\documentclass{article} \usepackage{amssymb} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %TCIDATA{OutputFilter=LATEX.DLL} %TCIDATA{Created=Monday, June 25, 2001 17:45:57} %TCIDATA{LastRevised=Saturday, June 01, 2002 20:27:05} %TCIDATA{} %TCIDATA{} %TCIDATA{CSTFile=On line bluem.cst} %TCIDATA{PageSetup=72,72,72,72,0} %TCIDATA{AllPages= %H=36 %F=36,\PARA{038
\QTR{small}{Matematick\U{e1} anal\U{fd}za I online - D\U{f4}kazy\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{Met\'{o}dy integr\'{a}lneho po\v{c}tu} \subsection{D\^{o}kaz substitu\v{c}nej met\'{o}dy} \textbf{D\^{o}kaz: }K funkcii $f:J\longrightarrow \mathbf{R}$ existuje primit% \'{\i}vna funkcia $F:J\longrightarrow \mathbf{R}$, tak\'{a} \v{z}e $% F^{\prime }(x)=f(x),\forall x\in J$. Potom zlo\v{z}en\'{a} funkcia $\left( F\circ g\right) :I\longrightarrow \mathbf{R}$ je diferencovate\v{l}n\'{a} a plat\'{\i} $(F\circ g)^{\prime }=(F^{\prime }\circ g)g^{\prime }=(f\circ g)g^{\prime }$ teda funkcia $F\circ g$ je primit\'{\i}vna funkcia ku funkcii $\left( f\circ g\right) g^{\prime }:I\longrightarrow \mathbf{R}$. Pod\v{l}a predpokladu aj funkcia $G:I\longrightarrow \mathbf{R}$ je primit\'{\i}vna funkcia k funkcii $\left( f\circ g\right) g^{\prime }:I\longrightarrow \mathbf{R}$ \v{c}o znamen\'{a}, \v{z}e $G(t)=F(g(t))+C,\,\forall t\in I$. Nech $g^{-1}:J\longrightarrow I$ je inverzn\'{a} bijekcia ku $% g:I\longrightarrow J$. Potom \[ (G\circ g^{-1})(x)=G\left( g^{-1}\left( x\right) \right) =(F\circ g)(g^{-1}(x))+C=F(x)+C,\,\forall x\in J. \] Preto \[ (G\circ g^{-1})^{\prime }(x)=(F(x)+C)^{\prime }=f(x),\,\forall x\in J.\blacksquare \] \begin{center} \begin{tabular}{|c|} \hline \hyperref{{\small Sp\"{a}\v{t}}}{}{}{M82.tex#2} \\ \hline \end{tabular} \end{center} \end{document}