Lösung 2.2:2d - Versionsgeschichte

Sekretariat1 um 12:24, 27. Aug. 2009

2009-08-27T12:24:56Z

← Nächstältere Version		Version vom 12:24, 27. Aug. 2009
Zeile 1:		Zeile 1:
-	Wir probieren die Substitution <math>u=1-x</math>,	+	Wir probieren die Substitution <math>u=1-x</math>.

	{{Abgesetzte Formel\|\|<math>\int\limits_0^1 \sqrt[3]{1-x}\,dx = \left\{ \begin{align}		{{Abgesetzte Formel\|\|<math>\int\limits_0^1 \sqrt[3]{1-x}\,dx = \left\{ \begin{align}
	u &= 1-x\\[5pt]		u &= 1-x\\[5pt]
	du &= (1-x)'\,dx = -\,dx		du &= (1-x)'\,dx = -\,dx
-	\end{align} \right\} = -\int\limits_1^0 \sqrt[3]{u}\,du\,~~\textrm{.}~~</math>}}	+	\end{align} \right\} = -\int\limits_1^0 \sqrt[3]{u}\,du\,</math>}}

-	~~Beachten Sie~~ die Änderungen in den Integrationsgrenzen. Wir können jetzt die obere und untere Grenzen wechseln, wenn wir gleichzeitig das Vorzeichen vom Integrand tauschen	+	Beachte die Änderungen in den Integrationsgrenzen. Wir können jetzt die obere und untere Grenzen wechseln, wenn wir gleichzeitig das Vorzeichen vom Integrand tauschen.

-	{{Abgesetzte Formel\|\|<math>-\int\limits_1^0 \sqrt[3]{u}\,du = +\int\limits_0^1 \sqrt[3]{u}\,du\,~~\textrm{.}~~</math>}}	+	{{Abgesetzte Formel\|\|<math>-\int\limits_1^0 \sqrt[3]{u}\,du = +\int\limits_0^1 \sqrt[3]{u}\,du\,</math>}}

	Wir erhalten		Wir erhalten

Moni um 09:47, 22. Aug. 2009

2009-08-22T09:47:14Z

← Nächstältere Version		Version vom 09:47, 22. Aug. 2009
Zeile 6:		Zeile 6:
	\end{align} \right\} = -\int\limits_1^0 \sqrt[3]{u}\,du\,\textrm{.}</math>}}		\end{align} \right\} = -\int\limits_1^0 \sqrt[3]{u}\,du\,\textrm{.}</math>}}

-	Beachten Sie die Änderungen in den Integrationsgrenzen. Wir können jetzt die obere und untere Grenzen ~~wächseln~~, wenn wir gleichzeitig das Vorzeichen vom Integrand tauschen,	+	Beachten Sie die Änderungen in den Integrationsgrenzen. Wir können jetzt die obere und untere Grenzen wechseln, wenn wir gleichzeitig das Vorzeichen vom Integrand tauschen

	{{Abgesetzte Formel\|\|<math>-\int\limits_1^0 \sqrt[3]{u}\,du = +\int\limits_0^1 \sqrt[3]{u}\,du\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>-\int\limits_1^0 \sqrt[3]{u}\,du = +\int\limits_0^1 \sqrt[3]{u}\,du\,\textrm{.}</math>}}

-	Wir erhalten,	+	Wir erhalten

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

Martin um 13:16, 5. Mai 2009

2009-05-05T13:16:10Z

← Nächstältere Version		Version vom 13:16, 5. Mai 2009
Zeile 1:		Zeile 1:
-	~~What makes the integral not entirely simple is the expression <math>1-x</math> under the root sign, so we try the substitution~~ <math>u=1-x</math>,	+	Wir probieren die Substitution <math>u=1-x</math>,

	{{Abgesetzte Formel\|\|<math>\int\limits_0^1 \sqrt[3]{1-x}\,dx = \left\{ \begin{align}		{{Abgesetzte Formel\|\|<math>\int\limits_0^1 \sqrt[3]{1-x}\,dx = \left\{ \begin{align}
Zeile 6:		Zeile 6:
	\end{align} \right\} = -\int\limits_1^0 \sqrt[3]{u}\,du\,\textrm{.}</math>}}		\end{align} \right\} = -\int\limits_1^0 \sqrt[3]{u}\,du\,\textrm{.}</math>}}

