Lösung 2.2:1a - Versionsgeschichte

Sekretariat1 um 11:50, 27. Aug. 2009

2009-08-27T11:50:54Z

Moni um 09:42, 22. Aug. 2009

2009-08-22T09:42:30Z

← Nächstältere Version		Version vom 09:42, 22. Aug. 2009
Zeile 13:		Zeile 13:
	{{Abgesetzte Formel\|\|<math>du = u'(x)\,dx = (3x-1)'\,dx = 3\,dx\,,</math>}}		{{Abgesetzte Formel\|\|<math>du = u'(x)\,dx = (3x-1)'\,dx = 3\,dx\,,</math>}}

-	und also ersetzen wir<math> dx </math> mit <math>\tfrac{1}{3}\du</math>.	+	und also ersetzen wir <math> dx </math> mit <math>\tfrac{1}{3} du</math>.

	Weiterhin entspricht die untere Integrationsgrenze <math>x=1</math>, <math>u=3\cdot 1-1=2</math>. Die obere Integrationsgrenze		Weiterhin entspricht die untere Integrationsgrenze <math>x=1</math>, <math>u=3\cdot 1-1=2</math>. Die obere Integrationsgrenze

Moni um 09:38, 22. Aug. 2009

2009-08-22T09:38:36Z

← Nächstältere Version		Version vom 09:38, 22. Aug. 2009
Zeile 1:		Zeile 1:
	Durch eine Substitution können wir ein kompliziertes Integral in vielen Fällen in ein einfacheres Integral bringen.		Durch eine Substitution können wir ein kompliziertes Integral in vielen Fällen in ein einfacheres Integral bringen.

-	Wenn wir die Substitution <math>u=u(x)</math>, machen, müssen wir folgendes bedenken:	+	Wenn wir die Substitution <math>u=u(x)</math> machen, müssen wir folgendes bedenken:

	# Das Integral muss mit der neuen Variable <math>u</math> umgeschrieben werden.		# Das Integral muss mit der neuen Variable <math>u</math> umgeschrieben werden.
-	# <math>dx</math>, muss mit <math>du</math> ersetzt werden, indem <math>du=u'(x)\,dx</math>.	+	# <math>dx</math> muss mit <math>du</math> ersetzt werden, indem <math>du=u'(x)\,dx</math>.
	# Die Integrationsgrenzen müssen an die neue Variable <math>u</math> angepasst werden.		# Die Integrationsgrenzen müssen an die neue Variable <math>u</math> angepasst werden.

Zeile 13:		Zeile 13:
	{{Abgesetzte Formel\|\|<math>du = u'(x)\,dx = (3x-1)'\,dx = 3\,dx\,,</math>}}		{{Abgesetzte Formel\|\|<math>du = u'(x)\,dx = (3x-1)'\,dx = 3\,dx\,,</math>}}

-	und also ersetzen wir<math>dx</math>mit <math>\tfrac{1}{3}\,du</math>.	+	und also ersetzen wir<math> dx </math> mit <math>\tfrac{1}{3}\du</math>.

	Weiterhin entspricht die untere Integrationsgrenze <math>x=1</math>, <math>u=3\cdot 1-1=2</math>. Die obere Integrationsgrenze		Weiterhin entspricht die untere Integrationsgrenze <math>x=1</math>, <math>u=3\cdot 1-1=2</math>. Die obere Integrationsgrenze

Martin um 12:37, 5. Mai 2009

2009-05-05T12:37:34Z

← Nächstältere Version		Version vom 12:37, 5. Mai 2009
Zeile 1:		Zeile 1:
-	~~A substitution of variables is often carried out so as to transform a complicated integral to one that is less complicated which one can either directly calculate or continue to work with~~.	+	Durch eine Substitution können wir ein kompliziertes Integral in vielen Fällen in ein einfacheres Integral bringen.

