Lösung 2.3:1c - Versionsgeschichte

Sekretariat1 um 07:43, 28. Aug. 2009

2009-08-28T07:43:50Z

← Nächstältere Version		Version vom 07:43, 28. Aug. 2009
Zeile 9:		Zeile 9:
	&= 2x\cdot (-\cos x) - \int 2\cdot (-\cos x)\,dx\\[5pt]		&= 2x\cdot (-\cos x) - \int 2\cdot (-\cos x)\,dx\\[5pt]
	&= -2x\cos x + 2\int \cos x\,dx\\[5pt]		&= -2x\cos x + 2\int \cos x\,dx\\[5pt]
-	&= -2x\cos x + 2\sin x + C~~\,\textrm{.}~~	+	&= -2x\cos x + 2\sin x + C
	\end{align}</math>}}		\end{align}</math>}}

Moni um 11:01, 22. Aug. 2009

2009-08-22T11:01:44Z

← Nächstältere Version		Version vom 11:01, 22. Aug. 2009
Zeile 1:		Zeile 1:
-	Wählen wir unsere Faktoren so dass wir <math>x^2</math> ableiten und <math>\cos x</math> integrieren, erhalten wir ein Integral mit ~~einen~~ linearen Term statt ~~einen~~ quadratischen.	+	Wählen wir unsere Faktoren so, dass wir <math>x^2</math> ableiten und <math>\cos x</math> integrieren, erhalten wir ein Integral mit einem linearen Term statt einem quadratischen.

	{{Abgesetzte Formel\|\|<math>\int x^2\cdot\cos x\,dx = x^2\cdot\sin x - \int 2x\cdot\sin x\,dx\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\int x^2\cdot\cos x\,dx = x^2\cdot\sin x - \int 2x\cdot\sin x\,dx\,\textrm{.}</math>}}

-	Das rechte Integral berechnen wir ähnlich wie das vorige. Wir leiten <math>2x</math>ab, und integrieren <math>\sin x</math>.	+	Das rechte Integral berechnen wir ähnlich wie das vorige. Wir leiten <math> 2x </math> ab und integrieren <math>\sin x</math>.

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

Martin um 15:53, 5. Mai 2009

2009-05-05T15:53:09Z

← Nächstältere Version		Version vom 15:53, 5. Mai 2009
Zeile 1:		Zeile 1:
-	~~The integrand consists of two factors,~~ so ~~integration by parts is a plausible method. The most obvious thing to do is to choose~~ <math>x^2</math> ~~as the factor that we will differentiate and~~ <math>\cos x</math> ~~as the factor that we will integrate~~. ~~Admittedly, the <math>x^2</math>-factor will not be differentiated away, but its exponent decreases by 1 and this makes the integral a little easier,~~	+	Wählen wir unsere Faktoren so dass wir <math>x^2</math> ableiten und <math>\cos x</math> integrieren, erhalten wir ein Integral mit einen linearen Term statt einen quadratischen.

	{{Abgesetzte Formel\|\|<math>\int x^2\cdot\cos x\,dx = x^2\cdot\sin x - \int 2x\cdot\sin x\,dx\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\int x^2\cdot\cos x\,dx = x^2\cdot\sin x - \int 2x\cdot\sin x\,dx\,\textrm{.}</math>}}

-	~~We can attack the integral on the right-hand side in the same way~~. ~~Let~~ <math>2x</math> ~~be the factor that we differentiate and~~ <math>\sin x</math> ~~the factor that we integrate~~. ~~This time, we have only one factor left,~~	+	Das rechte Integral berechnen wir ähnlich wie das vorige. Wir leiten <math>2x</math>ab, und integrieren <math>\sin x</math>.

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 12:		Zeile 12:
	\end{align}</math>}}		\end{align}</math>}}

-	~~All~~ in ~~all, we obtain~~	+	Alles in allem erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 20:		Zeile 20:
	\end{align}</math>}}		\end{align}</math>}}

-	~~For more difficult integrals~~, ~~it is quite normal to have to work step by step before getting the final answer~~.	+	'''Hinweis:''' Wenn man mehrere partielle Integrationen benötigt, rechnet man oft in Schritten bevor man die endgültige Antwort berechnet.

