%% 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=Sunday, February 13, 2005 18:10:47} %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 - Aplik\U{e1}cie integr\U{e1}lneho po\U{10d}tu - 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{Aplik\'{a}cie integr\'{a}lneho po\v{c}tu} \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{maindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{maindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{M9.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{M9.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{O9.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{O9.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{G1.tex}}% %BeginExpansion \msihyperref{Index}{}{}{G1.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \section{Cvi\v{c}enia} \textbf{Pr\'{\i}klad 1. }Vypo\v{c}\'{\i}tajte integr\'{a}l $% \int_{0}^{4}f(x)dx$, ak $f(x)=\left\{ \begin{tabular}{ccc} $x^{2}$ & pre & $x\in \langle 0,2)$ \\ $3$ & pre & $x\in \langle 2,3)$ \\ $4-x$ & pre & $x\in \langle 3,4\rangle $% \end{tabular}% \right. .$ \CustomNote{Prob_Solv_Hint}{$\int_{0}^{4}f(x)dx=\int_{0}^{2}f(x)dx+% \int_{2}^{3}f(x)dx+\int_{3}^{4}f(x)dx= $% \par $=\int_{0}^{2}x^{2}dx+\int_{2}^{3}3dx+\int_{3}^{4}(4-x)dx=\frac{37}{6}$.} \textbf{Pr\'{\i}klad 2.} Vypo\v{c}\'{\i}tajte integr\'{a}l $\int_{0}^{1+% \frac{\pi }{2}}f(x)dx$, ak $f(x)=\left\{ \begin{tabular}{ccc} $\sin x$ & pre & $x\in \langle 0,\frac{\pi }{4})$ \\ $x$ & pre & $x\in \langle \frac{\pi }{4},1)$ \\ $\cos {(x-1)}$ & pre & $x\in \langle 1,1+\frac{\pi }{2}\rangle $% \end{tabular}% \right. .$% \CustomNote{Answer}{$\frac{3-\sqrt{2}}{2}-\frac{\pi ^{2}}{32}$.} \textbf{Pr\'{\i}klad 3. }Vypo\v{c}\'{\i}tajte integr\'{a}l $\int_{0}^{1}x\ln xdx$.% \CustomNote{Prob_Solv_Hint}{$\int_{0}^{1}x\ln xdx=\lim_{c\rightarrow 0^{+}}\int_{c}^{1}x\ln xdx=$% \par $=\left| f(x)=\ln x,f^{\prime }(x)=\frac{1}{x},g^{\prime }(x)=x,g(x)=\frac{% x^{2}}{2}\right| =$% \par $=\lim_{c\rightarrow 0^{+}}\left\{ \left[ \frac{x^{2}}{2}\ln x\right] _{c}^{1}-\frac{1}{4}\left[ x^{2}\right] _{c}^{1}\right\} =-\frac{1}{4}$ Integr\'{a}l konverguje. \par {}} \textbf{Pr\'{\i}klad 4. }Vypo\v{c}\'{\i}tajte integr\'{a}l $\int_{1}^{2}% \frac{1}{\sqrt{x^{2}-1}}dx$.% \CustomNote{Prob_Solv_Hint}{% Pom\^{o}cka: najsk\^{o}r pou\v{z}ite substit\'{u}ciu $x=\frac{1}{\cos t}$ Integr\'{a}l diverguje.} \textbf{Pr\'{\i}klad 5. }Vypo\v{c}\'{\i}tajte integr\'{a}l $\int_{0}^{1}% \sqrt{\frac{1+x}{1-x}}dx$.