Lösung 2.1:4e - Versionsgeschichte

Sekretariat1 um 11:23, 27. Aug. 2009

Sekretariat1 — Thu, 27 Aug 2009 11:23:01 GMT

← Nächstältere Version		Version vom 11:23, 27. Aug. 2009
Zeile 7:		Zeile 7:
	Die Fläche des Gebietes ist		Die Fläche des Gebietes ist

-	{{Abgesetzte Formel\|\|<math>\text{Fläche} = \int\limits_a^b \bigl(x+2-x^2\bigr)\,dx\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\text{Fläche} = \int\limits_a^b \bigl(x+2-x^2\bigr)\,dx\,\textrm{,}</math>}}

-	wo <math>x=a</math> und <math>x=b</math> die Schnittstellen der beiden Kurven sind, die wir durch folgende Gleichung erhalten	+	wobei <math>x=a</math> und <math>x=b</math> die Schnittstellen der beiden Kurven sind, die wir durch folgende Gleichung erhalten

	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}		{{Abgesetzte Formel\|\|<math>\left\{\begin{align}
Zeile 16:		Zeile 16:
	\end{align} \right.</math>}}		\end{align} \right.</math>}}

-	Eliminieren wir <math>y</math>, erhalten wir ~~eine Gleichung~~ für <math>x</math>,	+	Eliminieren wir <math>y</math>, erhalten wir für <math>x</math> diese Gleichung

	{{Abgesetzte Formel\|\|<math>x^{2}=x+2\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x^{2}=x+2\,\textrm{.}</math>}}
Zeile 22:		Zeile 22:
	Hohlen wir alle ''x''-Terme zu einer Seite erhalten wir		Hohlen wir alle ''x''-Terme zu einer Seite erhalten wir

-	{{Abgesetzte Formel\|\|<math>x^2-x=2\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>x^2-x=2\,</math>.}}

-	~~und durch~~ quadratische Ergänzung ~~erhalten wir~~	+	Durch quadratische Ergänzung ergibt sich

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 31:		Zeile 31:
	\end{align}</math>}}		\end{align}</math>}}

-	~~wir~~ erhalten also die Wurzeln <math>x=\tfrac{1}{2}\pm \tfrac{3}{2}</math>, oder <math>x=-1</math> und <math>x=2\,</math>.	+	Wir erhalten also die Wurzeln <math>x=\tfrac{1}{2}\pm \tfrac{3}{2}</math>, oder <math>x=-1</math> und <math>x=2\,</math>.

	Die Fläche ist also		Die Fläche ist also

Moni um 22:44, 21. Aug. 2009

Moni — Fri, 21 Aug 2009 22:44:17 GMT

← Nächstältere Version		Version vom 22:44, 21. Aug. 2009
Zeile 1:		Zeile 1:
-	~~die~~ doppelte ~~ungleichung~~ bedeutet dass ''y'' zwischen den Kurven <math>y=x+2</math> und <math>y=x^2</math> liegt.	+	Die doppelte Ungleichung bedeutet, dass ''y'' zwischen den Kurven <math>y=x+2</math> und <math>y=x^2</math> liegt.

-	In der Figur unten ist das ~~gebiet~~ eingezeichnet.	+	In der Figur unten ist das Gebiet eingezeichnet.

	[[Image:2_1_4_e.gif\|center]]		[[Image:2_1_4_e.gif\|center]]
Zeile 7:		Zeile 7:
	Die Fläche des Gebietes ist		Die Fläche des Gebietes ist

-	{{Abgesetzte Formel\|\|<math>\text{~~Area~~} = \int\limits_a^b \bigl(x+2-x^2\bigr)\,dx\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\text{Fläche} = \int\limits_a^b \bigl(x+2-x^2\bigr)\,dx\,\textrm{.}</math>}}

-	wo <math>x=a</math> und <math>x=b</math> die Schnittstellen der beiden Kurven sind, die wir durch folgende Gleichung erhalten,	+	wo <math>x=a</math> und <math>x=b</math> die Schnittstellen der beiden Kurven sind, die wir durch folgende Gleichung erhalten

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

Martin um 19:28, 28. Apr. 2009

Martin — Tue, 28 Apr 2009 19:28:54 GMT

← Nächstältere Version		Version vom 19:28, 28. Apr. 2009
Zeile 1:		Zeile 1:
-	~~The double inequality means that we look for the area of the region which is bounded above in the~~ y~~-direction by the straight line~~ <math>y=x+2</math> ~~and from below by the parabola~~ <math>y=x^2</math>.	+	die doppelte ungleichung bedeutet dass ''y'' zwischen den Kurven <math>y=x+2</math> und <math>y=x^2</math> liegt.

-	~~If we sketch the line and the parabola, the region is given by the region shaded in the figure below~~.	+	In der Figur unten ist das gebiet eingezeichnet.

