Lösung 2.3:2d - Versionsgeschichte

Sekretariat1 um 11:34, 28. Aug. 2009

2009-08-28T11:34:27Z

← Nächstältere Version		Version vom 11:34, 28. Aug. 2009
Zeile 8:		Zeile 8:
	{{Abgesetzte Formel\|\|<math>1\cdot \ln x</math>}}		{{Abgesetzte Formel\|\|<math>1\cdot \ln x</math>}}

-	betrachten, ~~den~~ 1~~:er~~ integrieren und <math>\ln x</math> ableiten.	+	betrachten, <math>1</math> integrieren und <math>\ln x</math> ableiten.

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

Sekretariat1 um 08:20, 28. Aug. 2009

2009-08-28T08:20:57Z

← Nächstältere Version		Version vom 08:20, 28. Aug. 2009
Zeile 39:		Zeile 39:
	\end{align}</math>}}		\end{align}</math>}}

-	Dieses Integral berechnen wir durch partielle ~~Intagration~~	+	Dieses Integral berechnen wir durch partielle Integration

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

Sekretariat1 um 08:20, 28. Aug. 2009

2009-08-28T08:20:36Z

← Nächstältere Version		Version vom 08:20, 28. Aug. 2009
Zeile 1:		Zeile 1:
-	Wir werden das Problem mit zwei verschiedenen ~~Methode~~ lösen.	+	Wir werden das Problem mit zwei verschiedenen Methoden lösen.


	'''Methode 1''' (partielle Integration)		'''Methode 1''' (partielle Integration)

-	Am ersten ~~Anblich~~ scheint es unmöglich partielle Integration ~~auszuführen~~. Der Trick ist, dass wir den Integrand als ~~den~~ Produkt	+	Beim ersten Anblick scheint es unmöglich partielle Integration anzuwenden. Der Trick ist, dass wir den Integrand als das Produkt

-	{{Abgesetzte Formel\|\|<math>1\cdot \ln x~~\,\textrm{.}~~</math>}}	+	{{Abgesetzte Formel\|\|<math>1\cdot \ln x</math>}}

-	betrachten, den 1:er integrieren und <math>\ln x</math> ableiten,	+	betrachten, den 1:er integrieren und <math>\ln x</math> ableiten.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 14:		Zeile 14:
	&= x\cdot\ln x - \int x\cdot\frac{1}{x}\,dx\\[5pt]		&= x\cdot\ln x - \int x\cdot\frac{1}{x}\,dx\\[5pt]
	&= x\cdot\ln x - \int 1\,dx\\[5pt]		&= x\cdot\ln x - \int 1\,dx\\[5pt]
-	&= x\cdot\ln x - x + C~~\,\textrm{.}~~	+	&= x\cdot\ln x - x + C
	\end{align}</math>}}		\end{align}</math>}}

Zeile 24:		Zeile 24:
	{{Abgesetzte Formel\|\|<math>du = (\ln x)'\,dx = \frac{1}{x}\,dx</math>}}		{{Abgesetzte Formel\|\|<math>du = (\ln x)'\,dx = \frac{1}{x}\,dx</math>}}

-	und ~~nachdem~~ <math>u = \ln x</math>, ist <math>x=e^u</math> und dadurch erhalten wir	+	und da <math>u = \ln x</math>, ist <math>x=e^u</math> und dadurch erhalten wir

	{{Abgesetzte Formel\|\|<math>du = \frac{1}{e^u}\,dx\quad\Leftrightarrow\quad dx = e^u\,du\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>du = \frac{1}{e^u}\,dx\quad\Leftrightarrow\quad dx = e^u\,du\,\textrm{.}</math>}}
Zeile 39:		Zeile 39:
	\end{align}</math>}}		\end{align}</math>}}

-	Dieses Integral berechnen wir durch partielle Intagration,	+	Dieses Integral berechnen wir durch partielle Intagration

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 46:		Zeile 46:
	&= ue^u - \int e^u\,du\\[5pt]		&= ue^u - \int e^u\,du\\[5pt]
	&= ue^u - e^u + C\\[5pt]		&= ue^u - e^u + C\\[5pt]
-	&= (u-1)e^u + C~~\,,~~	+	&= (u-1)e^u + C
	\end{align}</math>}}		\end{align}</math>}}

