6x^{2}y^{2}-5y^{4}\right) dy,\,A=\left[ -2,-1\right] ,\,B=\left[ 3,0\right] .% \CustomNote{Answer}{$62.$}$ \end{enumerate} \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{maiindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{maiindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{Ma6.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{Ma6.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{O6.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{O6.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{Glos.tex}}% %BeginExpansion \msihyperref{Index}{}{}{Glos.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \rule{6.5in}{0.04in} \textsl{Matematick\'{a} anal\'{y}za II} \section{Krivkov\'{e} integr\'{a}ly} \end{document}