-	~~When we carry out a substitution of variables~~ <math>u=u(x)</math>, ~~there are three things which are affected in the integral~~:	+	Wenn wir die Substitution <math>u=u(x)</math>, machen, müssen wir folgendes bedenken:

-	# ~~the integral must be rewritten in terms of the new variable~~ <math>u</math>;	+	# Das Integral muss mit der neuen Variable <math>u</math> umgeschrieben werden.
-	# ~~the element of integration,~~ <math>dx</math>, ~~is replaced by~~ <math>du</math>, ~~according to the formula~~ <math>du=u'(x)\,dx</math>;	+	# <math>dx</math>, muss mit <math>du</math> ersetzt werden, indem <math>du=u'(x)\,dx</math>.
-	# ~~the limits of integration are for <math>x</math> and must be changed to limits of integration for the variable~~ <math>u</math>.	+	# Die Integrationsgrenzen müssen an die neue Variable <math>u</math> angepasst werden.

-	In ~~this case, we will perform the change of variables~~ <math>u=3x-1</math>, ~~mainly because the integrand~~ <math>1/(3x-1)^4</math> ~~will then be replaced by~~ <math>1/u^4</math>.	+	In diesem Fall machen wir die Substitution <math>u=3x-1</math>, nachdem <math>1/(3x-1)^4</math> mit <math>1/u^4</math> ersetzt wird, und dieser Ausdruck einfacher zu integrieren ist.

-	~~The relation between~~ <math>dx</math> ~~and~~ <math>du</math> ~~reads~~	+	Das Verhältnis zwischen <math>dx</math> und <math>du</math> lautet

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

-	~~which means that~~ <math>dx</math> ~~is replaced by~~ <math>\tfrac{1}{3}\,du</math>.	+	und also ersetzen wir<math>dx</math>mit <math>\tfrac{1}{3}\,du</math>.

-	~~Furthermore, when~~ <math>x=1</math> ~~in the lower limit of integration~~, ~~the corresponding ''u''-value becomes~~ <math>u=3\cdot 1-1=2</math>~~, and when~~	+	Weiterhin entspricht die untere Integrationsgrenze <math>x=1</math>, <math>u=3\cdot 1-1=2</math>. Die obere Integrationsgrenze
-	<math>x=2</math>, ~~we obtain the ''u''-value~~ <math>u=3\cdot 2-1=5\,</math>.	+	<math>x=2</math>, entspricht <math>u=3\cdot 2-1=5\,</math>.

-	~~One usually writes the whole substitution of variables as~~	+	Die Schreibweise für die Variabelsubstitution ist meistens

	{{Abgesetzte Formel\|\|<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \left\{ \begin{align}		{{Abgesetzte Formel\|\|<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \left\{ \begin{align}
Zeile 25:		Zeile 25:
	\end{align} \right\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}</math>}}		\end{align} \right\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}</math>}}

-	~~Sometimes, we are more brief and hide the details~~,	+	oder weniger detailliert,

	{{Abgesetzte Formel\|\|<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \bigl\{ u=3x-1 \bigr\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \bigl\{ u=3x-1 \bigr\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}</math>}}

-	~~After the substitution of variables, we have a standard integral which is easy to compute~~.	+	Nach der Substitution erhalten wir ein einfaches Integral. Die ganze Rechnung lautet;
-		+
-	~~In summary, the whole calculation is,~~	+

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

Tekbot: Solution 2.2:1a moved to Lösung 2.2:1a: Robot: moved page

2009-03-11T10:20:15Z

Solution 2.2:1a moved to Lösung 2.2:1a: Robot: moved page

← Nächstältere Version

Version vom 10:20, 11. Mär. 2009

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

2009-03-10T13:00:34Z

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

← Nächstältere Version		Version vom 13:00, 10. Mär. 2009
Zeile 11:		Zeile 11:
	The relation between <math>dx</math> and <math>du</math> reads		The relation between <math>dx</math> and <math>du</math> reads

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

	which means that <math>dx</math> is replaced by <math>\tfrac{1}{3}\,du</math>.		which means that <math>dx</math> is replaced by <math>\tfrac{1}{3}\,du</math>.
Zeile 20:		Zeile 20:
	One usually writes the whole substitution of variables as		One usually writes the whole substitution of variables as

-	{{~~Displayed math~~\|\|<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \left\{ \begin{align}	+	{{Abgesetzte Formel\|\|<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \left\{ \begin{align}
	u &= 3x-1\\[5pt]		u &= 3x-1\\[5pt]
	du &= 3\,dx		du &= 3\,dx
Zeile 27:		Zeile 27:
	Sometimes, we are more brief and hide the details,		Sometimes, we are more brief and hide the details,

