Lösung 2.2:4c - Versionsgeschichte

Sekretariat1 um 13:03, 27. Aug. 2009

Sekretariat1 — Thu, 27 Aug 2009 13:03:37 GMT

← Nächstältere Version		Version vom 13:03, 27. Aug. 2009
Zeile 3:		Zeile 3:
	{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}</math>}}		{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}</math>}}

-	Jetzt machen wir weiter wie in der vorigen Übung. Wir ziehen den Faktor 4 aus dem Nenner.	+	Jetzt machen wir weiter wie in der vorigen Übung. Wir ziehen den Faktor 4 aus dem Nenner

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

Sekretariat1 um 13:03, 27. Aug. 2009

Sekretariat1 — Thu, 27 Aug 2009 13:03:07 GMT

← Nächstältere Version		Version vom 13:03, 27. Aug. 2009
Zeile 1:		Zeile 1:
-	Wir führen quadratische Ergänzung im Nenner aus,	+	Wir führen die quadratische Ergänzung im Nenner aus.

-	{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4~~}\,\textrm{.~~}</math>}}	+	{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}</math>}}

-	Jetzt machen wir weiter wie in der vorigen Übung. Wir ziehen den Faktor 4 aus dem Nenner,	+	Jetzt machen wir weiter wie in der vorigen Übung. Wir ziehen den Faktor 4 aus dem Nenner.

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

Moni um 10:08, 22. Aug. 2009

Moni — Sat, 22 Aug 2009 10:08:47 GMT

← Nächstältere Version		Version vom 10:08, 22. Aug. 2009
Zeile 3:		Zeile 3:
	{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}</math>}}

-	Jetzt machen wir weiter wie in der vorigen Übung. Wir ziehen den Faktor 4 aus ~~den nenner~~,	+	Jetzt machen wir weiter wie in der vorigen Übung. Wir ziehen den Faktor 4 aus dem Nenner,

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

-	und ~~schreien~~ den quadratischen Term wie	+	und schreiben den quadratischen Term wie

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

Martin um 13:47, 5. Mai 2009

Martin — Tue, 05 May 2009 13:47:34 GMT

← Nächstältere Version		Version vom 13:47, 5. Mai 2009
Zeile 1:		Zeile 1:
-	~~The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b~~,	+	Wir führen quadratische Ergänzung im Nenner aus,

	{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}</math>}}

-	~~We take out a factor~~ 4 ~~from the denominator~~	+	Jetzt machen wir weiter wie in der vorigen Übung. Wir ziehen den Faktor 4 aus den nenner,

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

-	~~and rewrite the quadratic term as~~	+	und schreien den quadratischen Term wie

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

-	~~If we now substitute~~ <math>u = (x+2)/2</math>, ~~we obtain the integral in the exercise~~	+	Machen wir die Substitution <math>u = (x+2)/2</math>, erhalten wir das erwünschte Integral

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

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

Tekbot — Wed, 11 Mar 2009 10:25:15 GMT

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

← Nächstältere Version

Version vom 10:25, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 13:03:04 GMT

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

← Nächstältere Version		Version vom 13:03, 10. Mär. 2009
Zeile 1:		Zeile 1:
	The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,		The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,

-	{{~~Displayed math~~\|\|<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}</math>}}

	We take out a factor 4 from the denominator		We take out a factor 4 from the denominator

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

	and rewrite the quadratic term as		and rewrite the quadratic term as

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

	If we now substitute <math>u = (x+2)/2</math>, we obtain the integral in the exercise		If we now substitute <math>u = (x+2)/2</math>, we obtain the integral in the exercise

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

Tek um 15:27, 28. Okt. 2008

Tek — Tue, 28 Oct 2008 15:27:01 GMT

← Nächstältere Version		Version vom 15:27, 28. Okt. 2008
Zeile 1:		Zeile 1:
	The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,		The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,

		+	{{Displayed math\|\|<math>\int \frac{dx}{x^2+4x+8} = \int \frac{dx}{(x+2)^2-2^2+8} = \int \frac{dx}{(x+2)^2+4}\,\textrm{.}</math>}}

-	~~<math>\int{\frac{\,dx}{x^{2}+4x+8}}=\int{\frac{\,dx}{\left( x+2 \right)^{2}-2^{2}+8}}=\int{\frac{\,dx}{\left( x+2 \right)^{2}+4}}</math>~~	+	We take out a factor 4 from the denominator
-		+
-		+
-	We take out a factor	+
-	~~<math>\text{~~4~~}</math>~~	+
-	from the denominator	+
-		+
-		+
-	~~<math>\int{\frac{\,dx}{\left( x+2 \right)^{2}+4}}=\int{\frac{\,dx}{4\left( \frac{1}{4}\left( x+2 \right)^{2}+1 \right)}}=\frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}\left( x+2 \right)^{2}+1}}</math>~~	+

