Lösung 2.3:1b - Versionsgeschichte

Sekretariat1 um 13:55, 27. Aug. 2009

2009-08-27T13:55:50Z

← Nächstältere Version		Version vom 13:55, 27. Aug. 2009
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	Die Formel für partielle Integration lautet		Die Formel für partielle Integration lautet

-	{{Abgesetzte Formel\|\|<math>\int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx~~\,,~~</math>}}	+	{{Abgesetzte Formel\|\|<math>\int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx</math>.}}

-	~~und wir~~ wählen hier <math>f(x)=\sin x</math> und <math>g(x)=x+1</math>, nachdem die Ableitung von <math>g(x)</math> nur eine Konstante ist.	+	Wir wählen hier <math>f(x)=\sin x</math> und <math>g(x)=x+1</math>, nachdem die Ableitung von <math>g(x)</math> nur eine Konstante ist.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 9:		Zeile 9:
	&= (x+1)\cdot (-\cos x) - \int 1\cdot (-\cos x)\,dx\\[5pt]		&= (x+1)\cdot (-\cos x) - \int 1\cdot (-\cos x)\,dx\\[5pt]
	&= -(x+1)\cos x + \int \cos x\,dx\\[5pt]		&= -(x+1)\cos x + \int \cos x\,dx\\[5pt]
-	&= -(x+1)\cos x + \sin x + C~~\,\textrm{.}~~	+	&= -(x+1)\cos x + \sin x + C
	\end{align}</math>}}		\end{align}</math>}}

Martin um 15:47, 5. Mai 2009

2009-05-05T15:47:25Z

← Nächstältere Version		Version vom 15:47, 5. Mai 2009
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-	~~If we look at the formula for integration by parts,~~	+	Die Formel für partielle Integration lautet

	{{Abgesetzte Formel\|\|<math>\int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,</math>}}		{{Abgesetzte Formel\|\|<math>\int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,</math>}}

-	~~we see that if we choose~~ <math>f(x)=\sin x</math> ~~and~~ <math>g(x)=x+1</math>, ~~then the factor~~ <math>g(x)</math> ~~will be differentiated to a constant on the right-hand side of the integral~~. ~~Naturally, this presupposes that we can find a primitive function for <math>f(x)</math> (which we can) and that we can then integrate it. Let's try!~~	+	und wir wählen hier <math>f(x)=\sin x</math> und <math>g(x)=x+1</math>, nachdem die Ableitung von <math>g(x)</math> nur eine Konstante ist.

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

Tekbot: Solution 2.3:1b moved to Lösung 2.3:1b: Robot: moved page

2009-03-11T10:26:15Z

Solution 2.3:1b moved to Lösung 2.3:1b: Robot: moved page

← Nächstältere Version

Version vom 10:26, 11. Mär. 2009

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

2009-03-10T13:03:34Z

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

← Nächstältere Version		Version vom 13:03, 10. Mär. 2009
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	If we look at the formula for integration by parts,		If we look at the formula for integration by parts,

-	{{~~Displayed math~~\|\|<math>\int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,</math>}}

	we see that if we choose <math>f(x)=\sin x</math> and <math>g(x)=x+1</math>, then the factor <math>g(x)</math> will be differentiated to a constant on the right-hand side of the integral. Naturally, this presupposes that we can find a primitive function for <math>f(x)</math> (which we can) and that we can then integrate it. Let's try!		we see that if we choose <math>f(x)=\sin x</math> and <math>g(x)=x+1</math>, then the factor <math>g(x)</math> will be differentiated to a constant on the right-hand side of the integral. Naturally, this presupposes that we can find a primitive function for <math>f(x)</math> (which we can) and that we can then integrate it. Let's try!

