Lösung 2.1:2d - Versionsgeschichte

Sekretariat1 um 09:22, 21. Aug. 2009

Sekretariat1 — Fri, 21 Aug 2009 09:22:48 GMT

← Nächstältere Version		Version vom 09:22, 21. Aug. 2009
Zeile 1:		Zeile 1:
-	Wir schreiben <math>\sqrt{x}</math> wie <math>x^{1/2}</math>, und erhalten durch die Rechenregeln für Exponenten	+	Wir schreiben <math>\sqrt{x}</math> wie <math>x^{1/2}</math> und erhalten durch die Rechenregeln für Exponenten

	{{Abgesetzte Formel\|\|<math>\int\limits_1^4 \frac{\sqrt{x}}{x^2}\,dx = \int\limits_1^4 \frac{x^{1/2}}{x^2}\,dx = \int\limits_1^4 x^{1/2-2}\,dx = \int\limits_1^4 x^{-3/2}\,dx\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\int\limits_1^4 \frac{\sqrt{x}}{x^2}\,dx = \int\limits_1^4 \frac{x^{1/2}}{x^2}\,dx = \int\limits_1^4 x^{1/2-2}\,dx = \int\limits_1^4 x^{-3/2}\,dx\,\textrm{.}</math>}}

-	~~Jetzt verwenden wir dass die~~ Stammfunktion von <math>x^{n}</math>, <math>x^{n+1}/(n+1)</math> ~~ist,~~ und berechnen das Integral	+	Die Stammfunktion von <math>x^{n}</math> ist <math>x^{n+1}/(n+1)</math> und damit berechnen das Integral.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 13:		Zeile 13:
	&= -\frac{2}{\sqrt{4}} - \Bigl(-\frac{2}{\sqrt{1}}\Bigr)\\[5pt]		&= -\frac{2}{\sqrt{4}} - \Bigl(-\frac{2}{\sqrt{1}}\Bigr)\\[5pt]
	&= -\frac{2}{2}+2\\[5pt]		&= -\frac{2}{2}+2\\[5pt]
-	&= 1~~\,\textrm{.}~~	+	&= 1
	\end{align}</math>}}		\end{align}</math>}}

Martin um 16:50, 28. Apr. 2009

Martin — Tue, 28 Apr 2009 16:50:46 GMT

← Nächstältere Version		Version vom 16:50, 28. Apr. 2009
Zeile 1:		Zeile 1:
-	~~If we rewrite~~ <math>\sqrt{x}</math> as <math>x^{1/2}</math>, ~~the integrand can then be simplified using the power laws,~~	+	Wir schreiben <math>\sqrt{x}</math> wie <math>x^{1/2}</math>, und erhalten durch die Rechenregeln für Exponenten

	{{Abgesetzte Formel\|\|<math>\int\limits_1^4 \frac{\sqrt{x}}{x^2}\,dx = \int\limits_1^4 \frac{x^{1/2}}{x^2}\,dx = \int\limits_1^4 x^{1/2-2}\,dx = \int\limits_1^4 x^{-3/2}\,dx\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\int\limits_1^4 \frac{\sqrt{x}}{x^2}\,dx = \int\limits_1^4 \frac{x^{1/2}}{x^2}\,dx = \int\limits_1^4 x^{1/2-2}\,dx = \int\limits_1^4 x^{-3/2}\,dx\,\textrm{.}</math>}}

-	~~We can now use the fact that a primitive function for~~ <math>x^{n}</math> is <math>x^{n+1}/(n+1)</math> ~~and calculate the integral's value~~,	+	Jetzt verwenden wir dass die Stammfunktion von <math>x^{n}</math>, <math>x^{n+1}/(n+1)</math> ist, und berechnen das Integral

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

Tekbot: Solution 2.1:2d moved to Lösung 2.1:2d: Robot: moved page

Tekbot — Wed, 11 Mar 2009 10:16:15 GMT

Solution 2.1:2d moved to Lösung 2.1:2d: Robot: moved page

← Nächstältere Version

Version vom 10:16, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 12:58:34 GMT