-	~~Note how the new limits of integration go from 1 to 0 (and not the other way around!)~~. ~~It is possible to change the order of the limits if we change sign at the same time~~, ~~i.e.~~	+	Beachten Sie die Änderungen in den Integrationsgrenzen. Wir können jetzt die obere und untere Grenzen wächseln, wenn wir gleichzeitig das Vorzeichen vom Integrand tauschen,

	{{Abgesetzte Formel\|\|<math>-\int\limits_1^0 \sqrt[3]{u}\,du = +\int\limits_0^1 \sqrt[3]{u}\,du\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>-\int\limits_1^0 \sqrt[3]{u}\,du = +\int\limits_0^1 \sqrt[3]{u}\,du\,\textrm{.}</math>}}

-	~~All that is now left is routine calculations~~,	+	Wir erhalten,

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

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

2009-03-11T10:22:15Z

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

← Nächstältere Version

Version vom 10:22, 11. Mär. 2009

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

2009-03-10T13:01:34Z

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

← Nächstältere Version		Version vom 13:01, 10. Mär. 2009
Zeile 1:		Zeile 1:
	What makes the integral not entirely simple is the expression <math>1-x</math> under the root sign, so we try the substitution <math>u=1-x</math>,		What makes the integral not entirely simple is the expression <math>1-x</math> under the root sign, so we try the substitution <math>u=1-x</math>,

-	{{~~Displayed math~~\|\|<math>\int\limits_0^1 \sqrt[3]{1-x}\,dx = \left\{ \begin{align}	+	{{Abgesetzte Formel\|\|<math>\int\limits_0^1 \sqrt[3]{1-x}\,dx = \left\{ \begin{align}
	u &= 1-x\\[5pt]		u &= 1-x\\[5pt]
	du &= (1-x)'\,dx = -\,dx		du &= (1-x)'\,dx = -\,dx
Zeile 8:		Zeile 8:
	Note how the new limits of integration go from 1 to 0 (and not the other way around!). It is possible to change the order of the limits if we change sign at the same time, i.e.		Note how the new limits of integration go from 1 to 0 (and not the other way around!). It is possible to change the order of the limits if we change sign at the same time, i.e.

-	{{~~Displayed math~~\|\|<math>-\int\limits_1^0 \sqrt[3]{u}\,du = +\int\limits_0^1 \sqrt[3]{u}\,du\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>-\int\limits_1^0 \sqrt[3]{u}\,du = +\int\limits_0^1 \sqrt[3]{u}\,du\,\textrm{.}</math>}}

	All that is now left is routine calculations,		All that is now left is routine calculations,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int\limits_0^1 \sqrt[3]{u}\,du		\int\limits_0^1 \sqrt[3]{u}\,du
	&= \int\limits_0^1 u^{1/3}\,du		&= \int\limits_0^1 u^{1/3}\,du

Tek um 13:56, 28. Okt. 2008

2008-10-28T13:56:06Z

← Nächstältere Version		Version vom 13:56, 28. Okt. 2008
Zeile 1:		Zeile 1:
-	What makes the integral not entirely simple is the expression	+	What makes the integral not entirely simple is the expression <math>1-x</math> under the root sign, so we try the substitution <math>u=1-x</math>,
-	<math>~~\text{~~1}-x</math>	+
-	under the root sign, so we try the substitution	+
-	<math>u=~~\text{~~1}-x</math>,	+

		+	{{Displayed math\|\|<math>\int\limits_0^1 \sqrt[3]{1-x}\,dx = \left\{ \begin{align}
		+	u &= 1-x\\[5pt]
		+	du &= (1-x)'\,dx = -\,dx
		+	\end{align} \right\} = -\int\limits_1^0 \sqrt[3]{u}\,du\,\textrm{.}</math>}}

-	~~<math>\int\limits_{~~0~~}^{1}{\sqrt[3]{\text{1}-x}}\,dx=\left\{ \begin{matrix}~~	+	Note how the new limits of integration go from 1 to 0 (and not the other way around!). It is possible to change the order of the limits if we change sign at the same time, i.e.
-	~~u=\text{1}-x \\~~	+
-	~~du=\left~~( ~~\text{1}-x \right~~)~~^{\prime }\,dx=-\,dx \\~~	+
-	~~\end{matrix} \right\}=-\int\limits_{1}^{0}{\sqrt[3]{u}}\~~,~~du</math>~~	+