Moni um 11:22, 22. Aug. 2009

2009-08-22T11:22:38Z

← Nächstältere Version		Version vom 11:22, 22. Aug. 2009
Zeile 4:		Zeile 4:
	'''Methode 1''' (partielle Integration)		'''Methode 1''' (partielle Integration)

-	Am ersten Anblich scheint es unmöglich partielle Integration auszuführen. Der Trick ist dass wir den Integrand als den Produkt	+	Am ersten Anblich scheint es unmöglich partielle Integration auszuführen. Der Trick ist, dass wir den Integrand als den Produkt

-	{{Abgesetzte Formel\|\|<math>1\~~centerdot~~ \ln x\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>1\cdot \ln x\,\textrm{.}</math>}}

-	betrachten, ~~und~~ den 1:er integrieren, und <math>\ln x</math> ableiten,	+	betrachten, den 1:er integrieren und <math>\ln x</math> ableiten,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 24:		Zeile 24:
	{{Abgesetzte Formel\|\|<math>du = (\ln x)'\,dx = \frac{1}{x}\,dx</math>}}		{{Abgesetzte Formel\|\|<math>du = (\ln x)'\,dx = \frac{1}{x}\,dx</math>}}

-	und nachdem <math>u = \ln x</math>, ist <math>x=e^u</math>und dadurch erhalten wir	+	und nachdem <math>u = \ln x</math>, ist <math>x=e^u</math> und dadurch erhalten wir

	{{Abgesetzte Formel\|\|<math>du = \frac{1}{e^u}\,dx\quad\Leftrightarrow\quad dx = e^u\,du\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>du = \frac{1}{e^u}\,dx\quad\Leftrightarrow\quad dx = e^u\,du\,\textrm{.}</math>}}

Martin um 18:19, 5. Mai 2009

2009-05-05T18:19:54Z

← Nächstältere Version		Version vom 18:19, 5. Mai 2009
Zeile 1:		Zeile 1:
-	~~We shall solve the exercise in two different ways~~.	+	Wir werden das Problem mit zwei verschiedenen Methode lösen.


-	'''~~Method~~ 1''' (~~integration by parts~~)	+	'''Methode 1''' (partielle Integration)

-	~~At first sight, integration by parts seems impossible, but the trick is to see the integrand as the product~~	+	Am ersten Anblich scheint es unmöglich partielle Integration auszuführen. Der Trick ist dass wir den Integrand als den Produkt

	{{Abgesetzte Formel\|\|<math>1\centerdot \ln x\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>1\centerdot \ln x\,\textrm{.}</math>}}

-	~~We integrate the factor <math>~~1~~</math> and differentiate~~ <math>\ln x</math>,	+	betrachten, und den 1:er integrieren, und <math>\ln x</math> ableiten,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 18:		Zeile 18:


-	'''~~Method~~ 2''' (~~substitution~~)	+	'''Methode 2''' (Substitution und partielle Integration)

-	~~It seems difficult to find some suitable expression to substitute, so we try to substitute the whole expression~~ <math>u=\ln x\,</math>. ~~The problem we encounter is how we should handle the change from <math>dx</math> to <math>du</math>. With this substitution, the relation between <math>dx</math> and <math>du</math> becomes~~	+	Wir substituieren <math>u=\ln x\,</math>. So erhalten wir das Verhältnis

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

-	~~and because~~ <math>u = \ln x</math>, ~~then~~ <math>x=e^u</math> ~~and we have that~~	+	und nachdem <math>u = \ln x</math>, ist <math>x=e^u</math>und dadurch erhalten wir

	{{Abgesetzte Formel\|\|<math>du = \frac{1}{e^u}\,dx\quad\Leftrightarrow\quad dx = e^u\,du\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>du = \frac{1}{e^u}\,dx\quad\Leftrightarrow\quad dx = e^u\,du\,\textrm{.}</math>}}

-	~~Thus, the substitution becomes~~	+	Also haben wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 39:		Zeile 39:
	\end{align}</math>}}		\end{align}</math>}}

