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\begin{document}
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\center{\Large 14. MATEMATIKA A1 FELADATSOR}

\begin{enumerate}
\item Döntse el, hogy az alábbi improrius integrál konvergens vagy divergens. Amelyik konvergens, annak határozza meg az értékét!
\begin{enumerate}
\item $\int_1^{\infty }\frac{1}{x^2}dx$
\megold{$$
\int_1^{\infty }\frac{1}{x^2}dx=\lim_{d\to \infty}\int_1^d\frac{1}{x^2}dx=\lim_{d\to \infty}[\frac{-1}{x}]_1^d=\lim_{d\to \infty}(\frac{-1}{d}-(-1))=1,
$$
tehát konvergens.}
\item $\int_1^{\infty }\frac{1}{\sqrt{x}}dx$
\megold{$$
\int_1^{\infty }\frac{1}{\sqrt{x}}dx=\lim_{d\to \infty}\int_1^d\frac{1}{\sqrt{x}}dx=\lim_{d\to \infty}[\frac{\sqrt{x}}{\frac{1}{2}}]_1^d=\lim_{d\to \infty}(2\sqrt{d}-2)=+\infty ,
$$
tehát divergens.}
\item $\int_1^{\infty }\frac{1}{1+x^2}dx$
\megold{$$
\int_1^{\infty }\frac{1}{1+x^2}dx=\lim_{d\to \infty}\int_1^d\frac{1}{1+x^2}dx=\lim_{d\to \infty}[\arctg x]_1^d=\lim_{d\to \infty}(\arctg d-\arctg 1)=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4},
$$
tehát konvergens.}
\item $\int_{0,5}^{\infty }\frac{1}{1+4x^2}dx$
\megold{$$
\int_{0,5}^{\infty }\frac{1}{1+4x^2}dx=\lim_{d\to \infty}\int_{0,5}^d\frac{1}{1+(2x)^2}dx=\lim_{d\to \infty}[\frac{\arctg 2x}{2}]_{0,5}^d=\lim_{d\to \infty}(\frac{\arctg d}{2}-\frac{\arctg 1}{2})=\frac{\pi }{4}-\frac{\pi }{8}=\frac{\pi }{8},
$$
tehát konvergens}
\item $\int_1^{\infty }\frac{1}{x^2+x}dx$
\megold{$x^2+x=x(x+1)$, ezért $$\frac{1}{x^2+x}=\frac{A}{x}+\frac{B}{x+1}=\frac{A(x+1)+Bx}{x(x+1)}=\frac{(A+B)x+A}{x^2+x},$$ ezért $A+B=0$ és $A=1$, azaz $B=-1$. Igy
$$
\int \frac{1}{x^2+x}dx=\int \frac{1}{x}-\frac{1}{x+1}dx=\ln |x|-\ln |x+1|+c,
$$
ezért
$$
\int_1^{\infty }\frac{1}{x^2+x}dx=\lim_{d\to \infty }\int_1^d\frac{1}{x^2+x}dx=\lim_{d\to \infty }[\ln |x|-\ln |x+1|]_1^d=\lim_{d\to \infty }(\ln |d|-\ln |d+1|]-(\ln |1|-\ln |2|)=
$$
$$
\lim_{d\to \infty }\ln \frac{d}{d+1}-\ln \frac{1}{2}=\ln 1-\ln \frac{1}{2}=\ln 2,
$$
tehát konvergens.}
\item $\int_{-\infty }^0\frac{1}{1+16x^2}dx$
\megold{$$
\int_{-\infty }^0\frac{1}{1+16x^2}dx=\lim_{c\to -\infty} \int_c^{0}\frac{1}{1+(4x)^2}dx=\lim_{c\to -\infty}[\frac{\arctg 4x}{4}]_c^0=
$$
$$
\lim_{c\to -\infty}\frac{\arctg 0}{4}-\frac{\arctg 4c}{4}=0-(-\frac{\pi }{8})=\frac{\pi }{8},
$$
tehát konvergens.}
\item $\int_{-\infty }^0e^{x}dx$c
\megold{$$
\int_{-\infty }^0e^{x}dx=\lim_{c\to -\infty} \int_c^{0}e^{x}dx=\lim_{c\to -\infty}[e^{x}]_c^0=\lim_{c\to -\infty}(e^0-e^c)=1,
$$
tehát konvergens.}
\item $\int_{-\infty }^1xe^xdx$
\megold{Parciális integrálva $u=x$, $v'=e^x$ választással $u'=1$ és $v=e^x$, ezért
$$
\int xe^xdx=xe^x-\int e^xdx=xe^x-e^x+c,
$$
ezért
$$
\int_{-\infty }^1xe^{x}dx=\lim_{c\to -\infty}\int_c^1xe^{x}dx=\lim_{c\to -\infty}[xe^x-e^x]_c^1=\lim_{c\to -\infty}\left( (1e^1-e^1)-(ce^c-e^c)\right) ,
$$
mivel a L'Hospital-szabály alapján
$$
\lim_{c\to -\infty}ce^c=\lim_{c\to -\infty}\frac{c}{e^{-c}}=\frac{+\infty }{+\infty}=\lim_{c\to -\infty}\frac{1}{-e^{-c}}=\frac{1}{+\infty }=0,
$$
és $\lim_{c\to -\infty}e^c=0$, ezért
$$
\int_{-\infty }^1xe^{x}dx=\lim_{c\to -\infty}\left( (1e^1-e^1)-(ce^c-e^c)\right)=0,
$$
tehát konvergens.}
\item $\int_{-\infty }^{\infty }\frac{1}{1+9x^2}dx$
\megold{$$
\int_{-\infty }^{\infty }\frac{1}{1+9x^2}dx=\lim_{c\to -\infty,d\to +\infty}\int_c^d\frac{1}{1+(3x)^2}dx=\lim_{c\to -\infty,d\to +\infty}[\frac{\arctg3x}{3}]_c^d=
$$
$$
\lim_{c\to -\infty,d\to +\infty}\left( \frac{\arctg3d}{3}-\frac{\arctg3c}{3}\right) q
=\frac{\pi /2}{3}-\frac{-\pi /2}{3}=\frac{\pi }{3},
$$
tehát konvergens.}
\item $\int_{-\infty }^{\infty }\frac{1}{x^2+2x+2}dx$
\megold{$$
\int_{-\infty }^{\infty }\frac{1}{x^2+2x+2}dx=\lim_{c\to -\infty,d\to +\infty}\int_c^d\frac{1}{1+(x+1)^2}dx=\lim_{c\to -\infty,d\to +\infty}[\frac{\arctg(x+1)}{1}]_c^d=
$$
$$
\lim_{c\to -\infty,d\to +\infty}\left( \arctg(d+1)-\arctg(c+1)\right) =\frac{\pi }{2}-\frac{-\pi }{2}=\pi ,
$$
theát konvergens.}
\end{enumerate}
\item Döntse el, hogy az alábbi improrius integrál konvergens vagy divergens. Amelyik konvergens, annak határozza meg az értékét!