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

2009-03-11T10:26:35Z

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

← Nächstältere Version

Version vom 10:26, 11. Mär. 2009

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

2009-03-10T13:03:44Z

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

← Nächstältere Version		Version vom 13:03, 10. Mär. 2009
Zeile 1:		Zeile 1:
	The integrand consists of two factors, so integration by parts is a plausible method. The most obvious thing to do is to choose <math>x^2</math> as the factor that we will differentiate and <math>\cos x</math> as the factor that we will integrate. Admittedly, the <math>x^2</math>-factor will not be differentiated away, but its exponent decreases by 1 and this makes the integral a little easier,		The integrand consists of two factors, so integration by parts is a plausible method. The most obvious thing to do is to choose <math>x^2</math> as the factor that we will differentiate and <math>\cos x</math> as the factor that we will integrate. Admittedly, the <math>x^2</math>-factor will not be differentiated away, but its exponent decreases by 1 and this makes the integral a little easier,

-	{{~~Displayed math~~\|\|<math>\int x^2\cdot\cos x\,dx = x^2\cdot\sin x - \int 2x\cdot\sin x\,dx\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\int x^2\cdot\cos x\,dx = x^2\cdot\sin x - \int 2x\cdot\sin x\,dx\,\textrm{.}</math>}}

	We can attack the integral on the right-hand side in the same way. Let <math>2x</math> be the factor that we differentiate and <math>\sin x</math> the factor that we integrate. This time, we have only one factor left,		We can attack the integral on the right-hand side in the same way. Let <math>2x</math> be the factor that we differentiate and <math>\sin x</math> the factor that we integrate. This time, we have only one factor left,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int 2x\cdot \sin x\,dx		\int 2x\cdot \sin x\,dx
	&= 2x\cdot (-\cos x) - \int 2\cdot (-\cos x)\,dx\\[5pt]		&= 2x\cdot (-\cos x) - \int 2\cdot (-\cos x)\,dx\\[5pt]
Zeile 14:		Zeile 14:
	All in all, we obtain		All in all, we obtain

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\int x^2\cos x\,dx		\int x^2\cos x\,dx
	&= x^2\cdot\sin x - (-2x\cos x+2\sin x+C)\\[5pt]		&= x^2\cdot\sin x - (-2x\cos x+2\sin x+C)\\[5pt]

Tek um 08:22, 29. Okt. 2008

2008-10-29T08:22:07Z

← Nächstältere Version		Version vom 08:22, 29. Okt. 2008
Zeile 1:		Zeile 1:
-	The integrand consists of two factors, so ~~partial~~ integration is a plausible method. The most obvious thing to do is to choose	+	The integrand consists of two factors, so integration by parts is a plausible method. The most obvious thing to do is to choose <math>x^2</math> as the factor that we will differentiate and <math>\cos x</math> as the factor that we will integrate. Admittedly, the <math>x^2</math>-factor will not be differentiated away, but its exponent decreases by 1 and this makes the integral a little easier,
-	<math>x^{2}</math>	+
-	as the factor that we will differentiate and	+
-	<math>\cos x</math>	+
-	as the factor that we will integrate. Admittedly, the	+
-	<math>x^{2}</math>	+
-	-factor will not be differentiated away, but its exponent decreases by	+
-	~~<math>\text{~~1~~}</math>~~	+
-	and this makes the integral a little easier:	+

		+	{{Displayed math\|\|<math>\int x^2\cdot\cos x\,dx = x^2\cdot\sin x - \int 2x\cdot\sin x\,dx\,\textrm{.}</math>}}

-	~~<math>\int{x^{2}\cos x\,dx=x^{2}\centerdot \sin x-\int{2x\centerdot \sin x\,dx}}</math>~~	+	We can attack the integral on the right-hand side in the same way. Let <math>2x</math> be the factor that we differentiate and <math>\sin x</math> the factor that we integrate. This time, we have only one factor left,
-		+
-		+
-	We can attack the integral on the right-hand side in the same way. Let	+
-	<math>2x</math>	+
-	be the factor that we differentiate and	+
-	<math>\sin x</math>	+
-	the factor that we integrate. This time, we have only one factor left:	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \int{2x\centerdot \sin x\~~,~~dx}=2x\centerdot \left( -\cos x \right)-\int{2\centerdot }\left( -\cos x \right)\,dx \\~~	+
-	~~& =-2x\cos x+2\int{\cos x\,dx} \\~~	+
-	~~& =-2x\cos x+2\sin x+C \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\int 2x\cdot \sin x\,dx
		+	&= 2x\cdot (-\cos x) - \int 2\cdot (-\cos x)\,dx\\[5pt]
		+	&= -2x\cos x + 2\int \cos x\,dx\\[5pt]
		+	&= -2x\cos x + 2\sin x + C\,\textrm{.}
		+	\end{align}</math>}}