-	~~Now, we carry out an integration by parts~~,	+	Dieses Integral berechnen wir durch partielle Intagration,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 49:		Zeile 49:
	\end{align}</math>}}		\end{align}</math>}}

-	~~and the answer becomes~~	+	und wir erhalten

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

Tekbot: Solution 2.3:2d moved to Lösung 2.3:2d: Robot: moved page

2009-03-11T10:28:15Z

Solution 2.3:2d moved to Lösung 2.3:2d: Robot: moved page

← Nächstältere Version

Version vom 10:28, 11. Mär. 2009

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

2009-03-10T13:04:34Z

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

← Nächstältere Version		Version vom 13:04, 10. Mär. 2009
Zeile 6:		Zeile 6:
	At first sight, integration by parts seems impossible, but the trick is to see the integrand as the product		At first sight, integration by parts seems impossible, but the trick is to see the integrand as the product

-	{{~~Displayed math~~\|\|<math>1\centerdot \ln x\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>1\centerdot \ln x\,\textrm{.}</math>}}

	We integrate the factor <math>1</math> and differentiate <math>\ln x</math>,		We integrate the factor <math>1</math> and differentiate <math>\ln x</math>,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int 1\cdot\ln x\,dx		\int 1\cdot\ln x\,dx
	&= x\cdot\ln x - \int x\cdot\frac{1}{x}\,dx\\[5pt]		&= x\cdot\ln x - \int x\cdot\frac{1}{x}\,dx\\[5pt]
Zeile 22:		Zeile 22:
	It seems difficult to find some suitable expression to substitute, so we try to substitute the whole expression <math>u=\ln x\,</math>. The problem we encounter is how we should handle the change from <math>dx</math> to <math>du</math>. With this substitution, the relation between <math>dx</math> and <math>du</math> becomes		It seems difficult to find some suitable expression to substitute, so we try to substitute the whole expression <math>u=\ln x\,</math>. The problem we encounter is how we should handle the change from <math>dx</math> to <math>du</math>. With this substitution, the relation between <math>dx</math> and <math>du</math> becomes

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

	and because <math>u = \ln x</math>, then <math>x=e^u</math> and we have that		and because <math>u = \ln x</math>, then <math>x=e^u</math> and we have that

-	{{~~Displayed math~~\|\|<math>du = \frac{1}{e^u}\,dx\quad\Leftrightarrow\quad dx = e^u\,du\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>du = \frac{1}{e^u}\,dx\quad\Leftrightarrow\quad dx = e^u\,du\,\textrm{.}</math>}}

	Thus, the substitution becomes		Thus, the substitution becomes

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int \ln x\,dx		\int \ln x\,dx
	= \left\{\begin{align}		= \left\{\begin{align}
Zeile 41:		Zeile 41:
	Now, we carry out an integration by parts,		Now, we carry out an integration by parts,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int u\cdot e^u\,du		\int u\cdot e^u\,du
	&= u\cdot e^u - \int 1\cdot e^u\,du\\[5pt]		&= u\cdot e^u - \int 1\cdot e^u\,du\\[5pt]
Zeile 51:		Zeile 51:
	and the answer becomes		and the answer becomes

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int \ln x\,dx		\int \ln x\,dx
	&= (\ln x-1)e^{\ln x} + C\\[5pt]		&= (\ln x-1)e^{\ln x} + C\\[5pt]
	&= (\ln x-1)x + C\,\textrm{.}		&= (\ln x-1)x + C\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Tek um 09:13, 29. Okt. 2008

2008-10-29T09:13:31Z

← Nächstältere Version		Version vom 09:13, 29. Okt. 2008
Zeile 1:		Zeile 1:
	We shall solve the exercise in two different ways.		We shall solve the exercise in two different ways.

