Lösung 2.2:2b - Versionsgeschichte

Sekretariat1 um 12:21, 27. Aug. 2009

2009-08-27T12:21:27Z

← Nächstältere Version		Version vom 12:21, 27. Aug. 2009
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-	Mit der Substitution <math>u=2x+3</math>, erhalten wir das Intagral <math>e^u</math>. Wir müssen aber auch den Faktor <math>dx</math> berücksichtigen. In diesem Fall erhalten wir,	+	Mit der Substitution <math>u=2x+3</math> erhalten wir das Intagral <math>e^u</math>. Wir müssen aber auch den Faktor <math>dx</math> berücksichtigen. In diesem Fall erhalten wir

-	{{Abgesetzte Formel\|\|<math>du = (2x+3)'\,dx = 2\,dx</math>}}	+	{{Abgesetzte Formel\|\|<math>du = (2x+3)'\,dx = 2\,dx</math>.}}

-	Dies ändert den Integrand nur mit einer Konstante, ~~und~~ also erhalten wir	+	Dies ändert den Integrand nur mit einer Konstante, also erhalten wir

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

Martin um 13:06, 5. Mai 2009

2009-05-05T13:06:46Z

← Nächstältere Version		Version vom 13:06, 5. Mai 2009
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-	~~If we set~~ <math>u=2x+3</math>, ~~the integral simplifies to~~ <math>e^u</math>. ~~However, this is only part of the truth. We must in addition take account of the relation between the integration element~~ <math>dx</math> ~~and <math>du</math>, which can give undesired effects~~. In ~~this case, however~~, ~~we have~~	+	Mit der Substitution <math>u=2x+3</math>, erhalten wir das Intagral <math>e^u</math>. Wir müssen aber auch den Faktor <math>dx</math> berücksichtigen. In diesem Fall erhalten wir,

	{{Abgesetzte Formel\|\|<math>du = (2x+3)'\,dx = 2\,dx</math>}}		{{Abgesetzte Formel\|\|<math>du = (2x+3)'\,dx = 2\,dx</math>}}

-	~~which only affects by a constant factor, so the substitution <math>u = 2x+3</math> seems to work, in spite of everything~~,	+	Dies ändert den Integrand nur mit einer Konstante, und also erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 13:		Zeile 13:
	\end{align}</math>}}		\end{align}</math>}}

-		+	Hinweis: Eine weitere mögliche Substitution ist <math>u=e^{2x+3}</math>. Normalerweise funktioniert es aber nicht mit so "großen" Substitutionen.
-	~~Note~~: ~~Another possible substitution is~~ <math>u=e^{2x+3}</math> ~~which also happens to work (usually, such an extensive substitution almost always fails)~~.	+

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

2009-03-11T10:21:35Z

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

← Nächstältere Version

Version vom 10:21, 11. Mär. 2009

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

2009-03-10T13:01:14Z

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

← Nächstältere Version		Version vom 13:01, 10. Mär. 2009
Zeile 1:		Zeile 1:
	If we set <math>u=2x+3</math>, the integral simplifies to <math>e^u</math>. However, this is only part of the truth. We must in addition take account of the relation between the integration element <math>dx</math> and <math>du</math>, which can give undesired effects. In this case, however, we have		If we set <math>u=2x+3</math>, the integral simplifies to <math>e^u</math>. However, this is only part of the truth. We must in addition take account of the relation between the integration element <math>dx</math> and <math>du</math>, which can give undesired effects. In this case, however, we have

-	{{~~Displayed math~~\|\|<math>du = (2x+3)'\,dx = 2\,dx</math>}}	+	{{Abgesetzte Formel\|\|<math>du = (2x+3)'\,dx = 2\,dx</math>}}

	which only affects by a constant factor, so the substitution <math>u = 2x+3</math> seems to work, in spite of everything,		which only affects by a constant factor, so the substitution <math>u = 2x+3</math> seems to work, in spite of everything,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int\limits_0^{1/2} e^{2x+3}\,dx &= \left\{\begin{align}		\int\limits_0^{1/2} e^{2x+3}\,dx &= \left\{\begin{align}
	u &= 2x+3\\[5pt]		u &= 2x+3\\[5pt]

