Lösung 2.2:2c - Versionsgeschichte

Sekretariat1 um 12:23, 27. Aug. 2009

2009-08-27T12:23:20Z

← Nächstältere Version		Version vom 12:23, 27. Aug. 2009
Zeile 1:		Zeile 1:
-	Die Substitution <math>u=3x+1</math> ergibt einen einfacheren Integrand. ~~Nachdem~~ <math>u=3x+1</math> eine lineare Funktion ist, ist das Verhältnis zwischen <math>dx</math> und <math>du</math> nur eine Konstante	+	Die Substitution <math>u=3x+1</math> ergibt einen einfacheren Integrand. Da <math>u=3x+1</math> eine lineare Funktion ist, ist das Verhältnis zwischen <math>dx</math> und <math>du</math> nur eine Konstante.

-	{{Abgesetzte Formel\|\|<math>du = (3x+1)'\,dx = 3\,dx.</math>}}	+	{{Abgesetzte Formel\|\|<math>du = (3x+1)'\,dx = 3\,dx</math>}}

-	Wir erhalten,	+	Wir erhalten

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

Moni um 09:44, 22. Aug. 2009

2009-08-22T09:44:49Z

← Nächstältere Version		Version vom 09:44, 22. Aug. 2009
Zeile 1:		Zeile 1:
-	Die Substitution <math>u=3x+1</math> ~~würde~~ einen einfacheren Integrand ~~ergeben~~. Nachdem <math>u=3x+1</math> eine lineare Funktion ist, ist das Verhältnis zwischen <math>dx</math> und <math>du</math> nur eine Konstante.	+	Die Substitution <math>u=3x+1</math> ergibt einen einfacheren Integrand. Nachdem <math>u=3x+1</math> eine lineare Funktion ist, ist das Verhältnis zwischen <math>dx</math> und <math>du</math> nur eine Konstante

-	{{Abgesetzte Formel\|\|<math>du = (3x+1)'\,dx = 3\,dx~~\,,~~</math>}}	+	{{Abgesetzte Formel\|\|<math>du = (3x+1)'\,dx = 3\,dx.</math>}}

	Wir erhalten,		Wir erhalten,

Martin um 13:13, 5. Mai 2009

2009-05-05T13:13:38Z

← Nächstältere Version		Version vom 13:13, 5. Mai 2009
Zeile 1:		Zeile 1:
-	~~If we focus on the integrand, then the substitution~~ <math>u=3x+1</math> ~~seems suitable, since we then get <math>\sqrt{u}</math> which we can integrate~~. ~~There is also no risk involved in using a linear substitution such as~~ <math>u=3x+1</math>, ~~because the relation between~~ <math>dx</math> ~~and~~ <math>du</math> ~~will be a constant factor,~~	+	Die Substitution <math>u=3x+1</math> würde einen einfacheren Integrand ergeben. Nachdem <math>u=3x+1</math> eine lineare Funktion ist, ist das Verhältnis zwischen <math>dx</math> und <math>du</math> nur eine Konstante.

	{{Abgesetzte Formel\|\|<math>du = (3x+1)'\,dx = 3\,dx\,,</math>}}		{{Abgesetzte Formel\|\|<math>du = (3x+1)'\,dx = 3\,dx\,,</math>}}

-	~~which does not cause any problems.~~	+	Wir erhalten,
-		+
-	~~We obtain~~	+

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

Tekbot: Solution 2.2:2c moved to Lösung 2.2:2c: Robot: moved page

2009-03-11T10:21:55Z

Solution 2.2:2c moved to Lösung 2.2:2c: Robot: moved page

← Nächstältere Version

Version vom 10:21, 11. Mär. 2009

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

2009-03-10T13:01:24Z

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

← Nächstältere Version		Version vom 13:01, 10. Mär. 2009
Zeile 1:		Zeile 1:
	If we focus on the integrand, then the substitution <math>u=3x+1</math> seems suitable, since we then get <math>\sqrt{u}</math> which we can integrate. There is also no risk involved in using a linear substitution such as <math>u=3x+1</math>, because the relation between <math>dx</math> and <math>du</math> will be a constant factor,		If we focus on the integrand, then the substitution <math>u=3x+1</math> seems suitable, since we then get <math>\sqrt{u}</math> which we can integrate. There is also no risk involved in using a linear substitution such as <math>u=3x+1</math>, because the relation between <math>dx</math> and <math>du</math> will be a constant factor,

-	{{~~Displayed math~~\|\|<math>du = (3x+1)'\,dx = 3\,dx\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>du = (3x+1)'\,dx = 3\,dx\,,</math>}}

	which does not cause any problems.		which does not cause any problems.
Zeile 7:		Zeile 7:
	We obtain		We obtain

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int\limits_0^5 \sqrt{3x+1}\,dx		\int\limits_0^5 \sqrt{3x+1}\,dx
	&= \left\{\begin{align}		&= \left\{\begin{align}

Tek um 13:36, 28. Okt. 2008

2008-10-28T13:36:17Z

← Nächstältere Version		Version vom 13:36, 28. Okt. 2008
Zeile 1:		Zeile 1:
-	If we focus on the integrand, then the substitution	+	If we focus on the integrand, then the substitution <math>u=3x+1</math> seems suitable, since we then get <math>\sqrt{u}</math> which we can integrate. There is also no risk involved in using a linear substitution such as <math>u=3x+1</math>, because the relation between <math>dx</math> and <math>du</math> will be a constant factor,
-	<math>u=~~\text{3}x~~+~~\text{~~1}</math>	+
-	seems suitable, since we then get	+
-	<math>\sqrt{u}</math>	+
-	which we can integrate. There is also no risk involved in using a linear substitution such as	+
-	<math>u=~~\text{3}x~~+~~\text{~~1}</math>, because the relation between	+
-	<math>dx~~\text{ }~~</math>	+
-	and	+
-	<math>du</math>	+
-	will be a constant factor,	+
-		+
-		+
-	~~<math>du=\left( \text{3}x+\text{1} \right)^{\prime }\,dx=3\,dx</math>~~	+

