Lösung 2.2:2a - Versionsgeschichte

Sekretariat1 um 12:10, 27. Aug. 2009

2009-08-27T12:10:39Z

← Nächstältere Version		Version vom 12:10, 27. Aug. 2009
Zeile 1:		Zeile 1:
-	Wenn wir die Substition <math>u=5x</math> ausführen, erhalten wir ein schon bekanntes Integral,	+	Wenn wir die Substition <math>u=5x</math> ausführen, erhalten wir ein schon bekanntes Integral.

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

-	Hier haben wir <math>dx</math> mit <math>\tfrac{1}{5}\,du</math> ersetzt. Die Grenzen die wir erhalten sind <math>u=5\cdot 0=0</math>	+	Hier haben wir <math>dx</math> mit <math>\tfrac{1}{5}\,du</math> ersetzt. Die Grenzen, die wir erhalten sind <math>u=5\cdot 0=0</math>
	und <math>u=5\cdot \pi = 5\pi\,</math>.		und <math>u=5\cdot \pi = 5\pi\,</math>.

-	Wir erhalten das Integral,	+	Wir erhalten das Integral

	{{Abgesetzte Formel\|\|<math>\frac{1}{5}\int\limits_0^{5\pi} \cos u\,du = \frac{1}{5}\Bigl[\ \sin u\ \Bigr]_0^{5\pi} = \frac{1}{5}( \sin 5\pi -\sin 0) = \frac{1}{5}(0-0) = 0\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{1}{5}\int\limits_0^{5\pi} \cos u\,du = \frac{1}{5}\Bigl[\ \sin u\ \Bigr]_0^{5\pi} = \frac{1}{5}( \sin 5\pi -\sin 0) = \frac{1}{5}(0-0) = 0\,\textrm{.}</math>}}


-	Hinweis: Zeichnen wir die Graphe von <math>y=\cos 5x</math>, sehen wir ~~das~~ die gesamte Fläche oberhalb der ''x''-Achse genauso groß ist ~~wir~~ die gesamte Fläche unterhalb der ''x''-Achse ist.	+	Hinweis: Zeichnen wir die Graphe von <math>y=\cos 5x</math>, sehen wir, dass die gesamte Fläche oberhalb der ''x''-Achse genauso groß ist wie die gesamte Fläche unterhalb der ''x''-Achse ist.

	[[Image:2_2_2_a.gif\|center]]		[[Image:2_2_2_a.gif\|center]]

Martin um 12:51, 5. Mai 2009

2009-05-05T12:51:15Z

← Nächstältere Version		Version vom 12:51, 5. Mai 2009
Zeile 1:		Zeile 1:
-	~~The integral is a standard integral, with <math>5x</math> as the argument of the cosine function. If we therefore substitute~~ <math>u=5x</math>, ~~we obtain the “correct” argument of the cosine~~,	+	Wenn wir die Substition <math>u=5x</math> ausführen, erhalten wir ein schon bekanntes Integral,

	{{Abgesetzte Formel\|\|<math>\int\limits_0^{\pi} \cos 5x\,dx = \left\{\begin{align}		{{Abgesetzte Formel\|\|<math>\int\limits_0^{\pi} \cos 5x\,dx = \left\{\begin{align}
Zeile 6:		Zeile 6:
	\end{align}\right\} = \frac{1}{5}\int\limits_0^{5\pi} \cos u\,du\,\textrm{.}</math>}}		\end{align}\right\} = \frac{1}{5}\int\limits_0^{5\pi} \cos u\,du\,\textrm{.}</math>}}

-	~~As can be seen, the variable change replaced~~ <math>dx</math> by <math>\tfrac{1}{5}\,du</math> ~~and the new limits of integration become~~ <math>u=5\cdot 0=0</math>	+	Hier haben wir <math>dx</math> mit <math>\tfrac{1}{5}\,du</math> ersetzt. Die Grenzen die wir erhalten sind <math>u=5\cdot 0=0</math>
-	~~and~~ <math>u=5\cdot \pi = 5\pi\,</math>.	+	und <math>u=5\cdot \pi = 5\pi\,</math>.

-	~~Now, we have a standard integral which we can easily compute~~,	+	Wir erhalten das Integral,

	{{Abgesetzte Formel\|\|<math>\frac{1}{5}\int\limits_0^{5\pi} \cos u\,du = \frac{1}{5}\Bigl[\ \sin u\ \Bigr]_0^{5\pi} = \frac{1}{5}( \sin 5\pi -\sin 0) = \frac{1}{5}(0-0) = 0\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\frac{1}{5}\int\limits_0^{5\pi} \cos u\,du = \frac{1}{5}\Bigl[\ \sin u\ \Bigr]_0^{5\pi} = \frac{1}{5}( \sin 5\pi -\sin 0) = \frac{1}{5}(0-0) = 0\,\textrm{.}</math>}}


-	~~Note~~: ~~If we draw the graph of~~ <math>y=\cos 5x</math>, ~~we see also that the area between the curve and~~ ''x''-~~axis above the~~ ''x''-~~axis is the same as the area under the ''x''-axis~~.	+	Hinweis: Zeichnen wir die Graphe von <math>y=\cos 5x</math>, sehen wir das die gesamte Fläche oberhalb der ''x''-Achse genauso groß ist wir die gesamte Fläche unterhalb der ''x''-Achse ist.

