Lösung 2.2:4a - Versionsgeschichte

Sekretariat1 um 12:52, 27. Aug. 2009

2009-08-27T12:52:49Z

← Nächstältere Version		Version vom 12:52, 27. Aug. 2009
Zeile 1:		Zeile 1:
-	~~Unsere~~ Integral unterscheidet sich ~~nur~~ von dem ~~im Hinweis~~, ~~indem~~ der Nenner <math>x^2+4</math> statt <math>x^2+1</math> ist. Ziehen wir aber den Faktor 4 aus dem Nenner heraus, erhalten wir	+	Unser Integral unterscheidet sich von dem in der Aufgabe nur dadurch, dass der Nenner <math>x^2+4</math> statt <math>x^2+1</math> ist. Ziehen wir aber den Faktor 4 aus dem Nenner heraus, erhalten wir einen ähnlicheren Ausdruck.

	{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}</math>}}		{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}</math>}}

-	~~erhalten wir einen ähnlicheren Ausdruck.~~ Durch die Substitution <math>u=\tfrac{1}{2}x</math> erhalten wir	+	Durch die Substitution <math>u=\tfrac{1}{2}x</math> erhalten wir

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

Sekretariat1 um 12:48, 27. Aug. 2009

2009-08-27T12:48:10Z

← Nächstältere Version		Version vom 12:48, 27. Aug. 2009
Zeile 1:		Zeile 1:
-	~~Unseres~~ Integral unterscheidet sich nur von dem im Hinweis, indem der Nenner <math>x^2+4</math> statt <math>x^2+1</math> ist. Ziehen wir aber den Faktor 4 aus dem Nenner heraus, erhalten wir	+	Unsere Integral unterscheidet sich nur von dem im Hinweis, indem der Nenner <math>x^2+4</math> statt <math>x^2+1</math> ist. Ziehen wir aber den Faktor 4 aus dem Nenner heraus, erhalten wir

	{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}</math>}}		{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}</math>}}

Martin um 13:35, 5. Mai 2009

2009-05-05T13:35:10Z

← Nächstältere Version		Version vom 13:35, 5. Mai 2009
Zeile 1:		Zeile 1:
-	~~What makes our integral differ from that in the exercise´s text is that there is a term~~ <math>x^2+4</math> ~~instead of~~ <math>x^2+1</math>~~, but if we factor out the~~ 4 ~~from the denominator~~,	+	Unseres Integral unterscheidet sich nur von dem im Hinweis, indem der Nenner <math>x^2+4</math> statt <math>x^2+1</math> ist. Ziehen wir aber den Faktor 4 aus dem Nenner heraus, erhalten wir

	{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}</math>}}		{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}</math>}}

-	~~we obtain the correct second term in the denominator~~. ~~On the other hand, there is no longer <math>x^2</math> but <math>\tfrac{1}{4}x^2</math>, although we can get around this by substituting~~ <math>u=\tfrac{1}{2}x</math>,	+	erhalten wir einen ähnlicheren Ausdruck. Durch die Substitution <math>u=\tfrac{1}{2}x</math> erhalten wir

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

Tekbot: Solution 2.2:4a moved to Lösung 2.2:4a: Robot: moved page

2009-03-11T10:24:35Z

Solution 2.2:4a moved to Lösung 2.2:4a: Robot: moved page

← Nächstältere Version

Version vom 10:24, 11. Mär. 2009

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

2009-03-10T13:02:44Z

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

← Nächstältere Version		Version vom 13:02, 10. Mär. 2009
Zeile 1:		Zeile 1:
	What makes our integral differ from that in the exercise´s text is that there is a term <math>x^2+4</math> instead of <math>x^2+1</math>, but if we factor out the 4 from the denominator,		What makes our integral differ from that in the exercise´s text is that there is a term <math>x^2+4</math> instead of <math>x^2+1</math>, but if we factor out the 4 from the denominator,

-	{{~~Displayed math~~\|\|<math>\int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}</math>}}	+	{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}</math>}}

	we obtain the correct second term in the denominator. On the other hand, there is no longer <math>x^2</math> but <math>\tfrac{1}{4}x^2</math>, although we can get around this by substituting <math>u=\tfrac{1}{2}x</math>,		we obtain the correct second term in the denominator. On the other hand, there is no longer <math>x^2</math> but <math>\tfrac{1}{4}x^2</math>, although we can get around this by substituting <math>u=\tfrac{1}{2}x</math>,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}		\frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}
	&= \frac{1}{4}\int \frac{dx}{(x/2)^2+1}		&= \frac{1}{4}\int \frac{dx}{(x/2)^2+1}

Tek um 15:02, 28. Okt. 2008

2008-10-28T15:02:59Z

← Nächstältere Version		Version vom 15:02, 28. Okt. 2008
Zeile 1:		Zeile 1:
-	What makes our integral differ from that in the exercise´s text is that there is a term	+	What makes our integral differ from that in the exercise´s text is that there is a term <math>x^2+4</math> instead of <math>x^2+1</math>, but if we factor out the 4 from the denominator,
-	<math>x^{2}+4</math>	+
-	instead of	+
-	<math>x^{2}+1</math>, but if we factor out the 4 from the denominator,	+