-	{{~~Displayed math~~\|\|<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \bigl\{ u=3x-1 \bigr\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \bigl\{ u=3x-1 \bigr\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}</math>}}

	After the substitution of variables, we have a standard integral which is easy to compute.		After the substitution of variables, we have a standard integral which is easy to compute.
Zeile 33:		Zeile 33:
	In summary, the whole calculation is,		In summary, the whole calculation is,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int\limits_1^2 \frac{dx}{(3x-1)^4}		\int\limits_1^2 \frac{dx}{(3x-1)^4}
	&= \left\{\begin{align}		&= \left\{\begin{align}

Tek um 12:31, 28. Okt. 2008

2008-10-28T12:31:42Z

← Nächstältere Version		Version vom 12:31, 28. Okt. 2008
Zeile 1:		Zeile 1:
	A substitution of variables is often carried out so as to transform a complicated integral to one that is less complicated which one can either directly calculate or continue to work with.		A substitution of variables is often carried out so as to transform a complicated integral to one that is less complicated which one can either directly calculate or continue to work with.

-	When we carry out a substitution of variables	+	When we carry out a substitution of variables <math>u=u(x)</math>, there are three things which are affected in the integral:
-	<math>u=u~~\left~~( x ~~\right~~)</math>, there are three things which are affected in the integral:	+

-	1. the integral must be rewritten in terms of the new variable	+	# the integral must be rewritten in terms of the new variable <math>u</math>;
-	<math>u</math>;	+	# the element of integration, <math>dx</math>, is replaced by <math>du</math>, according to the formula <math>du=u'(x)\,dx</math>;
-	2. the element of integration,	+	# the limits of integration are for <math>x</math> and must be changed to limits of integration for the variable <math>u</math>.
-	<math>dx</math>, is replaced by	+
-	<math>du</math>, according to the formula	+
-	<math>du={u}'~~\left~~( x ~~\right~~)dx</math>;	+
-	3. the limits of integration are for	+
-	<math>x~~\text{ }~~</math>	+
-	and must be changed to limits of integration for the variable	+
-	<math>u</math>.	+

-	In this case, we will perform the change of variables	+	In this case, we will perform the change of variables <math>u=3x-1</math>, mainly because the integrand <math>1/(3x-1)^4</math> will then be replaced by <math>1/u^4</math>.
-	<math>u=3x-1</math>, mainly because the integrand	+
-	<math>~~\frac{~~1~~}{\left~~( 3x-1 ~~\right~~)^{4}}</math>	+
-	will then be replaced by	+
-	<math>~~\frac{~~1}{u^{4}}</math>.	+

-	The relation between	+	The relation between <math>dx</math> and <math>du</math> reads
-	<math>dx</math>	+
-	and	+
-	<math>du</math>	+
-	reads	+

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

-	~~<math>du={u}'\left( x \right)\,dx=\left( 3x-1 \right)^{\prime }\,dx=3\,dx</math>,~~	+	which means that <math>dx</math> is replaced by <math>\tfrac{1}{3}\,du</math>.
-		+
-	which means that	+
-	<math>dx</math>	+
-	is replaced by	+
-	<math>\~~frac~~{1}{3}\,du</math>.	+
-		+
-	~~Furthermore, when~~	+
-	~~<math>x=\text{1}</math>~~	+
-	~~in the lower limit of integration, the corresponding~~	+
-	~~<math>u</math>~~	+
-	~~-value becomes~~	+
-	~~<math>u=3\centerdot 1-1=2</math>, and when~~	+
-	~~<math>x=2</math>, we obtain the u-value~~	+
-	~~<math>u=3\centerdot 2-1=5</math>~~	+

		+	Furthermore, when <math>x=1</math> in the lower limit of integration, the corresponding ''u''-value becomes <math>u=3\cdot 1-1=2</math>, and when
		+	<math>x=2</math>, we obtain the ''u''-value <math>u=3\cdot 2-1=5\,</math>.

