Lösung 2.3:1c - Versionsgeschichte

Sekretariat1 um 07:43, 28. Aug. 2009

Sekretariat1 — Fri, 28 Aug 2009 07:43:50 GMT

← Nächstältere Version		Version vom 07:43, 28. Aug. 2009
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	&= 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

Moni — Sat, 22 Aug 2009 11:01:44 GMT

← 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

Martin — Tue, 05 May 2009 15:53:09 GMT

← 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

Tekbot — Wed, 11 Mar 2009 10:26:35 GMT

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)

Tekbot — Tue, 10 Mar 2009 13:03:44 GMT

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

Tek — Wed, 29 Oct 2008 08:22:07 GMT

← 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

Ian — Tue, 21 Oct 2008 13:40:54 GMT

← 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

Tekbot — Tue, 16 Sep 2008 14:47:01 GMT

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:)

Tekbot — Thu, 21 Aug 2008 08:03:14 GMT

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

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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

Oskar — Tue, 22 Apr 2008 09:35:26 GMT

← 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}}