% \CustomNote{Prob_Solv_Hint}{% Pom\^{o}cka: najsk\^{o}r pou\v{z}ite Eulerovu substit\'{u}ciu $t=\sqrt{\frac{% 1+x}{1-x}}$. Integr\'{a}l konverguje. $\frac{\pi }{2}+1.$} \textbf{Pr\'{\i}klad 6. } Vypo\v{c}\'{\i}tajte integr\'{a}l $% \int_{2}^{\infty }\frac{1}{x^{2}}dx$.% \CustomNote{Prob_Solv_Hint}{$\int_{2}^{\infty }\frac{1}{x^{2}}% dx=\lim_{c\rightarrow \infty }\int_{2}^{c}\frac{1}{x^{2}}dx=\lim_{c% \rightarrow \infty }\left[ -\frac{1}{x}\right] _{2}^{c}=$% \par $=\lim_{c\rightarrow \infty }\left[ \frac{1}{2}-\frac{1}{c}\right] =\frac{1}{% 2}$. Integr\'{a}l konverguje.} \textbf{Pr\'{\i}klad 7. } Vypo\v{c}\'{\i}tajte integr\'{a}l $% \int_{1}^{\infty }\frac{1}{x}dx$.% \CustomNote{Answer}{% Integr\'{a}l diverguje.} \textbf{Pr\'{\i}klad 8. } Vypo\v{c}\'{\i}tajte integr\'{a}l $\int_{\sqrt{2}% }^{\infty }\frac{1}{4+x^{2}}dx$.% \CustomNote{Answer}{% Integr\'{a}l konverguje. $\frac{\pi }{8}$} \textbf{Pr\'{\i}klad 9. } Vypo\v{c}\'{\i}tajte integr\'{a}l $\int_{-\infty }^{-\frac{1}{2}}\frac{1}{x^{2}+x+1}dx$.% \CustomNote{Answer}{% Integr\'{a}l konverguje. $\frac{\pi }{\sqrt{3}}$} \textbf{Pr\'{\i}klad 10. } Vypo\v{c}\'{\i}tajte integr\'{a}l $\int_{-\infty }^{\infty }\frac{1}{x^{2}+2x+2}dx$.% \CustomNote{Prob_Solv_Hint}{% Pom\^{o}cka: $\int_{-\infty }^{\infty }\frac{1}{x^{2}+2x+2}dx=\int_{-\infty }^{0}\frac{1}{x^{2}+2x+2}dx+\int_{0}^{\infty }\frac{1}{x^{2}+2x+2}dx$. Integr% \'{a}l konverguje. $\pi $} \textbf{Pr\'{\i}klad 11. } Vypo\v{c}\'{\i}tajte integr\'{a}l $% \int_{0}^{\infty }xe^{-2x}dx$.% \CustomNote{Answer}{% Integr\'{a}l konverguje. $\frac{1}{2}$} \textbf{Pr\'{\i}klad 12. } Vypo\v{c}\'{\i}tajte integr\'{a}l $% \int_{2}^{\infty }\frac{\ln x}{x}dx$.% \CustomNote{Answer}{% Integr\'{a}l diverguje.} \textbf{Pr\'{\i}klad 13. } Vypo\v{c}\'{\i}tajte integr\'{a}l $% \int_{0}^{\infty }\frac{x\ln x}{\left( 1+x^{2}\right) ^{2}}dx$.% \CustomNote{Prob_Solv_Hint}{$\int_{0}^{\infty }\frac{x\ln x}{\left( 1+x^{2}\right) ^{2}}dx=\int_{0}^{1}\frac{x\ln x}{\left( 1+x^{2}\right) ^{2}}% dx+\int_{1}^{\infty }\frac{x\ln x}{\left( 1+x^{2}\right) ^{2}}dx=$% \par $=\lim_{c\rightarrow 0^{+}}\frac{1}{2}\int_{c}^{1}\frac{2x\ln x}{\left( 1+x^{2}\right) ^{2}}dx+\lim_{d\rightarrow \infty }\frac{1}{2}\int_{1}^{d}% \frac{2x\ln x}{\left( 1+x^{2}\right) ^{2}}dx=$% \par $=\left| f(x)=\ln x,f^{\prime }(x)=\frac{1}{x},g^{\prime }(x)=\frac{2x}{% (1+x^{2})^{2}},g(x)=-\frac{1}{1+x^{2}}\right| =$% \par $=\lim_{c\rightarrow 0^{+}}\frac{1}{2}\left\{ \left[ -\frac{\ln x}{1+x^{2}}% \right] _{c}^{1}+\int_{c}^{1}\frac{1}{x(1+x^{2})}dx\right\} +$% \par $+\lim_{d\rightarrow \infty }\frac{1}{2}\left\{ \left[ -\frac{\ln x}{1+x^{2}}% \right] _{1}^{d}+\int_{1}^{d}\frac{1}{x(1+x^{2})}dx\right\} $, preto\v{z}e plat\'{\i} $\lim_{c\rightarrow 0^{+}}\frac{\ln c}{1+c^{2}}=-\infty $ integr% \'{a}l diverguje.