	One usually writes the whole substitution of variables as		One usually writes the whole substitution of variables as

		+	{{Displayed math\|\|<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \left\{ \begin{align}
		+	u &= 3x-1\\[5pt]
		+	du &= 3\,dx
		+	\end{align} \right\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}</math>}}

-	~~<math>\int\limits_{1}^{2}{\frac{\,dx}{\left( 3x-1 \right)^{4}}}=\left\{ \begin{matrix}~~	+	Sometimes, we are more brief and hide the details,
-	~~u=3x-1 \\~~	+
-	~~du=3\,dx \\~~	+
-	~~\end{matrix} \right\}=\int\limits_{2}^{5}{\frac{\frac{1}{3}\,du}{u^{4}}}</math>~~	+
-		+
-		+
-	Sometimes, we are more brief and hide the details:	+
-		+
-		+
-	~~<math>\int\limits_{1}^{2}{\frac{\,dx}{\left( 3x-1 \right)^{4}}}=\left\{ u=3x-1 \right\}=\int\limits_{2}^{5}{\frac{\frac{1}{3}\~~,~~du}{u^{4}}}</math>~~	+

		+	{{Displayed math\|\|<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \bigl\{ u=3x-1 \bigr\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}</math>}}

	After the substitution of variables, we have a standard integral which is easy to compute.		After the substitution of variables, we have a standard integral which is easy to compute.

-	In summary, the whole calculation is:	+	In summary, the whole calculation is,
-		+

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& \int\~~limits_{1}~~^{2}{\frac{\,dx}{~~\left~~( 3x-1 ~~\right~~)^{4}}}=\left\{ \begin{~~matrix~~}	+	\int\limits_1^2 \frac{dx}{(3x-1)^4}
-	u=3x-1 \\	+	&= \left\{\begin{align}
-	du=3\,dx \\	+	u &= 3x-1\\[5pt]
-	\end{~~matrix~~} \right\}=\int\~~limits_{2}~~^{5}{\frac{\~~frac~~{1}{3}\,du}{u^{4}~~}} \\~~	+	du &= 3\,dx
-	& =\frac{1}{3}\int\~~limits_{2}~~^{5~~}{\,~~u^{-4}du}=\~~left~~[ \frac{u^{-4+1}}{-4+1} \~~right~~]~~_{2}~~^{5~~} \\~~	+	\end{align}\right\}
-	& =-\frac{1}{9}\~~left~~[ \frac{1}{u^{3}} \~~right~~]~~_{2}~~^{5}=-\frac{1}{9}\~~left~~( \frac{1}{5^{3}}-\frac{1}{2^{3}} \~~right~~) \\	+	= \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}
-	& =-\frac{1}{9}\~~centerdot~~ \frac{2^{3}-5^{3}}{2^{3}\~~centerdot~~ 5^{3}}=\frac{117}{3^{2}\~~centerdot~~ 2^{3}\~~centerdot~~ 5^{3}~~} \\~~	+	= \frac{1}{3}\int\limits_2^5 u^{-4}\,du\\[5pt]
-	& =\frac{3^{2}\~~centerdot~~ 13}{3^{2}\~~centerdot~~ 2^{3}\~~centerdot~~ 5^{3}}=\frac{13}{2^{3}\~~centerdot~~ 5^{3}}=\frac{13}{1000} \\	+	&= \frac{1}{3}\Bigl[\ \frac{u^{-4+1}}{-4+1}\ \Bigr]_2^5
-	\end{align}</math>.	+	= -\frac{1}{9}\Bigl[\ \frac{1}{u^3}\ \Bigr]_2^5
		+	= -\frac{1}{9}\Bigl(\frac{1}{5^3} - \frac{1}{2^3} \Bigr)
		+	= -\frac{1}{9}\cdot\frac{2^3-5^3}{2^3\cdot 5^3}\\[5pt]
		+	&= \frac{117}{3^2\cdot 2^3\cdot 5^3}
		+	= \frac{3^2\cdot 13}{3^2\cdot 2^3\cdot 5^3}
		+	= \frac{13}{2^3\cdot 5^3}
		+	= \frac{13}{1000}\,\textrm{.}
		+	\end{align}</math>}}