		+	{{Displayed math\|\|<math>du = (3x+1)'\,dx = 3\,dx\,,</math>}}

	which does not cause any problems.		which does not cause any problems.
Zeile 18:		Zeile 7:
	We obtain		We obtain

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\int\limits_0^5 \sqrt{3x+1}\,dx
-	& \int\~~limits_{0}~~^{5}{\sqrt{3x+1}}\,dx=\left\{ \begin{~~matrix~~}	+	&= \left\{\begin{align}
-	u=~~\text{3}x~~+~~\text{~~1} \\	+	u &= 3x+1\\[5pt]
-	du=3\,dx \\	+	du &= 3\,dx
-	\end{~~matrix~~} \right\}=\frac{1}{3}\int\~~limits_{1}~~^{16}{\sqrt{u}}\,du \\	+	\end{align}\right\} = \frac{1}{3}\int\limits_1^{16} \sqrt{u}\,du\\[5pt]
-	& =\frac{1}{3}\int\~~limits_{1}~~^{16}{u^{{1}/{2~~}\;}~~}\,du=\frac{1}{3}\~~left~~[ \frac{u^~~{\frac~~{1}{2}+1}}{\~~frac~~{1}{2}+1} \~~right~~]~~_{1}~~^{16} \\	+	&= \frac{1}{3}\int\limits_1^{16} u^{1/2}\,du = \frac{1}{3}\biggl[\ \frac{u^{1/2+1}}{\tfrac{1}{2}+1}\ \biggr]_1^{16}\\[5pt]
-	& =\frac{1}{3}\~~left~~[ \frac{2}{3}u\sqrt{u} \~~right~~]~~_{1}~~^{16}=\frac{2}{9}\~~left~~( 16\sqrt{16}-1\sqrt{1} \~~right~~) \\	+	&= \frac{1}{3}\Bigl[\ \frac{2}{3}u\sqrt{u}\ \Bigr]_1^{16} = \frac{2}{9}\bigl( 16\sqrt{16}-1\sqrt{1} \bigr)\\[8pt]
-	& =\frac{2}{9}\~~left~~( 16\~~centerdot~~ 4-1 \~~right~~)=\frac{2\~~centerdot~~ 63}{9}=14 \\	+	&= \frac{2}{9}\bigl( 16\cdot 4-1 \bigr) = \frac{2\cdot 63}{9} = 14\,\textrm{.}
-	\end{align}</math>	+	\end{align}</math>}}

Ian um 12:39, 20. Okt. 2008

2008-10-20T12:39:12Z

← Nächstältere Version		Version vom 12:39, 20. Okt. 2008
Zeile 11:		Zeile 11:


-	<math>du=\left( \text{3}x+\text{1} \right)^{1}\,dx=3\,dx</math>	+	<math>du=\left( \text{3}x+\text{1} \right)^{\prime }\,dx=3\,dx</math>

Ian um 12:10, 20. Okt. 2008

2008-10-20T12:10:28Z

← Nächstältere Version		Version vom 12:10, 20. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START}~~}	+	If we focus on the integrand, then the substitution
-	<~~center~~> ~~[[Image:2_2_2c~~.~~gif]]~~ </~~center~~>	+	<math>u=\text{3}x+\text{1}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	seems suitable, since we then get
		+	<math>\sqrt{u}</math>
		+	which we can integrate. There is also no risk involved in using a linear substitution such as
		+	<math>u=\text{3}x+\text{1}</math>, because the relation between
		+	<math>dx\text{ }</math>
		+	and
		+	<math>du</math>
		+	will be a constant factor,
		+
		+
		+	<math>du=\left( \text{3}x+\text{1} \right)^{1}\,dx=3\,dx</math>
		+
		+
		+	which does not cause any problems.
		+
		+	We obtain
		+
		+
		+	<math>\begin{align}
		+	& \int\limits_{0}^{5}{\sqrt{3x+1}}\,dx=\left\{ \begin{matrix}
		+	u=\text{3}x+\text{1} \\
		+	du=3\,dx \\
		+	\end{matrix} \right\}=\frac{1}{3}\int\limits_{1}^{16}{\sqrt{u}}\,du \\
		+	& =\frac{1}{3}\int\limits_{1}^{16}{u^{{1}/{2}\;}}\,du=\frac{1}{3}\left[ \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right]_{1}^{16} \\
		+	& =\frac{1}{3}\left[ \frac{2}{3}u\sqrt{u} \right]_{1}^{16}=\frac{2}{9}\left( 16\sqrt{16}-1\sqrt{1} \right) \\
		+	& =\frac{2}{9}\left( 16\centerdot 4-1 \right)=\frac{2\centerdot 63}{9}=14 \\
		+	\end{align}</math>

Tekbot: Lösning 2.2:2c moved to Solution 2.2:2c: Robot: moved page

2008-09-16T14:42:21Z

Lösning 2.2:2c moved to Solution 2.2:2c: Robot: moved page

← Nächstältere Version

Version vom 14:42, 16. Sep. 2008

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

2008-08-21T08:00:54Z

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

← Nächstältere Version		Version vom 08:00, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_2_2c.gif]] </center>	+	<center> [[Image:2_2_2c.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}