-	Method 1 (partial integration)

-	~~As first sight, partial~~ integration ~~seems impossible, but the trick is to see the integrand as the product~~	+	'''Method 1''' (integration by parts)

		+	At first sight, integration by parts seems impossible, but the trick is to see the integrand as the product

-	<math>1\centerdot \ln x</math>	+	{{Displayed math\|\|<math>1\centerdot \ln x\,\textrm{.}</math>}}

		+	We integrate the factor <math>1</math> and differentiate <math>\ln x</math>,

-	~~We integrate the factor~~	+	{{Displayed math\|\|<math>\begin{align}
-	<math>\~~text~~{1}~~</math>~~	+	\int 1\cdot\ln x\,dx
-	~~and differentiate~~	+	&= x\cdot\ln x - \int x\cdot\frac{1}{x}\,dx\\[5pt]
-	~~<math>~~\ln x</math>,	+	&= x\cdot\ln x - \int 1\,dx\\[5pt]
		+	&= x\cdot\ln x - x + C\,\textrm{.}
		+	\end{align}</math>}}


-	~~<math>\begin{align}~~	+	'''Method 2''' (substitution)
-	~~& \int{1\centerdot \ln x\,dx}=x\centerdot \ln x-\int{x\centerdot \frac{1}{x}\,dx} \\~~	+
-	~~& =x\centerdot \ln x-\int{1\,dx} \\~~	+
-	~~& =x\centerdot \ln x-x+C \\~~	+
-	~~\end{align}</math>~~	+

		+	It seems difficult to find some suitable expression to substitute, so we try to substitute the whole expression <math>u=\ln x\,</math>. The problem we encounter is how we should handle the change from <math>dx</math> to <math>du</math>. With this substitution, the relation between <math>dx</math> and <math>du</math> becomes

-	~~Method 2~~ (~~substitution~~)	+	{{Displayed math\|\|<math>du = (\ln x)'\,dx = \frac{1}{x}\,dx</math>}}

-	~~It seems difficult to find some suitable expression to substitute, so we try to substitute the whole expression~~	+	and because <math>u = \ln x</math>, then <math>x=e^u</math> and we have that
-	~~<math>u=\text{ln }x</math>. The problem we encounter is how we should handle the change from~~	+
-	~~<math>dx</math>~~	+
-	to	+
-	~~<math>du</math>. With this substitution, the relation between~~	+
-	~~<math>dx</math>~~	+
-	~~and~~	+
-	~~<math>du</math>~~	+
-	~~becomes~~	+
-		+
-		+
-	~~<math>du=\left( \ln x \right)^{\prime }\,dx=\frac{1}{x}\,dx</math>~~	+
-		+
-		+
-	and because	+
-	<math>u=\~~text{~~ln }x</math>, then	+
-	<math>x=e^{u}</math>	+
-	and we have that	+
-		+
-		+
-	~~<math>du=\frac{1}{e^{u}\,}\,dx\quad \Leftrightarrow \quad dx=e^{u}\,du</math>~~	+

		+	{{Displayed math\|\|<math>du = \frac{1}{e^u}\,dx\quad\Leftrightarrow\quad dx = e^u\,du\,\textrm{.}</math>}}

	Thus, the substitution becomes		Thus, the substitution becomes

		+	{{Displayed math\|\|<math>\begin{align}
		+	\int \ln x\,dx
		+	= \left\{\begin{align}
		+	u &= \ln x\\[5pt]
		+	dx &= e^u\,du
		+	\end{align}\right\}
		+	= \int ue^u\,du\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	Now, we carry out an integration by parts,
-	~~& \int{\ln x\,dx}=\left\{ \begin{matrix}~~	+
-	~~u=\text{ln }x \\~~	+
-	~~dx=e^{u}\,du \\~~	+
-	~~\end{matrix} \right\} \\~~	+
-	~~& =\int{ue^{u}\,du} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	Now, we carry out ~~a partial~~ integration	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \int{ue^{u}\,du}=u\centerdot e^{u}-\int{1\centerdot e^{u}\,du} \\~~	+
-	~~& =u\centerdot e^{u}-\int{e^{u}\~~,~~du} \\~~	+
-	~~& =u\centerdot e^{u}-e^{u}+C \\~~	+
-	~~& =\left( u-1 \right)e^{u}+C \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\int u\cdot e^u\,du
		+	&= u\cdot e^u - \int 1\cdot e^u\,du\\[5pt]
		+	&= ue^u - \int e^u\,du\\[5pt]
		+	&= ue^u - e^u + C\\[5pt]
		+	&= (u-1)e^u + C\,,
		+	\end{align}</math>}}