Tek um 13:28, 28. Okt. 2008

2008-10-28T13:28:20Z

← Nächstältere Version		Version vom 13:28, 28. Okt. 2008
Zeile 1:		Zeile 1:
-	If we set	+	If we set <math>u=2x+3</math>, the integral simplifies to <math>e^u</math>. However, this is only part of the truth. We must in addition take account of the relation between the integration element <math>dx</math> and <math>du</math>, which can give undesired effects. In this case, however, we have
-	<math>u=~~\text{2}x~~+~~\text{~~3}</math>	+
-	, the integral simplifies to	+
-	<math>e^{u}</math>. However, this is only part of the truth. We must in addition take account of the relation between the integration element	+
-	<math>dx~~\text{ }~~</math>	+
-	and	+
-	<math>du</math>, which can give undesired effects. In this case, however, we have	+

		+	{{Displayed math\|\|<math>du = (2x+3)'\,dx = 2\,dx</math>}}

-	<math>du=~~\left( \text{2}x~~+~~\text{~~3~~} \right)^{\prime }\,dx=2\,dx~~</math>	+	which only affects by a constant factor, so the substitution <math>u = 2x+3</math> seems to work, in spite of everything,

		+	{{Displayed math\|\|<math>\begin{align}
		+	\int\limits_0^{1/2} e^{2x+3}\,dx &= \left\{\begin{align}
		+	u &= 2x+3\\[5pt]
		+	du &= 2\,dx
		+	\end{align}\right\} = \frac{1}{2}\int\limits_3^4 e^u\,du\\[5pt]
		+	&= \frac{1}{2}\Bigl[\ e^u\ \Bigr]_3^4 = \frac{1}{2}\bigl(e^4-e^3\bigr)\,\textrm{.}
		+	\end{align}</math>}}

-	which only affects by a constant factor, so the substitution
-	<math>u=\text{2}x+\text{3}</math>
-	seems to work, in spite of everything:

-		+	Note: Another possible substitution is <math>u=e^{2x+3}</math> which also happens to work (usually, such an extensive substitution almost always fails).
-	~~<math>\begin{align}~~	+
-	~~& \int\limits_{0}^{{1}/{2}\;}{e^{\text{2}x+\text{3}}}\,dx=\left\{ \begin{matrix}~~	+
-	~~u=\text{2}x+\text{3} \\~~	+
-	~~du=2\,dx \\~~	+
-	~~\end{matrix} \right\}=\frac{1}{2}\int\limits_{3}^{4}{e^{u}\,du} \\~~	+
-	~~& =\frac{1}{2}\left[ e^{u} \right]_{3}^{4}=\frac{1}{2}\left( e^{4}-e^{3} \right) \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	~~NOTE~~: Another possible substitution is	+
-	<math>u=e^{2x+3}</math>	+
-	which also happens to work (usually, such an extensive substitution almost always fails).	+

Ian um 12:36, 20. Okt. 2008

2008-10-20T12:36:51Z

← Nächstältere Version		Version vom 12:36, 20. Okt. 2008
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-	{{~~NAVCONTENT_START~~}}	+	If we set
-	<~~center~~> ~~[[Image:2_2_2b~~.~~gif]]~~ </~~center~~>	+	<math>u=\text{2}x+\text{3}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	, the integral simplifies to
		+	<math>e^{u}</math>. However, this is only part of the truth. We must in addition take account of the relation between the integration element
		+	<math>dx\text{ }</math>
		+	and
		+	<math>du</math>, which can give undesired effects. In this case, however, we have
		+
		+
		+	<math>du=\left( \text{2}x+\text{3} \right)^{\prime }\,dx=2\,dx</math>
		+
		+
		+	which only affects by a constant factor, so the substitution
		+	<math>u=\text{2}x+\text{3}</math>
		+	seems to work, in spite of everything:
		+
		+
		+	<math>\begin{align}
		+	& \int\limits_{0}^{{1}/{2}\;}{e^{\text{2}x+\text{3}}}\,dx=\left\{ \begin{matrix}
		+	u=\text{2}x+\text{3} \\
		+	du=2\,dx \\
		+	\end{matrix} \right\}=\frac{1}{2}\int\limits_{3}^{4}{e^{u}\,du} \\
		+	& =\frac{1}{2}\left[ e^{u} \right]_{3}^{4}=\frac{1}{2}\left( e^{4}-e^{3} \right) \\
		+	\end{align}</math>
		+
		+
		+	NOTE: Another possible substitution is
		+	<math>u=e^{2x+3}</math>
		+	which also happens to work (usually, such an extensive substitution almost always fails).

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

2008-09-16T14:42:01Z

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

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Version vom 14:42, 16. Sep. 2008

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

2008-08-21T08:00:44Z

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

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2008-04-07T09:32:11Z

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