Lösung 2.2:4b - Versionsgeschichte

Sekretariat1 um 13:01, 27. Aug. 2009

Sekretariat1 — Thu, 27 Aug 2009 13:01:44 GMT

← Nächstältere Version		Version vom 13:01, 27. Aug. 2009
Zeile 1:		Zeile 1:
-	Es wäre möglich die Substitution <math>u=x-1</math> zu machen, aber dies würde nicht das Problem mit dem Term 3 lösen. Wir ziehen ~~statt dessen~~ den Faktor 3 aus den Nenner	+	Es wäre möglich, die Substitution <math>u=x-1</math> zu machen, aber dies würde nicht das Problem mit dem Term 3 lösen. Wir ziehen stattdessen den Faktor 3 aus den Nenner

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

-	und schreiben den Faktor <math>\tfrac{1}{3}</math> in das Quadrat <math>(x-1)^2</math>,	+	und schreiben den Faktor <math>\tfrac{1}{3}</math> in das Quadrat <math>(x-1)^2</math>.

-	{{Abgesetzte Formel\|\|<math>\frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1~~}\,\textrm{.~~}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}</math>}}

	Jetzt machen wir die Substitution <math>u = (x-1)/\!\sqrt{3}</math> und erhalten		Jetzt machen wir die Substitution <math>u = (x-1)/\!\sqrt{3}</math> und erhalten

Moni um 10:06, 22. Aug. 2009

Moni — Sat, 22 Aug 2009 10:06:41 GMT

← Nächstältere Version		Version vom 10:06, 22. Aug. 2009
Zeile 1:		Zeile 1:
-	Es wäre möglich die Substitution <math>u=x-1</math> zu machen, aber dies würde nicht das Problem mit ~~den~~ Term 3 lösen. Wir ziehen statt dessen den Faktor 3 aus den Nenner	+	Es wäre möglich die Substitution <math>u=x-1</math> zu machen, aber dies würde nicht das Problem mit dem Term 3 lösen. Wir ziehen statt dessen den Faktor 3 aus den Nenner

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

-	und ~~schieben~~ den Faktor <math>\tfrac{1}{3}</math> in ~~die Quadrate~~ <math>(x-1)^2</math>,	+	und schreiben den Faktor <math>\tfrac{1}{3}</math> in das Quadrat <math>(x-1)^2</math>,

	{{Abgesetzte Formel\|\|<math>\frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}</math>}}

Martin um 13:45, 5. Mai 2009

Martin — Tue, 05 May 2009 13:45:02 GMT

← Nächstältere Version		Version vom 13:45, 5. Mai 2009
Zeile 1:		Zeile 1:
-	~~We could substitute~~ <math>u=x-1</math>, ~~but we would then still have the problem of the second term,~~ 3~~, in the denominator~~. ~~Instead, we take out a factor~~ 3 ~~from the denominator,~~	+	Es wäre möglich die Substitution <math>u=x-1</math> zu machen, aber dies würde nicht das Problem mit den Term 3 lösen. Wir ziehen statt dessen den Faktor 3 aus den Nenner

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

-	~~and move a factor~~ <math>\tfrac{1}{3}</math> ~~into the square~~ <math>(x-1)^2</math>,	+	und schieben den Faktor <math>\tfrac{1}{3}</math> in die Quadrate <math>(x-1)^2</math>,

	{{Abgesetzte Formel\|\|<math>\frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}</math>}}

-	~~Now, we substitute~~ <math>u = (x-1)/\!\sqrt{3}</math> ~~and get rid of all the problems at once,~~	+	Jetzt machen wir die Substitution <math>u = (x-1)/\!\sqrt{3}</math> und erhalten

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

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

Tekbot — Wed, 11 Mar 2009 10:24:55 GMT

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

← Nächstältere Version

Version vom 10:24, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 13:02:54 GMT