Ian um 09:21, 19. Okt. 2008

2008-10-19T09:21:40Z

← Nächstältere Version		Version vom 09:21, 19. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START}~~}	+	A substitution of variables is often carried out so as to transform a complicated integral to one that is less complicated which one can either directly calculate or continue to work with.
-	<~~center~~> ~~[[Image:2_2_1a~~-1(3)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	When we carry out a substitution of variables
-	{{~~NAVCONTENT_START~~}}	+	<math>u=u\left( x \right)</math>, there are three things which are affected in the integral:
-	<~~center~~> ~~[[Image:2_2_1a~~-2(3~~).gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	1. the integral must be rewritten in terms of the new variable
-	{{~~NAVCONTENT_START~~}}	+	<math>u</math>;
-	<~~center~~> ~~[[Image~~:~~2_2_1a~~-3(3)~~.gif~~]] ~~</center>~~	+	2. the element of integration,
-	{{~~NAVCONTENT_STOP~~}}	+	<math>dx</math>, is replaced by
		+	<math>du</math>, according to the formula
		+	<math>du={u}'\left( x \right)dx</math>;
		+	3. the limits of integration are for
		+	<math>x\text{ }</math>
		+	and must be changed to limits of integration for the variable
		+	<math>u</math>.
		+
		+	In this case, we will perform the change of variables
		+	<math>u=3x-1</math>, mainly because the integrand
		+	<math>\frac{1}{\left( 3x-1 \right)^{4}}</math>
		+	will then be replaced by
		+	<math>\frac{1}{u^{4}}</math>.
		+
		+	The relation between
		+	<math>dx</math>
		+	and
		+	<math>du</math>
		+	reads
		+
		+
		+	<math>du={u}'\left( x \right)\,dx=\left( 3x-1 \right)^{\prime }\,dx=3\,dx</math>,
		+
		+	which means that
		+	<math>dx</math>
		+	is replaced by
		+	<math>\frac{1}{3}\,du</math>.
		+
		+	Furthermore, when
		+	<math>x=\text{1}</math>
		+	in the lower limit of integration, the corresponding
		+	<math>u</math>
		+	-value becomes
		+	<math>u=3\centerdot 1-1=2</math>, and when
		+	<math>x=2</math>, we obtain the u-value
		+	<math>u=3\centerdot 2-1=5</math>
		+
		+
		+	One usually writes the whole substitution of variables as
		+
		+
		+	<math>\int\limits_{1}^{2}{\frac{\,dx}{\left( 3x-1 \right)^{4}}}=\left\{ \begin{matrix}
		+	u=3x-1 \\
		+	du=3\,dx \\
		+	\end{matrix} \right\}=\int\limits_{2}^{5}{\frac{\frac{1}{3}\,du}{u^{4}}}</math>
		+
		+
		+	Sometimes, we are more brief and hide the details:
		+
		+
		+	<math>\int\limits_{1}^{2}{\frac{\,dx}{\left( 3x-1 \right)^{4}}}=\left\{ u=3x-1 \right\}=\int\limits_{2}^{5}{\frac{\frac{1}{3}\,du}{u^{4}}}</math>
		+
		+
		+	After the substitution of variables, we have a standard integral which is easy to compute.
		+
		+	In summary, the whole calculation is:
		+
		+
		+	<math>\begin{align}
		+	& \int\limits_{1}^{2}{\frac{\,dx}{\left( 3x-1 \right)^{4}}}=\left\{ \begin{matrix}
		+	u=3x-1 \\
		+	du=3\,dx \\
		+	\end{matrix} \right\}=\int\limits_{2}^{5}{\frac{\frac{1}{3}\,du}{u^{4}}} \\
		+	& =\frac{1}{3}\int\limits_{2}^{5}{\,u^{-4}du}=\left[ \frac{u^{-4+1}}{-4+1} \right]_{2}^{5} \\
		+	& =-\frac{1}{9}\left[ \frac{1}{u^{3}} \right]_{2}^{5}=-\frac{1}{9}\left( \frac{1}{5^{3}}-\frac{1}{2^{3}} \right) \\
		+	& =-\frac{1}{9}\centerdot \frac{2^{3}-5^{3}}{2^{3}\centerdot 5^{3}}=\frac{117}{3^{2}\centerdot 2^{3}\centerdot 5^{3}} \\
		+	& =\frac{3^{2}\centerdot 13}{3^{2}\centerdot 2^{3}\centerdot 5^{3}}=\frac{13}{2^{3}\centerdot 5^{3}}=\frac{13}{1000} \\
		+	\end{align}</math>.