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int (x+1)\sin x\,dx		\int (x+1)\sin x\,dx
	&= (x+1)\cdot (-\cos x) - \int 1\cdot (-\cos x)\,dx\\[5pt]		&= (x+1)\cdot (-\cos x) - \int 1\cdot (-\cos x)\,dx\\[5pt]

Tek um 08:13, 29. Okt. 2008

2008-10-29T08:13:58Z

← Nächstältere Version		Version vom 08:13, 29. Okt. 2008
Zeile 1:		Zeile 1:
-	we look at the formula for ~~partial~~ integration,	+	If we look at the formula for integration by parts,

		+	{{Displayed math\|\|<math>\int f(x)g(x)\,dx = F(x)g(x) - \int F(x)g'(x)\,dx\,,</math>}}

-	<math>~~\int{~~f~~\left~~( x ~~\right~~)}g\~~left~~( x ~~\right~~)~~\,dx~~=~~F\left(~~ x ~~\right)~~g~~\left~~( x ~~\right~~)-~~\int{F\left~~( x ~~\right~~)~~{g}'\left( x \right)\,dx}~~</math>	+	we see that if we choose <math>f(x)=\sin x</math> and <math>g(x)=x+1</math>, then the factor <math>g(x)</math> will be differentiated to a constant on the right-hand side of the integral. Naturally, this presupposes that we can find a primitive function for <math>f(x)</math> (which we can) and that we can then integrate it. Let's try!

-		+	{{Displayed math\|\|<math>\begin{align}
-	~~we see that if we choose~~	+	\int (x+1)\sin x\,dx
-	~~<math>f\left( x \right)=\text~~{~~sin }x\text~~{ ~~}</~~math>	+	&= (x+1)\cdot (-\cos x) - \int 1\cdot (-\cos x)\,dx\\[5pt]
-	~~and~~	+	&= -(x+1)\cos x + \int \cos x\,dx\\[5pt]
-	~~<math>g\left( x \right)=x+\text{1}</math>, then the factor~~	+	&= -(x+1)\cos x + \sin x + C\,\textrm{.}
-	~~<math>g\left( x \right)</math>~~	+	\end{align}</math>}}
-	~~will be differentiated to a constant on the right-hand side of the integral. Naturally, this presupposes that we can find a primitive function for~~	+
-	~~<math>f\left( x \right)</math>~~	+
-	~~(which we can) and that we can then integrate it. Let's try!~~	+
-		+
-		+
-	<math>\begin{align}	+
-	& \int~~{\left~~( x+1 ~~\right~~)}\sin x\,dx=~~\left~~( x+1 ~~\right~~)\~~centerdot \left~~( -\cos x ~~\right~~)-\int{1\~~centerdot }\left~~( -\cos x ~~\right~~)\,dx \\	+
-	& =-~~\left~~( x+1 ~~\right~~)\cos x+\int{\cos x}\,dx \\	+
-	& =-~~\left~~( x+1 ~~\right~~)\cos x+\sin x+C \\	+
-	\end{align}</math>	+

Ian um 13:32, 21. Okt. 2008

2008-10-21T13:32:02Z

← Nächstältere Version		Version vom 13:32, 21. Okt. 2008
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-	{{~~NAVCONTENT_START~~}}	+	we look at the formula for partial integration,
-	<~~center~~> ~~[[Image:2_3_1b~~.~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\int{f\left( x \right)}g\left( x \right)\,dx=F\left( x \right)g\left( x \right)-\int{F\left( x \right){g}'\left( x \right)\,dx}</math>
		+
		+
		+	we see that if we choose
		+	<math>f\left( x \right)=\text{sin }x\text{ }</math>
		+	and
		+	<math>g\left( x \right)=x+\text{1}</math>, then the factor
		+	<math>g\left( x \right)</math>
		+	will be differentiated to a constant on the right-hand side of the integral. Naturally, this presupposes that we can find a primitive function for
		+	<math>f\left( x \right)</math>
		+	(which we can) and that we can then integrate it. Let's try!
		+
		+
		+	<math>\begin{align}
		+	& \int{\left( x+1 \right)}\sin x\,dx=\left( x+1 \right)\centerdot \left( -\cos x \right)-\int{1\centerdot }\left( -\cos x \right)\,dx \\
		+	& =-\left( x+1 \right)\cos x+\int{\cos x}\,dx \\
		+	& =-\left( x+1 \right)\cos x+\sin x+C \\
		+	\end{align}</math>

Tekbot: Lösning 2.3:1b moved to Solution 2.3:1b: Robot: moved page

2008-09-16T14:46:41Z

Lösning 2.3:1b moved to Solution 2.3:1b: Robot: moved page

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Version vom 14:46, 16. Sep. 2008

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

2008-08-21T08:03:04Z

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

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2008-04-07T10:03:23Z

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