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

← Nächstältere Version		Version vom 12:58, 10. Mär. 2009
Zeile 1:		Zeile 1:
	If we rewrite <math>\sqrt{x}</math> as <math>x^{1/2}</math>, the integrand can then be simplified using the power laws,		If we rewrite <math>\sqrt{x}</math> as <math>x^{1/2}</math>, the integrand can then be simplified using the power laws,

-	{{~~Displayed math~~\|\|<math>\int\limits_1^4 \frac{\sqrt{x}}{x^2}\,dx = \int\limits_1^4 \frac{x^{1/2}}{x^2}\,dx = \int\limits_1^4 x^{1/2-2}\,dx = \int\limits_1^4 x^{-3/2}\,dx\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\int\limits_1^4 \frac{\sqrt{x}}{x^2}\,dx = \int\limits_1^4 \frac{x^{1/2}}{x^2}\,dx = \int\limits_1^4 x^{1/2-2}\,dx = \int\limits_1^4 x^{-3/2}\,dx\,\textrm{.}</math>}}

	We can now use the fact that a primitive function for <math>x^{n}</math> is <math>x^{n+1}/(n+1)</math> and calculate the integral's value,		We can now use the fact that a primitive function for <math>x^{n}</math> is <math>x^{n+1}/(n+1)</math> and calculate the integral's value,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int\limits_1^4 x^{-3/2}\,dx		\int\limits_1^4 x^{-3/2}\,dx
	&= \Bigl[\ \frac{x^{-3/2+1}}{-3/2+1}\ \Bigr]_1^4\\[5pt]		&= \Bigl[\ \frac{x^{-3/2+1}}{-3/2+1}\ \Bigr]_1^4\\[5pt]

Tek um 13:09, 21. Okt. 2008

Tek — Tue, 21 Oct 2008 13:09:40 GMT

← Nächstältere Version		Version vom 13:09, 21. Okt. 2008
Zeile 1:		Zeile 1:
-	If we rewrite	+	If we rewrite <math>\sqrt{x}</math> as <math>x^{1/2}</math>, the integrand can then be simplified using the power laws,
-	<math>\sqrt{x}</math>	+
-	as	+
-	<math>x^{{1}/{2~~}\;~~}</math>, the integrand can then be simplified using the power laws:	+

		+	{{Displayed math\|\|<math>\int\limits_1^4 \frac{\sqrt{x}}{x^2}\,dx = \int\limits_1^4 \frac{x^{1/2}}{x^2}\,dx = \int\limits_1^4 x^{1/2-2}\,dx = \int\limits_1^4 x^{-3/2}\,dx\,\textrm{.}</math>}}

-	<math>~~\int\limits_{1}^{4}{\frac{\sqrt{x}}{~~x^{2}~~}}\,dx=\int\limits_{1}^{4}{\frac{~~x^{~~\frac{~~1}~~{2}}}{x^{2}}}\,dx=\int\limits_{~~1~~}^{4}{x^{\frac{1}{2}-2}}\,dx=\int\limits_{1}^{4}{x^{-\frac{3}{2}}}\,dx~~</math>	+	We can now use the fact that a primitive function for <math>x^{n}</math> is <math>x^{n+1}/(n+1)</math> and calculate the integral's value,