} \textbf{Pr\'{\i}klad 14. } Vypo\v{c}\'{\i}tajte integr\'{a}l $% \int_{1}^{\infty }\frac{x^{3}+1}{x^{4}}dx$.% \CustomNote{Answer}{% Integr\'{a}l diverguje.} \textbf{Pr\'{\i}klad 15. } Vypo\v{c}\'{\i}tajte plo\v{s}n\'{y} obsah rovinnej oblasti ohrani\v{c}enej \v{c}iarami $y=x^{2}-x,\ y=x$. Na\v{c}% rtnite obr\'{a}zok!$% \CustomNote{Prob_Solv_Hint}{$x^{2}-2x=x\ \Rightarrow x_{1}=0,\ x_{2}=3$, $% P=\int_{0}^{3}\left| x-(x^{2}-2x)\right| dx=\int_{0}^{3}\left( 3x-x^{2}\right) dx=$% \par $=\left[ \frac{3}{2}x^{2}-\frac{x^{3}}{3}\right] _{0}^{3}=\frac{9}{2}$}$ \textbf{Pr\'{\i}klad 16. } Vypo\v{c}\'{\i}tajte plo\v{s}n\'{y} obsah rovinnej oblasti ohrani\v{c}enej \v{c}iarami $y=\frac{27}{x^{2}+9},\ y=\frac{% x^{2}}{6}$. Na\v{c}rtnite obr\'{a}zok!% \CustomNote{Prob_Solv_Hint}{$P=\int_{-3}^{3}\left| \frac{27}{x^{2}+9}-\frac{% x^{2}}{6}\right| dx=\frac{9\pi }{2}$.} \textbf{Pr\'{\i}klad 17. } Vypo\v{c}\'{\i}tajte plo\v{s}n\'{y} obsah rovinnej oblasti ohrani\v{c}enej \v{c}iarami $y=x\ln x,\ x=\frac{1}{2},\ x=2,\ o_{x}$. Na\v{c}rtnite obr\'{a}zok!% \CustomNote{Prob_Solv_Hint}{$P=\int_{\frac{1}{2}}^{2}\left| x\ln x\right| dx=-\int_{\frac{1}{2}}^{1}x\ln xdx+\int_{1}^{2}x\ln xdx=\frac{15}{8}\ln 2-% \frac{25}{16}$.} \textbf{Pr\'{\i}klad 18. } Vypo\v{c}\'{\i}tajte plo\v{s}n\'{y} obsah rovinnej oblasti ohrani\v{c}enej \v{c}iarami $yx=4,\ x+y=5$. Na\v{c}rtnite obr\'{a}zok!% \CustomNote{Prob_Solv_Hint}{$P=\int_{1}^{4}\left| 5-x-\frac{4}{x}\right| dx=% \frac{15}{2}-2\ln 8$.} \textbf{Pr\'{\i}klad 19. } Vypo\v{c}\'{\i}tajte plo\v{s}n\'{y} obsah rovinnej oblasti ohrani\v{c}enej \v{c}iarami $y=0,\ y=xe^{-2x}$ na $\langle 0,\frac{1}{2}\rangle $. Na\v{c}rtnite obr\'{a}zok!% \CustomNote{Prob_Solv_Hint}{$P=\int_{0}^{\frac{1}{2}}xe^{-2x}dx=\frac{1}{4}-% \frac{1}{2e}$.} \textbf{Pr\'{\i}klad 20. } Vypo\v{c}\'{\i}tajte objem telesa, ktor\'{e} vznikne rot\'{a}ciou oblasti ur\v{c}enej \v{c}iarami $y=\frac{2x}{\pi },\ y=\sin x,\ x\geq 0$ okolo osi $o_{x}$. Na\v{c}rtnite obr\'{a}zok! $% \CustomNote{Prob_Solv_Hint}{$V=\pi \int_{0}^{\frac{\pi }{2}}\left[ \sin ^{2}x-\frac{4x^{2}}{{\pi }^{2}}\right] dx=\frac{{\pi }^{2}}{12}$.