\begin{enumerate}
\item $\int_0^1\frac{1}{\sqrt{x}}dx$
\megold{$$
\int_0^1\frac{1}{\sqrt{x}}dx=\lim_{a\to 0+}\int_a^1x^{-1/2}dx=\lim_{a\to 0+}[\frac{x^{1/2}}{1/2}]_a^1=\lim_{a\to 0+}2-2\sqrt{a}=2,
$$
tehát konvergens.}
\item $\int_0^1\frac{1}{x^3}dx$
\megold{$$
\int_0^1\frac{1}{x^3}dx=\lim_{a\to 0+}\int_a^1x^{-3}dx=\lim_{a\to 0+}[\frac{x^{-2}}{-2}]_a^1=\lim_{a\to 0+}\frac{-0,5}{1^2}-\frac{-0,5}{a^2}=\infty,
$$
tehát divergens.}
\item $\int_1^2\frac{1}{x-1}dx$
\megold{$$\int_1^2\frac{1}{x-1}dxx=\lim_{a\to 1+}\int_a^1\frac{1}{x-1}dx=\lim_{a\to 0+}[\ln |x-1|]_a^2=\lim_{a\to 0+}\ln 1-\ln |a-1|=\infty,
$$
tehát divergens.}
\item $\int_{-1}^0\frac{1}{\sqrt[3]{x}}dx$
\megold{$$
\int_{-1}^0\frac{1}{\sqrt[3]{x}}dx=\lim_{b\to 0-}\int_{-1}^bx^{-1/3}dx=\lim_{b\to 0-}[\frac{x^{2/3}}{2/3}]_b^{-1}=\lim_{b\to 0-}\frac{3}{2}(-1)^{2/3}-\frac{3}{2}b^{2/3}=\frac{3}{2},
$$
tehát konvergens.}
\item $\int_{-3}^0\frac{1}{x^2}dx$
\megold{$$
\int_{-3}^0\frac{1}{x^2}dx=\lim_{b\to 0-}\int_{-3}^bx^{-2}dx=\lim_{a\to 0+}[\frac{x^{-1}}{-1}]_{-3}^b=\lim_{b\to 0-}\frac{-1}{b}-\frac{-1}{-3}=+\infty ,
$$
tehát divergens.}
\item $\int_{-1}^1\frac{1}{\sqrt{1-x^2}}dx$
\megold{$$
\int_{-1}^1\frac{1}{\sqrt{1-x^2}}dx=\lim_{a\to -1+, b\to 1-}\int_a^b\frac{1}{\sqrt{1-x^2}}dx=\lim_{a\to -1+,b\to 1-}[\arcsin x]_a^b$$ $$\lim_{a\to -1+, b\to 1-}\left( \arcsin b-\arcsin a \right)=\frac{\pi }{2}-\frac{-\pi }{2}=\pi,
$$
tehát konvergens.
}
\item $\int_{-2}^2\frac{2}{\sqrt{4-x^2}}dx$
\megold{$$
\int_{-2}^2\frac{1}{\sqrt{4-x^2}}dx=\lim_{a\to -2+, b\to 2-}\int_a^b\frac{1}{\sqrt{4-x^2}}dx=\lim_{a\to -2+, b\to 2-}\int_a^b\frac{1}{\sqrt{4\left( 1- \frac{x^2}{4}\right) }}dx=$$
$$
\lim_{a\to -2+, b\to 2-}\int_a^b\frac{1}{2}\frac{1}{\sqrt{1-\left( \frac{x}{2}\right) ^2}}dx=\lim_{a\to -1+,b\to 1-}[\frac{1}{2}\frac{\arcsin \frac{x}{2}}{\frac{1}{2}}]_a^b=\lim_{a\to -2+, b\to 2-}\left( \arcsin\frac{b}{2}-\arcsin\frac{a}{2} \right)=$$ $$\frac{\pi }{2}-\frac{-\pi }{2}=\pi,
$$
tehát konvergens.
}
\end{enumerate}
\end{enumerate}


\end{document} 