	[[Image:2_2_2_a.gif\|center]]		[[Image:2_2_2_a.gif\|center]]

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

2009-03-11T10:21:15Z

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

← Nächstältere Version

Version vom 10:21, 11. Mär. 2009

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

2009-03-10T13:01:04Z

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

← Nächstältere Version		Version vom 13:01, 10. Mär. 2009
Zeile 1:		Zeile 1:
	The integral is a standard integral, with <math>5x</math> as the argument of the cosine function. If we therefore substitute <math>u=5x</math>, we obtain the “correct” argument of the cosine,		The integral is a standard integral, with <math>5x</math> as the argument of the cosine function. If we therefore substitute <math>u=5x</math>, we obtain the “correct” argument of the cosine,

-	{{~~Displayed math~~\|\|<math>\int\limits_0^{\pi} \cos 5x\,dx = \left\{\begin{align}	+	{{Abgesetzte Formel\|\|<math>\int\limits_0^{\pi} \cos 5x\,dx = \left\{\begin{align}
	u &= 5x\\[5pt]		u &= 5x\\[5pt]
	du &= (5x)'\,dx = 5\,dx		du &= (5x)'\,dx = 5\,dx
Zeile 11:		Zeile 11:
	Now, we have a standard integral which we can easily compute,		Now, we have a standard integral which we can easily compute,

-	{{~~Displayed math~~\|\|<math>\frac{1}{5}\int\limits_0^{5\pi} \cos u\,du = \frac{1}{5}\Bigl[\ \sin u\ \Bigr]_0^{5\pi} = \frac{1}{5}( \sin 5\pi -\sin 0) = \frac{1}{5}(0-0) = 0\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\frac{1}{5}\int\limits_0^{5\pi} \cos u\,du = \frac{1}{5}\Bigl[\ \sin u\ \Bigr]_0^{5\pi} = \frac{1}{5}( \sin 5\pi -\sin 0) = \frac{1}{5}(0-0) = 0\,\textrm{.}</math>}}

Tek um 12:57, 28. Okt. 2008

2008-10-28T12:57:14Z

← Nächstältere Version		Version vom 12:57, 28. Okt. 2008
Zeile 1:		Zeile 1:
-	The integral is a standard integral, with	+	The integral is a standard integral, with <math>5x</math> as the argument of the cosine function. If we therefore substitute <math>u=5x</math>, we obtain the “correct” argument of the cosine,
-	<math>~~\text{5}x~~</math>	+
-	as the argument of the cosine function. If we therefore substitute	+
-	<math>u=~~\text{5}x~~</math>, we obtain the “correct” argument of the cosine,	+

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

-	<math>\~~begin~~{~~align~~}	+	As can be seen, the variable change replaced <math>dx</math> by <math>\tfrac{1}{5}\,du</math> and the new limits of integration become <math>u=5\cdot 0=0</math>
-	~~& \int\limits_~~{0}~~^{\pi }{\cos 5x~~\,~~dx=\left\{ \begin{array}{*{35}l}~~	+	and <math>u=5\cdot \pi = 5\pi\,</math>.
-	u=~~\text{~~5~~}x \~~\	+
-	~~du=\left( 5x \right)^{\prime }\,dx~~=5\~~,dx~~ \\	+
-	~~\end{array} \right\}} \\~~	+
-	& =~~\frac{1}{5}\int\limits_{0}^{~~5\pi ~~}{\cos u~~\,~~du} \\~~	+
-	~~\end{align}~~</math>	+

-	~~As can be seen~~, ~~the variable change replaced~~	+	Now, we have a standard integral which we can easily compute,
-	~~<math>dx</math>~~	+
-	by	+
-	~~<math>\frac{1}{5}\~~,~~du</math>~~	+
-	~~and the new limits of integration become~~	+
-	~~<math>u=5\centerdot 0=0</math>~~	+
-	~~and~~	+
-	~~<math>u=5\centerdot \pi =5\pi </math>.~~	+

-	~~Now~~, ~~we have a standard integral which can easily compute:~~	+	{{Displayed math\|\|<math>\frac{1}{5}\int\limits_0^{5\pi} \cos u\,du = \frac{1}{5}\Bigl[\ \sin u\ \Bigr]_0^{5\pi} = \frac{1}{5}( \sin 5\pi -\sin 0) = \frac{1}{5}(0-0) = 0\,\textrm{.}</math>}}