}$ \textbf{Pr\'{\i}klad 21. } Zistite plochu kolm\'{e}ho rezu a vypo\v{c}\'{\i}% tajte objem gule s polomerom $r$. \CustomNote{Prob_Solv_Hint}{% Gu\v{l}u umiestnime tak, \v{z}e jej stred je toto\v{z}n\'{y} so stredom s% \'{u}radnicovej s\'{u}stavy. Kolm\'{y} rez je kru\v{z}nica s polomerom $r(x)$% , kde $x$ je vzdialenos\v{t} kolm\'{e}ho rezu od stredu gule. Plat\'{\i} $% r(x)=\sqrt{r^{2}-x^{2}}$. Plocha kolm\'{e}ho rezu je $A(x)=\pi r^{2}(x)=\pi (r^{2}-x^{2})$. Potom $V=\int_{-r}^{r}\pi (r^{2}-x^{2})dx=\frac{4}{3}\pi r^{3}$.} \textbf{Pr\'{\i}klad 22. } Zistite plochu kolm\'{e}ho rezu a vypo\v{c}\'{\i}% tajte objem pravideln\'{e}ho \v{s}tvorbok\'{e}ho ihlana so z\'{a}klad\v{n}ou d\'{l}\v{z}ky $a$ a v\'{y}\v{s}kou $h$. \CustomNote{Prob_Solv_Hint}{% Ihlan umiestnime tak, \v{z}e jeho vrchol d\'{a}me do stredu s\'{u}radnicovej s\'{u}stavy a v\'{y}\v{s}ka je toto\v{z}n\'{a} s osou $o_{x}$. Kolm\'{y} rez je \v{s}tvorec so stranou $a(x)$, kde $x$ je vzdialenos\v{t} kolm\'{e}ho rezu od vrcholu ihlana. Plat\'{\i} $a(x)=\frac{a}{h}x$. Plocha kolm\'{e}ho rezu je $A(x)=a^{2}(x)=\frac{a^{2}x^{2}}{h^{2}}$. Potom $V=\int_{0}^{h}\frac{% a^{2}x^{2}}{h^{2}}dx=\frac{1}{3}a^{2}h$.} \textbf{Pr\'{\i}klad 23. } Vypo\v{c}\'{\i}tajte objem telesa, ktor\'{e} vznikne rot\'{a}ciou rovinnej oblasti ohrani\v{c}enej \v{c}iarami $% y=1-x^{2},\ y=x^{2}$ okolo osi $o_{x}$.$% \CustomNote{Answer}{$V=\frac{2\sqrt{2}\pi }{3}$.}$ \textbf{Pr\'{\i}klad 24. } Vypo\v{c}\'{\i}tajte d\'{l}\v{z}ku krivky $y=2% \sqrt{x}$, $1\leq x\leq 2$. $% \CustomNote{Answer}{$\sqrt{6}-\sqrt{2}+\frac{1}{2}\ln {\frac{2\sqrt{6}+5}{2% \sqrt{2}+3}}$.}$ \textbf{Pr\'{\i}klad 25. } Vypo\v{c}\'{\i}tajte d\'{l}\v{z}ku krivky $y=\ln {% \frac{e^{x}+1}{e^{x}-1}},\ x\in \langle \ln 2,\ln 5\rangle $. $% \CustomNote{Answer}{$\ln {\frac{16}{5}}$.}$ \begin{center} \begin{tabular}{|c|c|c|c|} \hline \textbf{% %TCIMACRO{\hyperref{Obsah}{}{}{maindex.tex}}% %BeginExpansion \msihyperref{Obsah}{}{}{maindex.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Obsah kapitoly}{}{}{M9.tex}}% %BeginExpansion \msihyperref{Obsah kapitoly}{}{}{M9.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Ot\'{a}zky}{}{}{O9.tex}}% %BeginExpansion \msihyperref{Ot\'{a}zky}{}{}{O9.tex}% %EndExpansion } & \textbf{% %TCIMACRO{\hyperref{Index}{}{}{G1.tex}}% %BeginExpansion \msihyperref{Index}{}{}{G1.tex}% %EndExpansion } \\ \hline \end{tabular} \end{center} \rule{6.5in}{0.04in} \textsl{Matematick\'{a} anal\'{y}za I} \section{Aplik\'{a}cie integr\'{a}lneho po\v{c}tu} \end{document}