		+	{{Displayed math\|\|<math>-\int\limits_1^0 \sqrt[3]{u}\,du = +\int\limits_0^1 \sqrt[3]{u}\,du\,\textrm{.}</math>}}

-	~~Note how the new limits of integration go from~~	+	All that is now left is routine calculations,
-	~~<math>\text{1}</math>~~	+
-	to	+
-	~~<math>0</math>~~	+
-	~~(and not the other way around!). It~~ is ~~possible to change the order of the limits we change sign at the same time~~, ~~i.e.~~	+

-		+	{{Displayed math\|\|<math>\begin{align}
-	~~<math>-\int\limits_~~{~~1}^~~{~~0}{\sqrt[3]{u}}\,du=+\int\limits_{0}^{1}{\sqrt[3]{u}}\,du</~~math>	+	\int\limits_0^1 \sqrt[3]{u}\,du
-		+	&= \int\limits_0^1 u^{1/3}\,du
-		+	= \biggl[\ \frac{u^{1/3+1}}{\tfrac{1}{3}+1}\ \biggr]_0^1\\[5pt]
-	~~All that is now left is routine calculations:~~	+	&= \frac{3}{4}\Bigl[\ u^{4/3}\ \Bigr]_0^1
-		+	= \frac{3}{4}\bigl( 1^{4/3}-0^{4/3} \bigr) = \frac{3}{4}\,\textrm{.}
-		+	\end{align}</math>}}
-	<math>\begin{align}	+
-	& \int\~~limits_{0}~~^{1}{\sqrt[3]{u}}\,du=\int\~~limits_{0}~~^{1}{u^~~{\frac~~{1}{3}}}\,du=\~~left~~[ \frac{u^~~{\frac~~{1}{3}+1}}{\~~frac~~{1}{3}+1} \~~right~~]~~_{0}~~^{1} \\	+
-	& =\frac{3}{4}\~~left~~[ u^~~{\frac~~{4}{3}} \~~right~~]~~_{0}~~^{1}=\frac{3}{4}\~~left~~( 1^~~{\frac~~{4}{3}}-0^~~{\frac~~{4}{3}} \~~right~~)=\frac{3}{4} \\	+
-	\end{align}</math>	+

Ian um 12:34, 20. Okt. 2008

2008-10-20T12:34:09Z

← Nächstältere Version		Version vom 12:34, 20. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START}~~}	+	What makes the integral not entirely simple is the expression
-	<~~center~~> [[~~Image:2_2_2d~~.~~gif~~]] </~~center~~>	+	<math>\text{1}-x</math>
-	{{~~NAVCONTENT_STOP~~}}	+	under the root sign, so we try the substitution
		+	<math>u=\text{1}-x</math>,
		+
		+
		+	<math>\int\limits_{0}^{1}{\sqrt[3]{\text{1}-x}}\,dx=\left\{ \begin{matrix}
		+	u=\text{1}-x \\
		+	du=\left( \text{1}-x \right)^{\prime }\,dx=-\,dx \\
		+	\end{matrix} \right\}=-\int\limits_{1}^{0}{\sqrt[3]{u}}\,du</math>
		+
		+
		+	Note how the new limits of integration go from
		+	<math>\text{1}</math>
		+	to
		+	<math>0</math>
		+	(and not the other way around!). It is possible to change the order of the limits we change sign at the same time, i.e.
		+
		+
		+	<math>-\int\limits_{1}^{0}{\sqrt[3]{u}}\,du=+\int\limits_{0}^{1}{\sqrt[3]{u}}\,du</math>
		+
		+
		+	All that is now left is routine calculations:
		+
		+
		+	<math>\begin{align}
		+	& \int\limits_{0}^{1}{\sqrt[3]{u}}\,du=\int\limits_{0}^{1}{u^{\frac{1}{3}}}\,du=\left[ \frac{u^{\frac{1}{3}+1}}{\frac{1}{3}+1} \right]_{0}^{1} \\
		+	& =\frac{3}{4}\left[ u^{\frac{4}{3}} \right]_{0}^{1}=\frac{3}{4}\left( 1^{\frac{4}{3}}-0^{\frac{4}{3}} \right)=\frac{3}{4} \\
		+	\end{align}</math>

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

2008-09-16T14:42:41Z

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

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Version vom 14:42, 16. Sep. 2008

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

2008-08-21T08:01:04Z

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

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2008-04-07T09:33:28Z

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