	All in all, we obtain		All in all, we obtain

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\int x^2\cos x\,dx
-	& \int{x^{2}\cos x\,dx=x^{2}\~~centerdot~~ \sin x-~~\left~~( -2x\cos x+2\sin x+C ~~\right~~)} \\	+	&= x^2\cdot\sin x - (-2x\cos x+2\sin x+C)\\[5pt]
-	& =x^{2~~}\centerdot~~ \sin x+2x\cos x-2\sin x+C \\	+	&= x^2\sin x + 2x\cos x - 2\sin x + C\,\textrm{.}
-	\end{align}</math>	+	\end{align}</math>}}
-		+

	For more difficult integrals, it is quite normal to have to work step by step before getting the final answer.		For more difficult integrals, it is quite normal to have to work step by step before getting the final answer.

Ian um 13:40, 21. Okt. 2008

2008-10-21T13:40:54Z

← Nächstältere Version		Version vom 13:40, 21. Okt. 2008
Zeile 1:		Zeile 1:
-	{~~{NAVCONTENT_START}~~}	+	The integrand consists of two factors, so partial integration is a plausible method. The most obvious thing to do is to choose
-	<~~center~~> ~~[[Image:2_3_1c~~-1~~(2).gif]]~~ </~~center~~>	+	<math>x^{2}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	as the factor that we will differentiate and
-	{~~{NAVCONTENT_START~~}}	+	<math>\cos x</math>
-	<~~center~~> ~~[[Image~~:~~2_3_1c~~-2(2)~~.gif]]~~ </~~center~~>	+	as the factor that we will integrate. Admittedly, the
-	{{~~NAVCONTENT_STOP~~}}	+	<math>x^{2}</math>
		+	-factor will not be differentiated away, but its exponent decreases by
		+	<math>\text{1}</math>
		+	and this makes the integral a little easier:
		+
		+
		+	<math>\int{x^{2}\cos x\,dx=x^{2}\centerdot \sin x-\int{2x\centerdot \sin x\,dx}}</math>
		+
		+
		+	We can attack the integral on the right-hand side in the same way. Let
		+	<math>2x</math>
		+	be the factor that we differentiate and
		+	<math>\sin x</math>
		+	the factor that we integrate. This time, we have only one factor left:
		+
		+
		+	<math>\begin{align}
		+	& \int{2x\centerdot \sin x\,dx}=2x\centerdot \left( -\cos x \right)-\int{2\centerdot }\left( -\cos x \right)\,dx \\
		+	& =-2x\cos x+2\int{\cos x\,dx} \\
		+	& =-2x\cos x+2\sin x+C \\
		+	\end{align}</math>
		+
		+
		+	All in all, we obtain
		+
		+
		+	<math>\begin{align}
		+	& \int{x^{2}\cos x\,dx=x^{2}\centerdot \sin x-\left( -2x\cos x+2\sin x+C \right)} \\
		+	& =x^{2}\centerdot \sin x+2x\cos x-2\sin x+C \\
		+	\end{align}</math>
		+
		+
		+	For more difficult integrals, it is quite normal to have to work step by step before getting the final answer.

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

2008-09-16T14:47:01Z

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

← Nächstältere Version

Version vom 14:47, 16. Sep. 2008

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

2008-08-21T08:03:14Z

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

← Nächstältere Version		Version vom 08:03, 21. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_3_1c-1(2).gif]] </center>	+	<center> [[Image:2_3_1c-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_3_1c-2(2).gif]] </center>	+	<center> [[Image:2_3_1c-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar um 09:35, 22. Apr. 2008

2008-04-22T09:35:26Z

← Nächstältere Version		Version vom 09:35, 22. Apr. 2008
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	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[Bild:2_3_1c- 2(2).gif]] </center>	+	<center> [[Bild:2_3_1c-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}