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

← Nächstältere Version		Version vom 13:02, 10. Mär. 2009
Zeile 1:		Zeile 1:
	We could substitute <math>u=x-1</math>, but we would then still have the problem of the second term, 3, in the denominator. Instead, we take out a factor 3 from the denominator,		We could substitute <math>u=x-1</math>, but we would then still have the problem of the second term, 3, in the denominator. Instead, we take out a factor 3 from the denominator,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int \frac{dx}{(x-1)^2+3}		\int \frac{dx}{(x-1)^2+3}
	&= \int \frac{dx}{3\bigl(\tfrac{1}{3}(x-1)^2+1\bigr)}\\[5pt]		&= \int \frac{dx}{3\bigl(\tfrac{1}{3}(x-1)^2+1\bigr)}\\[5pt]
Zeile 9:		Zeile 9:
	and move a factor <math>\tfrac{1}{3}</math> into the square <math>(x-1)^2</math>,		and move a factor <math>\tfrac{1}{3}</math> into the square <math>(x-1)^2</math>,

-	{{~~Displayed math~~\|\|<math>\frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}</math>}}

	Now, we substitute <math>u = (x-1)/\!\sqrt{3}</math> and get rid of all the problems at once,		Now, we substitute <math>u = (x-1)/\!\sqrt{3}</math> and get rid of all the problems at once,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}		\frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}
	&= \left\{\begin{align}		&= \left\{\begin{align}

Tek um 15:15, 28. Okt. 2008

Tek — Tue, 28 Oct 2008 15:15:27 GMT

← Nächstältere Version		Version vom 15:15, 28. Okt. 2008
Zeile 1:		Zeile 1:
-	We could substitute	+	We could substitute <math>u=x-1</math>, but we would then still have the problem of the second term, 3, in the denominator. Instead, we take out a factor 3 from the denominator,
-	<math>u=x-~~\text{~~1}</math>, but we would then still have the problem of the second term,	+
-	~~<math>\text{~~3~~}</math>~~	+
-	in the denominator. Instead, we take out a factor	+
-	~~<math>\text{~~3~~}</math>~~	+
-	from the denominator,	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\int \frac{dx}{(x-1)^2+3}
		+	&= \int \frac{dx}{3\bigl(\tfrac{1}{3}(x-1)^2+1\bigr)}\\[5pt]
		+	&= \frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1}
		+	\end{align}</math>}}

-	<math>\~~begin~~{~~align}~~	+	and move a factor <math>\tfrac{1}{3}</math> into the square <math>(x-1)^2</math>,
-	~~& \int{\frac{\,dx}{\left( x-~~1 ~~\right)^{2}+3}}=\int{\frac{\,dx~~}{3~~\left( \frac{1~~}~~{3}\left~~( x-1 ~~\right~~)^{2~~}+1 \right)}} \\~~	+
-	~~& =\frac{1}{3}\int{\frac{\,dx}{\frac{1}{3}\left( x-1 \right)^{2}+1}} \\~~	+
-	~~\end{align}~~</math>	+

-	~~and move a factor~~	+	{{Displayed math\|\|<math>\frac{1}{3}\int \frac{dx}{\tfrac{1}{3}(x-1)^2+1} = \frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}\,\textrm{.}</math>}}
-	<math>\frac{1}{3}~~</math>~~	+
-	~~into the square~~	+
-	~~<math>~~\~~left~~( x-1 ~~\right~~)^{2}</math>,	+

		+	Now, we substitute <math>u = (x-1)/\!\sqrt{3}</math> and get rid of all the problems at once,