-	~~<math>\begin{align}~~	+	Note: If we draw the graph of <math>y=\cos 5x</math>, we see also that the area between the curve and ''x''-axis above the ''x''-axis is the same as the area under the ''x''-axis.
-	~~& \frac{1}{5}\int\limits_{0}^{5\pi }{\cos u\,du}=\frac{1}{5}\left[ \sin u \right]_{0}^{5\pi } \\~~	+
-	~~& =\frac{1}{5}\left( \sin 5\pi -\sin 0 \right)=\frac{1}{5}\left( 0-0 \right)=0 \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	~~NOTE~~: if we draw the graph ~~for~~	+
-	<math>y=\cos 5x</math>, we see also that the area between the curve and	+
-	~~<math>~~x~~</math>~~	+
-	-axis above the	+
-	~~<math>~~x~~</math>~~	+
-	-axis is the same as the area under the	+
-	~~<math>~~x~~</math>~~	+
-	-axis.	+
-		+

	[[Image:2_2_2_a.gif\|center]]		[[Image:2_2_2_a.gif\|center]]

Ian um 10:07, 19. Okt. 2008

2008-10-19T10:07:10Z

← Nächstältere Version		Version vom 10:07, 19. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START}~~}	+	The integral is a standard integral, with
-	<~~center~~> ~~[[Image:2_2_2a-1~~(2)~~.gif]]~~ </~~center~~>	+	<math>\text{5}x</math>
-	{{~~NAVCONTENT_STOP~~}}	+	as the argument of the cosine function. If we therefore substitute
-	{{~~NAVCONTENT_START~~}}	+	<math>u=\text{5}x</math>, we obtain the “correct” argument of the cosine,
-	<~~center~~> ~~[[Image~~:~~2_2_2a~~-~~2(2).gif]]~~ </~~center~~>	+
-	~~{{NAVCONTENT_STOP}}~~	+
		+	<math>\begin{align}
		+	& \int\limits_{0}^{\pi }{\cos 5x\,dx=\left\{ \begin{array}{*{35}l}
		+	u=\text{5}x \\
		+	du=\left( 5x \right)^{\prime }\,dx=5\,dx \\
		+	\end{array} \right\}} \\
		+	& =\frac{1}{5}\int\limits_{0}^{5\pi }{\cos u\,du} \\
		+	\end{align}</math>
		+
		+	As can be seen, the variable change replaced
		+	<math>dx</math>
		+	by
		+	<math>\frac{1}{5}\,du</math>
		+	and the new limits of integration become
		+	<math>u=5\centerdot 0=0</math>
		+	and
		+	<math>u=5\centerdot \pi =5\pi </math>.
		+
		+	Now, we have a standard integral which can easily compute:
		+
		+
		+	<math>\begin{align}
		+	& \frac{1}{5}\int\limits_{0}^{5\pi }{\cos u\,du}=\frac{1}{5}\left[ \sin u \right]_{0}^{5\pi } \\
		+	& =\frac{1}{5}\left( \sin 5\pi -\sin 0 \right)=\frac{1}{5}\left( 0-0 \right)=0 \\
		+	\end{align}</math>
		+
		+
		+	NOTE: if we draw the graph for
		+	<math>y=\cos 5x</math>, we see also that the area between the curve and
		+	<math>x</math>
		+	-axis above the
		+	<math>x</math>
		+	-axis is the same as the area under the
		+	<math>x</math>
		+	-axis.


	[[Image:2_2_2_a.gif\|center]]		[[Image:2_2_2_a.gif\|center]]

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

2008-09-16T14:41:41Z

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

← Nächstältere Version

Version vom 14:41, 16. Sep. 2008

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

2008-08-21T08:00:34Z

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

← Nächstältere Version		Version vom 08:00, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_2_2a-1(2).gif]] </center>	+	<center> [[Image:2_2_2a-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_2_2a-2(2).gif]] </center>	+	<center> [[Image:2_2_2a-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}


-	[[~~Bild~~:2_2_2_a.gif\|center]]	+	[[Image:2_2_2_a.gif\|center]]

Madde um 16:39, 19. Aug. 2008

2008-08-19T16:39:34Z

← Nächstältere Version		Version vom 16:39, 19. Aug. 2008
Zeile 5:		Zeile 5:
	<center> [[Bild:2_2_2a-2(2).gif]] </center>		<center> [[Bild:2_2_2a-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+
		+
		+	[[Bild:2_2_2_a.gif\|center]]

Oskar um 09:23, 7. Apr. 2008

2008-04-07T09:23:50Z

← Nächstältere Version		Version vom 09:23, 7. Apr. 2008
Zeile 3:		Zeile 3:
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[Bild:2_2_2a-1(2).gif]] </center>	+	<center> [[Bild:2_2_2a-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}