Lösung 2.1:2c - Versionsgeschichte

Sekretariat1 um 09:18, 21. Aug. 2009

2009-08-21T09:18:19Z

← Nächstältere Version		Version vom 09:18, 21. Aug. 2009
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-	Wir erinnern uns daran dass <math>\sqrt{x} = x^{1/2}</math>, und erhalten	+	Wir erinnern uns daran, dass <math>\sqrt{x} = x^{1/2}</math> und erhalten

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 7:		Zeile 7:
	\end{align}</math>}}		\end{align}</math>}}

-	Dies ist ein Standardintegral mit nur <math>x^n</math>-Terme, wo <math>n=1/2</math> und <math>n=-1/2\,</math>, in den beiden Termen ist.	+	Dies ist ein Standardintegral mit nur <math>x^n</math>-Terme wo <math>n=1/2</math> und <math>n=-1/2\,</math>, in den beiden Termen ist.

	Wir erhalten das Integral		Wir erhalten das Integral

Martin um 16:48, 28. Apr. 2009

2009-04-28T16:48:56Z

← Nächstältere Version		Version vom 16:48, 28. Apr. 2009
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-	~~If we recall that~~ <math>\sqrt{x} = x^{1/2}</math>, ~~the integral can be written as~~	+	Wir erinnern uns daran dass <math>\sqrt{x} = x^{1/2}</math>, und erhalten

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 7:		Zeile 7:
	\end{align}</math>}}		\end{align}</math>}}

-	~~This is a standard integral in which the integrand consists of two terms looking like~~ <math>x^n</math>, ~~where~~ <math>n=1/2</math> ~~and~~ <math>n=-1/2\,</math>, ~~respectively~~.	+	Dies ist ein Standardintegral mit nur <math>x^n</math>-Terme, wo <math>n=1/2</math> und <math>n=-1/2\,</math>, in den beiden Termen ist.

-	~~We obtain~~	+	Wir erhalten das Integral

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}

Tekbot: Solution 2.1:2c moved to Lösung 2.1:2c: Robot: moved page

2009-03-11T10:15:55Z

Solution 2.1:2c moved to Lösung 2.1:2c: Robot: moved page

← Nächstältere Version

Version vom 10:15, 11. Mär. 2009

Tekbot: Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

2009-03-10T12:58:24Z

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 12:58, 10. Mär. 2009
Zeile 1:		Zeile 1:
	If we recall that <math>\sqrt{x} = x^{1/2}</math>, the integral can be written as		If we recall that <math>\sqrt{x} = x^{1/2}</math>, the integral can be written as

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int\limits_{4}^{9} \bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)\,dx		\int\limits_{4}^{9} \bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)\,dx
	&= \int\limits_{4}^{9}\Bigl( x^{1/2}-\frac{1}{x^{1/2}}\Bigr)\,dx\\[5pt]		&= \int\limits_{4}^{9}\Bigl( x^{1/2}-\frac{1}{x^{1/2}}\Bigr)\,dx\\[5pt]
Zeile 11:		Zeile 11:
	We obtain		We obtain

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int\limits_{4}^{9} \bigl( x^{1/2}-x^{-1/2}\bigr)\,dx		\int\limits_{4}^{9} \bigl( x^{1/2}-x^{-1/2}\bigr)\,dx
	&= \Bigl[\ \frac{x^{1/2+1}}{1/2+1} - \frac{x^{-1/2+1}}{-1/2+1}\ \Bigr]_{4}^{9}\\[5pt]		&= \Bigl[\ \frac{x^{1/2+1}}{1/2+1} - \frac{x^{-1/2+1}}{-1/2+1}\ \Bigr]_{4}^{9}\\[5pt]

Tek um 12:59, 21. Okt. 2008

2008-10-21T12:59:00Z

← Nächstältere Version		Version vom 12:59, 21. Okt. 2008
Zeile 1:		Zeile 1:
-	If we recall that	+	If we recall that <math>\sqrt{x} = x^{1/2}</math>, the integral can be written as
-	<math>\sqrt{x}=x^{{1}/{2~~}\;~~}</math>, the integral can be written as	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\int\limits_{4}^{9} \bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)\,dx
		+	&= \int\limits_{4}^{9}\Bigl( x^{1/2}-\frac{1}{x^{1/2}}\Bigr)\,dx\\[5pt]
		+	&= \int\limits_{4}^{9}\bigl(x^{1/2} - x^{-1/2}\bigr)\,dx\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	This is a standard integral in which the integrand consists of two terms looking like <math>x^n</math>, where <math>n=1/2</math> and <math>n=-1/2\,</math>, respectively.
-	~~& \int\limits_{4}^{9}{\left( \sqrt{x}-\frac{1}{\sqrt{x}} \right)}\,dx=\int\limits_{4}^{9}{\left( x^{{1}/{2}\;}-\frac{1}{x^{{1}/{2}\;}} \right)}\,dx \\~~	+
-	~~& =\int\limits_{4}^{9}{\left( x^{{1}/{2}\;}-x^{{-1}/{2}\;} \right)}\,dx \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	This is a standard integral in which the integrand consists of two terms looking like	+
-	<math>x^{n}</math>, where	+
-	<math>n=~~\frac{~~1}{2}</math>	+
-	and	+
-	<math>n=-~~\frac{~~1}{2}</math>.	+