-		+	{{Displayed math\|\|<math>\begin{align}
-	~~We can now use the fact that a primitive function for~~	+	\int\limits_1^4 x^{-3/2}\,dx
-	~~<math>x^~~{~~n}</math>~~	+	&= \Bigl[\ \frac{x^{-3/2+1}}{-3/2+1}\ \Bigr]_1^4\\[5pt]
-	is	+	&= \Bigl[\ \frac{x^{-1/2}}{-1/2}\ \Bigr]_1^4\\[5pt]
-	~~<math>\frac~~{~~x^{n+1}}{n+1}</~~math>	+	&= \Bigl[\ -2\frac{1}{x^{1/2}}\ \Bigr]_1^4\\[5pt]
-	~~and calculate the integral's value:~~	+	&= \Bigl[\ -\frac{2}{\sqrt{x}}\ \Bigr]_1^4\\[5pt]
-		+	&= -\frac{2}{\sqrt{4}} - \Bigl(-\frac{2}{\sqrt{1}}\Bigr)\\[5pt]
-		+	&= -\frac{2}{2}+2\\[5pt]
-	<math>\begin{align}	+	&= 1\,\textrm{.}
-	& \int\~~limits_{1}~~^{4}{x^{-~~\frac{~~3}{2}}}\,dx=\~~left~~[ \frac{x^{-~~\frac{~~3}{2}+1}}{^{-~~\frac{~~3}{2}+1}} \~~right~~]~~_{1}~~^{4} \\	+	\end{align}</math>}}
-	& =\~~left~~[ \frac{x^{-~~\frac{~~1}{2}}}{^{-~~\frac{~~1}{2}}} \~~right~~]~~_{1}~~^{4}=\~~left~~[ -2\frac{1}{x^{{1}/{2~~}\;~~}} \~~right~~]~~_{1}~~^{4} \\	+
-	& =\~~left~~[ -\frac{2}{\sqrt{x}} \~~right~~]~~_{1}~~^{4}=-\frac{2}{\sqrt{4}}-\~~left~~( -\frac{2}{\sqrt{1}} \~~right~~) \\	+
-	& =-\frac{2}{2}+2=1 \\	+
-	\end{align}</math>	+

Ian um 13:27, 17. Okt. 2008

Ian — Fri, 17 Oct 2008 13:27:26 GMT

← Nächstältere Version		Version vom 13:27, 17. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	If we rewrite
-	<~~center~~> [[~~Image:2_1_2d.gif~~]~~] <~~/~~center>~~	+	<math>\sqrt{x}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	as
		+	<math>x^{{1}/{2}\;}</math>, the integrand can then be simplified using the power laws:
		+
		+
		+	<math>\int\limits_{1}^{4}{\frac{\sqrt{x}}{x^{2}}}\,dx=\int\limits_{1}^{4}{\frac{x^{\frac{1}{2}}}{x^{2}}}\,dx=\int\limits_{1}^{4}{x^{\frac{1}{2}-2}}\,dx=\int\limits_{1}^{4}{x^{-\frac{3}{2}}}\,dx</math>
		+
		+
		+	We can now use the fact that a primitive function for
		+	<math>x^{n}</math>
		+	is
		+	<math>\frac{x^{n+1}}{n+1}</math>
		+	and calculate the integral's value:
		+
		+
		+	<math>\begin{align}
		+	& \int\limits_{1}^{4}{x^{-\frac{3}{2}}}\,dx=\left[ \frac{x^{-\frac{3}{2}+1}}{^{-\frac{3}{2}+1}} \right]_{1}^{4} \\
		+	& =\left[ \frac{x^{-\frac{1}{2}}}{^{-\frac{1}{2}}} \right]_{1}^{4}=\left[ -2\frac{1}{x^{{1}/{2}\;}} \right]_{1}^{4} \\
		+	& =\left[ -\frac{2}{\sqrt{x}} \right]_{1}^{4}=-\frac{2}{\sqrt{4}}-\left( -\frac{2}{\sqrt{1}} \right) \\
		+	& =-\frac{2}{2}+2=1 \\
		+	\end{align}</math>

Tekbot: Lösning 2.1:2d moved to Solution 2.1:2d: Robot: moved page

Tekbot — Tue, 16 Sep 2008 14:36:41 GMT

Lösning 2.1:2d moved to Solution 2.1:2d: Robot: moved page

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Version vom 14:36, 16. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 07:58:04 GMT

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

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Oskar — Mon, 07 Apr 2008 08:10:09 GMT

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