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

	and rewrite the quadratic term as		and rewrite the quadratic term as

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

-	~~<math>\frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}\left( x+2 \right)^{2}+1}}=\frac{1}{4}\int{\frac{\,dx}{\left( \frac{x+2}{2} \right)^{2}+1}}</math>~~	+	If we now substitute <math>u = (x+2)/2</math>, we obtain the integral in the exercise
-		+
-		+
-	If we now substitute	+
-	<math>u=~~\frac{~~x+2}{2}</math>, we obtain the integral in the exercise	+
-		+

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& \frac{1}{4}\int{\frac{\,dx}{\~~left~~( \~~frac~~{x+2}{2} \~~right~~)^{2}+1}}=\left\{ \begin{~~matrix~~}	+	\frac{1}{4}\int \frac{dx}{\Bigl(\dfrac{x+2}{2}\Bigr)^2+1}
-	u=~~\frac{~~x+2}{2} \\	+	&= \left\{\begin{align}
-	du=~~\frac{\,~~dx}{2~~} \\~~	+	u &= (x+2)/2\\[5pt]
-	\end{~~matrix~~} \right\} \\	+	du &= dx/2
-	& =\frac{1}{4}\int{\frac{\,~~2dx~~}{u^{2}+1}}=\frac{1}{2}\int{\frac{~~\,dx~~}{u^{2}+1}} \\	+	\end{align}\right\}\\[5pt]
-	& =\frac{1}{2}\arctan u+C \\	+	&= \frac{1}{4}\int \frac{2\,du}{u^2+1}\\[5pt]
-	& =\frac{1}{2}\arctan \frac{x+2}{2}+C \\	+	&= \frac{1}{2}\int \frac{du}{u^2+1}\\[5pt]
-	\end{align}</math>	+	&= \frac{1}{2}\arctan u + C\\[5pt]
		+	&= \frac{1}{2}\arctan \frac{x+2}{2} + C\,\textrm{.}
		+	\end{align}</math>}}

Ian um 12:51, 21. Okt. 2008

Ian — Tue, 21 Oct 2008 12:51:57 GMT

← Nächstältere Version		Version vom 12:51, 21. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	The trick is to complete the square in the denominator so that we obtain the same expression as in exercise b,
-	<~~center~~> ~~[[Image:2_2_4c-~~1(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>\int{\frac{\,dx}{x^{2}+4x+8}}=\int{\frac{\,dx}{\left( x+2 \right)^{2}-2^{2}+8}}=\int{\frac{\,dx}{\left( x+2 \right)^{2}+4}}</math>
-	<~~center~~> ~~[[Image:2_2_4c-~~2(2~~).gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	We take out a factor
		+	<math>\text{4}</math>
		+	from the denominator
		+
		+
		+	<math>\int{\frac{\,dx}{\left( x+2 \right)^{2}+4}}=\int{\frac{\,dx}{4\left( \frac{1}{4}\left( x+2 \right)^{2}+1 \right)}}=\frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}\left( x+2 \right)^{2}+1}}</math>
		+
		+
		+	and rewrite the quadratic term as
		+
		+
		+	<math>\frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}\left( x+2 \right)^{2}+1}}=\frac{1}{4}\int{\frac{\,dx}{\left( \frac{x+2}{2} \right)^{2}+1}}</math>
		+
		+
		+	If we now substitute
		+	<math>u=\frac{x+2}{2}</math>, we obtain the integral in the exercise
		+
		+
		+	<math>\begin{align}
		+	& \frac{1}{4}\int{\frac{\,dx}{\left( \frac{x+2}{2} \right)^{2}+1}}=\left\{ \begin{matrix}
		+	u=\frac{x+2}{2} \\
		+	du=\frac{\,dx}{2} \\
		+	\end{matrix} \right\} \\
		+	& =\frac{1}{4}\int{\frac{\,2dx}{u^{2}+1}}=\frac{1}{2}\int{\frac{\,dx}{u^{2}+1}} \\
		+	& =\frac{1}{2}\arctan u+C \\
		+	& =\frac{1}{2}\arctan \frac{x+2}{2}+C \\
		+	\end{align}</math>

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

Tekbot — Tue, 16 Sep 2008 14:45:41 GMT

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

← Nächstältere Version

Version vom 14:45, 16. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 08:02:34 GMT

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

← Nächstältere Version		Version vom 08:02, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_2_4c-1(2).gif]] </center>	+	<center> [[Image:2_2_4c-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_2_4c-2(2).gif]] </center>	+	<center> [[Image:2_2_4c-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}