	We obtain		We obtain

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\int\limits_{4}^{9} \bigl( x^{1/2}-x^{-1/2}\bigr)\,dx
-	& \int\limits_{4}^{9}{\~~left~~( x^{{1}/{2~~}\;~~}-x^{{-1}/{2}\~~;} \right~~)}\,dx=\~~left~~[ \frac{x^~~{\frac~~{1}{2}+1}}~~{\frac~~{1}{2}+1}-\frac{x^{-~~\frac{~~1}{2}+1}}{-~~\frac{~~1}{2}+1} \~~right~~]_{4}^{9} \\	+	&= \Bigl[\ \frac{x^{1/2+1}}{1/2+1} - \frac{x^{-1/2+1}}{-1/2+1}\ \Bigr]_{4}^{9}\\[5pt]
-	& =\~~left~~[ \frac{x^{1+~~\frac{~~1}{2}}}~~{\frac~~{3}{2}}-\frac{x^~~{\frac~~{1}{2}}}~~{\frac~~{1}{2}} \~~right~~]_{4}^{9} \\	+	&= \Bigl[\ \frac{x^{1+1/2}}{3/2} - \frac{x^{1/2}}{1/2}\ \Bigr]_{4}^{9}\\[5pt]
-	& =\~~left~~[ \frac{2}{3}x\sqrt{x}-2\sqrt{x} \~~right~~]_{4}^{9} \\	+	&= \Bigl[\ \frac{2}{3}x\sqrt{x} - 2\sqrt{x}\ \Bigr]_{4}^{9}\\[5pt]
-	& =\frac{2}{3}\~~centerdot~~ 9\~~centerdot~~ \sqrt{9}-2\sqrt{9}-\~~left~~( \frac{2}{3}\~~centerdot~~ 4\~~centerdot~~ \sqrt{4}-2\sqrt{4} \~~right~~) \\	+	&= \frac{2}{3}\cdot 9\cdot\sqrt{9} - 2\sqrt{9} - \Bigl(\frac{2}{3}\cdot 4\cdot \sqrt{4}-2\sqrt{4} \Bigr)\\[5pt]
-	& =\frac{2}{3}93-2\~~centerdot~~ 3-\~~left~~( \frac{2}{3}\~~centerdot~~ 4\~~centerdot~~ 2-2\~~centerdot~~ 2 \~~right~~) \\	+	&= \frac{2}{3}\cdot 9\cdot 3 - 2\cdot 3 - \Bigl( \frac{2}{3}\cdot 4\cdot 2 - 2\cdot 2 \Bigr)\\[5pt]
-	& =18-6-\frac{16}{3}+4=16-\frac{16}{3} \\	+	&= 18-6-\frac{16}{3}+4\\[5pt]
-	& =\frac{16\~~centerdot~~ 3-16}{3}=\frac{32}{3} \\	+	&= 16-\frac{16}{3}\\[5pt]
-	\end{align}</math>	+	&= \frac{16\cdot 3-16}{3}\\[5pt]
		+	&= \frac{32}{3}\,\textrm{.}
		+	\end{align}</math>}}

Ian um 13:14, 17. Okt. 2008

2008-10-17T13:14:46Z

← Nächstältere Version		Version vom 13:14, 17. Okt. 2008
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-	{{~~NAVCONTENT_START~~}}	+	If we recall that
-	<~~center~~> [[~~Image:2_1_2c.gif~~]] ~~</center>~~	+	<math>\sqrt{x}=x^{{1}/{2}\;}</math>, the integral can be written as
-	{{~~NAVCONTENT_STOP~~}}	+
		+
		+	<math>\begin{align}
		+	& \int\limits_{4}^{9}{\left( \sqrt{x}-\frac{1}{\sqrt{x}} \right)}\,dx=\int\limits_{4}^{9}{\left( x^{{1}/{2}\;}-\frac{1}{x^{{1}/{2}\;}} \right)}\,dx \\
		+	& =\int\limits_{4}^{9}{\left( x^{{1}/{2}\;}-x^{{-1}/{2}\;} \right)}\,dx \\
		+	\end{align}</math>
		+
		+
		+	This is a standard integral in which the integrand consists of two terms looking like
		+	<math>x^{n}</math>, where
		+	<math>n=\frac{1}{2}</math>
		+	and
		+	<math>n=-\frac{1}{2}</math>.
		+
		+	We obtain
		+
		+
		+	<math>\begin{align}
		+	& \int\limits_{4}^{9}{\left( x^{{1}/{2}\;}-x^{{-1}/{2}\;} \right)}\,dx=\left[ \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} \right]_{4}^{9} \\
		+	& =\left[ \frac{x^{1+\frac{1}{2}}}{\frac{3}{2}}-\frac{x^{\frac{1}{2}}}{\frac{1}{2}} \right]_{4}^{9} \\
		+	& =\left[ \frac{2}{3}x\sqrt{x}-2\sqrt{x} \right]_{4}^{9} \\
		+	& =\frac{2}{3}\centerdot 9\centerdot \sqrt{9}-2\sqrt{9}-\left( \frac{2}{3}\centerdot 4\centerdot \sqrt{4}-2\sqrt{4} \right) \\
		+	& =\frac{2}{3}93-2\centerdot 3-\left( \frac{2}{3}\centerdot 4\centerdot 2-2\centerdot 2 \right) \\
		+	& =18-6-\frac{16}{3}+4=16-\frac{16}{3} \\
		+	& =\frac{16\centerdot 3-16}{3}=\frac{32}{3} \\
		+	\end{align}</math>

Tekbot: Lösning 2.1:2c moved to Solution 2.1:2c: Robot: moved page

2008-09-16T14:36:21Z

Lösning 2.1:2c moved to Solution 2.1:2c: Robot: moved page

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Version vom 14:36, 16. Sep. 2008

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2008-08-21T07:57:54Z

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2008-04-07T08:09:29Z

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