	and the answer becomes		and the answer becomes

-		+	{{Displayed math\|\|<math>\begin{align}
-		+	\int \ln x\,dx
-	<math>\begin{align}	+	&= (\ln x-1)e^{\ln x} + C\\[5pt]
-	& \int{\ln x\,dx}=~~\left~~( \ln x-1 ~~\right~~)e^{\ln x}+C \\	+	&= (\ln x-1)x + C\,\textrm{.}
-	& =~~\left~~( \ln x-1 ~~\right~~)x+C \\	+	\end{align}</math>}}
-	\end{align}</math>	+

Ian um 14:32, 22. Okt. 2008

2008-10-22T14:32:11Z

← Nächstältere Version		Version vom 14:32, 22. Okt. 2008
Zeile 1:		Zeile 1:
-	{~~{NAVCONTENT_START}~~}	+	We shall solve the exercise in two different ways.
-	<~~center~~> ~~[[Image:2_3_2d~~-1(2).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	Method 1 (partial integration)
-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image:2_3_2d~~-2(2)~~.gif]]~~ </~~center~~>	+	As first sight, partial integration seems impossible, but the trick is to see the integrand as the product
-	{{~~NAVCONTENT_STOP~~}}	+
		+
		+	<math>1\centerdot \ln x</math>
		+
		+
		+	We integrate the factor
		+	<math>\text{1}</math>
		+	and differentiate
		+	<math>\ln x</math>,
		+
		+
		+	<math>\begin{align}
		+	& \int{1\centerdot \ln x\,dx}=x\centerdot \ln x-\int{x\centerdot \frac{1}{x}\,dx} \\
		+	& =x\centerdot \ln x-\int{1\,dx} \\
		+	& =x\centerdot \ln x-x+C \\
		+	\end{align}</math>
		+
		+
		+	Method 2 (substitution)
		+
		+	It seems difficult to find some suitable expression to substitute, so we try to substitute the whole expression
		+	<math>u=\text{ln }x</math>. The problem we encounter is how we should handle the change from
		+	<math>dx</math>
		+	to
		+	<math>du</math>. With this substitution, the relation between
		+	<math>dx</math>
		+	and
		+	<math>du</math>
		+	becomes
		+
		+
		+	<math>du=\left( \ln x \right)^{\prime }\,dx=\frac{1}{x}\,dx</math>
		+
		+
		+	and because
		+	<math>u=\text{ln }x</math>, then
		+	<math>x=e^{u}</math>
		+	and we have that
		+
		+
		+	<math>du=\frac{1}{e^{u}\,}\,dx\quad \Leftrightarrow \quad dx=e^{u}\,du</math>
		+
		+
		+	Thus, the substitution becomes
		+
		+
		+	<math>\begin{align}
		+	& \int{\ln x\,dx}=\left\{ \begin{matrix}
		+	u=\text{ln }x \\
		+	dx=e^{u}\,du \\
		+	\end{matrix} \right\} \\
		+	& =\int{ue^{u}\,du} \\
		+	\end{align}</math>
		+
		+
		+	Now, we carry out a partial integration
		+
		+
		+	<math>\begin{align}
		+	& \int{ue^{u}\,du}=u\centerdot e^{u}-\int{1\centerdot e^{u}\,du} \\
		+	& =u\centerdot e^{u}-\int{e^{u}\,du} \\
		+	& =u\centerdot e^{u}-e^{u}+C \\
		+	& =\left( u-1 \right)e^{u}+C \\
		+	\end{align}</math>
		+
		+
		+	and the answer becomes
		+
		+
		+
		+	<math>\begin{align}
		+	& \int{\ln x\,dx}=\left( \ln x-1 \right)e^{\ln x}+C \\
		+	& =\left( \ln x-1 \right)x+C \\
		+	\end{align}</math>

Tekbot: Lösning 2.3:2d moved to Solution 2.3:2d: Robot: moved page

2008-09-16T14:48:41Z

Lösning 2.3:2d moved to Solution 2.3:2d: Robot: moved page

← Nächstältere Version

Version vom 14:48, 16. Sep. 2008