Tekbot: Lösning 2.2:1a moved to Solution 2.2:1a: Robot: moved page

2008-09-16T14:40:41Z

Lösning 2.2:1a moved to Solution 2.2:1a: Robot: moved page

← Nächstältere Version

Version vom 14:40, 16. Sep. 2008

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

2008-08-21T08:00:04Z

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_1a-1(3).gif]] </center>	+	<center> [[Image:2_2_1a-1(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_2_1a-2(3).gif]] </center>	+	<center> [[Image:2_2_1a-2(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_2_1a-3(3).gif]] </center>	+	<center> [[Image:2_2_1a-3(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

← Nächstältere Version		Version vom 11:50, 27. Aug. 2009
Zeile 1:		Zeile 1:
	Durch eine Substitution können wir ein kompliziertes Integral in vielen Fällen in ein einfacheres Integral bringen.		Durch eine Substitution können wir ein kompliziertes Integral in vielen Fällen in ein einfacheres Integral bringen.

-	Wenn wir die Substitution <math>u=u(x)</math> ~~machen~~, müssen wir folgendes bedenken:	+	Wenn wir die Substitution <math>u=u(x)</math> durchführen, müssen wir folgendes bedenken:

	# Das Integral muss mit der neuen Variable <math>u</math> umgeschrieben werden.		# Das Integral muss mit der neuen Variable <math>u</math> umgeschrieben werden.
Zeile 7:		Zeile 7:
	# Die Integrationsgrenzen müssen an die neue Variable <math>u</math> angepasst werden.		# Die Integrationsgrenzen müssen an die neue Variable <math>u</math> angepasst werden.

-	In diesem Fall machen wir die Substitution <math>u=3x-1</math>, ~~nachdem~~ <math>1/(3x-1)^4</math> mit <math>1/u^4</math> ersetzt wird, und dieser Ausdruck einfacher zu integrieren ist.	+	In diesem Fall machen wir die Substitution <math>u=3x-1</math>, da <math>1/(3x-1)^4</math> mit <math>1/u^4</math> ersetzt wird und dieser Ausdruck einfacher zu integrieren ist.

	Das Verhältnis zwischen <math>dx</math> und <math>du</math> lautet		Das Verhältnis zwischen <math>dx</math> und <math>du</math> lautet
Zeile 13:		Zeile 13:
	{{Abgesetzte Formel\|\|<math>du = u'(x)\,dx = (3x-1)'\,dx = 3\,dx\,,</math>}}		{{Abgesetzte Formel\|\|<math>du = u'(x)\,dx = (3x-1)'\,dx = 3\,dx\,,</math>}}

-	~~und~~ also ersetzen wir <math> dx </math> mit <math>\tfrac{1}{3} du</math>.	+	also ersetzen wir <math> dx </math> mit <math>\tfrac{1}{3} du</math>.

	Weiterhin entspricht die untere Integrationsgrenze <math>x=1</math>, <math>u=3\cdot 1-1=2</math>. Die obere Integrationsgrenze		Weiterhin entspricht die untere Integrationsgrenze <math>x=1</math>, <math>u=3\cdot 1-1=2</math>. Die obere Integrationsgrenze
-	<math>x=2</math>, entspricht <math>u=3\cdot 2-1=5\,</math>.	+	<math>x=2</math> entspricht <math>u=3\cdot 2-1=5\,</math>.

	Die Schreibweise für die Variabelsubstitution ist meistens		Die Schreibweise für die Variabelsubstitution ist meistens
Zeile 23:		Zeile 23:
	u &= 3x-1\\[5pt]		u &= 3x-1\\[5pt]
	du &= 3\,dx		du &= 3\,dx
-	\end{align} \right\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,~~\textrm{.}~~</math>}}	+	\end{align} \right\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,</math>}}

-	oder weniger detailliert,	+	oder weniger detailliert

	{{Abgesetzte Formel\|\|<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \bigl\{ u=3x-1 \bigr\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\int\limits_1^2 \frac{dx}{(3x-1)^4} = \bigl\{ u=3x-1 \bigr\} = \int\limits_2^5 \frac{\tfrac{1}{3}\,du}{u^4}\,\textrm{.}</math>}}

-	Nach der Substitution erhalten wir ein einfaches Integral. Die ganze Rechnung lautet;	+	Nach der Substitution erhalten wir ein einfaches Integral. Die ganze Rechnung lautet

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