-	~~<math>\frac~~{1}{~~3}\int{\frac{\,dx}{\frac{1}{3}\left( x-1 \right)^{2}+1}}=\frac{1}{3}\int{\frac{\,dx}{\left( \frac{x-1}{\sqrt{3}} \right)^{2}+1}}</~~math>	+	{{Displayed math\|\|<math>\begin{align}
-		+	\frac{1}{3}\int \frac{dx}{\Bigl(\dfrac{x-1}{\sqrt{3}}\Bigr)^2+1}
-	~~Now, we substitute~~	+	&= \left\{\begin{align}
-	~~<math>u=\frac{x-1}{\sqrt{3}}</math>~~	+	u &= (x-1)/\!\sqrt{3}\\[5pt]
-	~~and get rid of all the problems at once:~~	+	du &= dx/\!\sqrt{3}
-		+	\end{align}\right\}\\[5pt]
-		+	&= \frac{1}{3}\int \frac{\sqrt{3}\,du}{u^2+1}\\[5pt]
-	<math>\begin{align}	+	&= \frac{\sqrt{3}}{3}\int \frac{du}{u^2+1}\\[5pt]
-	& \frac{1}{3}\int{\frac{\,dx}{\~~left~~( \~~frac~~{x-1}{\sqrt{3}} \~~right~~)^{2}+1}}=\left\{ \begin{~~matrix~~}	+	&= \frac{1}{\sqrt{3}}\arctan u + C\\[5pt]
-	u=~~\frac{~~x-1}{\sqrt{3}} \\	+	&= \frac{1}{\sqrt{3}}\arctan \frac{x-1}{\sqrt{3}} + C\,\textrm{.}
-	du=\~~frac{\,dx}{~~\sqrt{3}~~} \\~~	+	\end{align}</math>}}
-	\end{~~matrix~~} \right\} \\	+
-	& =\frac{1}{3}\int{\frac{\sqrt{3}\,du}{u^{2}+1}}=\frac{\sqrt{3}}{3}\int{\frac{\,du}{u^{2}+1}} \\	+
-	& =\frac{1}{\sqrt{3}}\arctan u+C \\	+
-	& =\frac{1}{\sqrt{3}}\arctan \frac{x-1}{\sqrt{3}}+C \\	+
-	\end{align}</math>	+

Ian um 12:38, 21. Okt. 2008

Ian — Tue, 21 Oct 2008 12:38:02 GMT

← Nächstältere Version		Version vom 12:38, 21. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	We could substitute
-	<~~center~~> ~~[[Image:2_2_4b~~-1(2)~~.gif]] </center>~~	+	<math>u=x-\text{1}</math>, but we would then still have the problem of the second term,
-	{{~~NAVCONTENT_STOP~~}}	+	<math>\text{3}</math>
-	{{~~NAVCONTENT_START~~}}	+	in the denominator. Instead, we take out a factor
-	<~~center~~> ~~[[Image:2_2_4b~~-2(2)~~.gif]]~~ </~~center~~>	+	<math>\text{3}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	from the denominator,
		+
		+
		+	<math>\begin{align}
		+	& \int{\frac{\,dx}{\left( x-1 \right)^{2}+3}}=\int{\frac{\,dx}{3\left( \frac{1}{3}\left( x-1 \right)^{2}+1 \right)}} \\
		+	& =\frac{1}{3}\int{\frac{\,dx}{\frac{1}{3}\left( x-1 \right)^{2}+1}} \\
		+	\end{align}</math>
		+
		+	and move a factor
		+	<math>\frac{1}{3}</math>
		+	into the square
		+	<math>\left( x-1 \right)^{2}</math>,
		+
		+
		+	<math>\frac{1}{3}\int{\frac{\,dx}{\frac{1}{3}\left( x-1 \right)^{2}+1}}=\frac{1}{3}\int{\frac{\,dx}{\left( \frac{x-1}{\sqrt{3}} \right)^{2}+1}}</math>
		+
		+	Now, we substitute
		+	<math>u=\frac{x-1}{\sqrt{3}}</math>
		+	and get rid of all the problems at once:
		+
		+
		+	<math>\begin{align}
		+	& \frac{1}{3}\int{\frac{\,dx}{\left( \frac{x-1}{\sqrt{3}} \right)^{2}+1}}=\left\{ \begin{matrix}
		+	u=\frac{x-1}{\sqrt{3}} \\
		+	du=\frac{\,dx}{\sqrt{3}} \\
		+	\end{matrix} \right\} \\
		+	& =\frac{1}{3}\int{\frac{\sqrt{3}\,du}{u^{2}+1}}=\frac{\sqrt{3}}{3}\int{\frac{\,du}{u^{2}+1}} \\
		+	& =\frac{1}{\sqrt{3}}\arctan u+C \\
		+	& =\frac{1}{\sqrt{3}}\arctan \frac{x-1}{\sqrt{3}}+C \\
		+	\end{align}</math>

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

Tekbot — Tue, 16 Sep 2008 14:45:21 GMT

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

← Nächstältere Version

Version vom 14:45, 16. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 08:02:24 GMT

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

← Nächstältere Version		Version vom 08:02, 21. Aug. 2008
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	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
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Oskar — Mon, 07 Apr 2008 09:25:07 GMT

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