		+	{{Displayed math\|\|<math>\int \frac{dx}{x^2+4} = \int \frac{dx}{4\bigl(\tfrac{1}{4}x^2+1\bigr)} = \frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}\,\textrm{,}</math>}}

-	<math>~~\int{\frac{\,dx}{~~x^{2~~}+4}}=~~\~~int{\frac{\,dx}{4\left( \frac~~{1}{4}x^{2~~}+1 \right)}}~~=\~~frac~~{1}{~~4}\int{\frac{\,dx}{\frac{1}{4~~}x~~^{2}+1}}~~</math>,	+	we obtain the correct second term in the denominator. On the other hand, there is no longer <math>x^2</math> but <math>\tfrac{1}{4}x^2</math>, although we can get around this by substituting <math>u=\tfrac{1}{2}x</math>,

-	~~we obtain the correct second term in the denominator. On the other hand, there is no longer~~	+	{{Displayed math\|\|<math>\begin{align}
-	~~<math>x^~~{~~2}</math>~~	+	\frac{1}{4}\int \frac{dx}{\tfrac{1}{4}x^2+1}
-	~~but~~	+	&= \frac{1}{4}\int \frac{dx}{(x/2)^2+1}
-	~~<math>\frac~~{~~1}{4}x^{2}</~~math~~>, although we can get around this by substituting~~	+	= \left\{\begin{align}
-	~~<math>u=\frac{1}{2}x</math>,~~	+	u &= x/2\\[5pt]
-		+	du &= \tfrac{1}{2}\,dx
-		+	\end{align}\right\}\\[5pt]
-	<math>\begin{align}	+	&= \frac{1}{4}\int \frac{2\,du}{u^2+1}
-	& \frac{1}{4}\int{\frac{\,dx}{\~~frac~~{1}{4}x^{2}+1}}=\frac{1}{4}\int{\frac{\,dx}{~~\left~~( ~~\frac{~~x}{2~~} \right~~)^{2}+1}}=\left\{ \begin{~~matrix~~}	+	= \frac{1}{2}\int\frac{du}{u^2+1}\\[5pt]
-	u=~~\frac{1}{~~2}x \\	+	&= \frac{1}{2}\arctan u + C
-	du=\~~frac~~{1}{2}\,dx \\	+	= \frac{1}{2}\arctan \frac{x}{2} + C\,\textrm{.}
-	\end{~~matrix~~} \right\} \\	+	\end{align}</math>}}
-	& =\frac{1}{4}\int{\frac{2\,du}{u^{2}+1}}=\frac{1}{2}\int{\frac{\,du}{u^{2}+1}} \\	+
-	& =\frac{1}{2}\arctan u+C=\frac{1}{2}\arctan \frac{x}{2}+C \\	+
-	\end{align}</math>	+

Ian um 12:22, 21. Okt. 2008

2008-10-21T12:22:53Z

← Nächstältere Version		Version vom 12:22, 21. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START}~~}	+	What makes our integral differ from that in the exercise´s text is that there is a term
-	<~~center~~> ~~[[Image:2_2_4a~~.~~gif]]~~ </~~center~~>	+	<math>x^{2}+4</math>
-	{{~~NAVCONTENT_STOP~~}}	+	instead of
		+	<math>x^{2}+1</math>, but if we factor out the 4 from the denominator,
		+
		+
		+	<math>\int{\frac{\,dx}{x^{2}+4}}=\int{\frac{\,dx}{4\left( \frac{1}{4}x^{2}+1 \right)}}=\frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}x^{2}+1}}</math>,
		+
		+	we obtain the correct second term in the denominator. On the other hand, there is no longer
		+	<math>x^{2}</math>
		+	but
		+	<math>\frac{1}{4}x^{2}</math>, although we can get around this by substituting
		+	<math>u=\frac{1}{2}x</math>,
		+
		+
		+	<math>\begin{align}
		+	& \frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}x^{2}+1}}=\frac{1}{4}\int{\frac{\,dx}{\left( \frac{x}{2} \right)^{2}+1}}=\left\{ \begin{matrix}
		+	u=\frac{1}{2}x \\
		+	du=\frac{1}{2}\,dx \\
		+	\end{matrix} \right\} \\
		+	& =\frac{1}{4}\int{\frac{2\,du}{u^{2}+1}}=\frac{1}{2}\int{\frac{\,du}{u^{2}+1}} \\
		+	& =\frac{1}{2}\arctan u+C=\frac{1}{2}\arctan \frac{x}{2}+C \\
		+	\end{align}</math>

Tekbot: Lösning 2.2:4a moved to Solution 2.2:4a: Robot: moved page

2008-09-16T14:45:01Z

Lösning 2.2:4a moved to Solution 2.2:4a: Robot: moved page

← Nächstältere Version

Version vom 14:45, 16. Sep. 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

2008-08-21T08:02:14Z

Robot: Automated text replacement (-[[Bild: +[[Image:)

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2008-04-07T09:27:13Z

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