	[[Image:2_1_4_e.gif\|center]]		[[Image:2_1_4_e.gif\|center]]

-	~~As soon as we have determined the ''x''-coordinates of the points of intersection,~~	+	Die Fläche des Gebietes ist
-	~~<math>x=a</math> and <math>x=b</math>, between the line and the parabola, we can calculate the area as the integral of the difference between the curves' ''y''-values,~~	+

	{{Abgesetzte Formel\|\|<math>\text{Area} = \int\limits_a^b \bigl(x+2-x^2\bigr)\,dx\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\text{Area} = \int\limits_a^b \bigl(x+2-x^2\bigr)\,dx\,\textrm{.}</math>}}

-	~~The curves' points of intersection are those points which lie on both curves~~, ~~i.e. which satisfy both curves' equations~~	+	wo <math>x=a</math> und <math>x=b</math> die Schnittstellen der beiden Kurven sind, die wir durch folgende Gleichung erhalten,

	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}		{{Abgesetzte Formel\|\|<math>\left\{\begin{align}
Zeile 17:		Zeile 16:
	\end{align} \right.</math>}}		\end{align} \right.</math>}}

-	~~By eliminating~~ <math>y</math>, ~~we obtain an equation for~~ <math>x</math>,	+	Eliminieren wir <math>y</math>, erhalten wir eine Gleichung für <math>x</math>,

	{{Abgesetzte Formel\|\|<math>x^{2}=x+2\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x^{2}=x+2\,\textrm{.}</math>}}

-	~~If we move all~~ ''x''-~~terms to the left-hand side,~~	+	Hohlen wir alle ''x''-Terme zu einer Seite erhalten wir

	{{Abgesetzte Formel\|\|<math>x^2-x=2\,,</math>}}		{{Abgesetzte Formel\|\|<math>x^2-x=2\,,</math>}}

-	~~and complete the square, we obtain~~	+	und durch quadratische Ergänzung erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 32:		Zeile 31:
	\end{align}</math>}}		\end{align}</math>}}

-	~~Taking the root then gives that~~ <math>x=\tfrac{1}{2}\pm \tfrac{3}{2}</math>~~. In other words~~, <math>x=-1</math> ~~and~~ <math>x=2\,</math>.	+	wir erhalten also die Wurzeln <math>x=\tfrac{1}{2}\pm \tfrac{3}{2}</math>, oder <math>x=-1</math> und <math>x=2\,</math>.

-	~~The area of the region is now given by~~	+	Die Fläche ist also

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
-	\text{~~Area~~}	+	\text{Fläche}
	&= \int\limits_{-1}^2 \bigl(x+2-x^2\bigr)\,dx\\[5pt]		&= \int\limits_{-1}^2 \bigl(x+2-x^2\bigr)\,dx\\[5pt]
	&= \Bigl[\ \frac{x^2}{2} + 2x - \frac{x^3}{3}\ \Bigr]_{-1}^2\\[5pt]		&= \Bigl[\ \frac{x^2}{2} + 2x - \frac{x^3}{3}\ \Bigr]_{-1}^2\\[5pt]

Tekbot: Solution 2.1:4e moved to Lösung 2.1:4e: Robot: moved page

Tekbot — Wed, 11 Mar 2009 10:19:15 GMT

Solution 2.1:4e moved to Lösung 2.1:4e: Robot: moved page

← Nächstältere Version

Version vom 10:19, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 13:00:04 GMT

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

← Nächstältere Version		Version vom 13:00, 10. Mär. 2009
Zeile 8:		Zeile 8:
	<math>x=a</math> and <math>x=b</math>, between the line and the parabola, we can calculate the area as the integral of the difference between the curves' ''y''-values,		<math>x=a</math> and <math>x=b</math>, between the line and the parabola, we can calculate the area as the integral of the difference between the curves' ''y''-values,

-	{{~~Displayed math~~\|\|<math>\text{Area} = \int\limits_a^b \bigl(x+2-x^2\bigr)\,dx\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\text{Area} = \int\limits_a^b \bigl(x+2-x^2\bigr)\,dx\,\textrm{.}</math>}}

	The curves' points of intersection are those points which lie on both curves, i.e. which satisfy both curves' equations		The curves' points of intersection are those points which lie on both curves, i.e. which satisfy both curves' equations

-	{{~~Displayed math~~\|\|<math>\left\{\begin{align}	+	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}
	y &= x+2\,,\\[5pt]		y &= x+2\,,\\[5pt]
	y &= x^2\,\textrm{.}		y &= x^2\,\textrm{.}
Zeile 19:		Zeile 19:
	By eliminating <math>y</math>, we obtain an equation for <math>x</math>,		By eliminating <math>y</math>, we obtain an equation for <math>x</math>,

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

	If we move all ''x''-terms to the left-hand side,		If we move all ''x''-terms to the left-hand side,

-	{{~~Displayed math~~\|\|<math>x^2-x=2\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>x^2-x=2\,,</math>}}

	and complete the square, we obtain		and complete the square, we obtain

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\Bigl(x-\frac{1}{2}\Bigr)^2 - \Bigl(\frac{1}{2}\Bigr)^2 &= 2\\[5pt]		\Bigl(x-\frac{1}{2}\Bigr)^2 - \Bigl(\frac{1}{2}\Bigr)^2 &= 2\\[5pt]
	\Bigl(x-\frac{1}{2}\Bigr)^2 &= \frac{9}{4}\,\textrm{.}		\Bigl(x-\frac{1}{2}\Bigr)^2 &= \frac{9}{4}\,\textrm{.}
Zeile 36:		Zeile 36:
	The area of the region is now given by		The area of the region is now given by

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\text{Area}		\text{Area}
	&= \int\limits_{-1}^2 \bigl(x+2-x^2\bigr)\,dx\\[5pt]		&= \int\limits_{-1}^2 \bigl(x+2-x^2\bigr)\,dx\\[5pt]

Tek um 09:18, 28. Okt. 2008

Tek — Tue, 28 Oct 2008 09:18:45 GMT

← Nächstältere Version		Version vom 09:18, 28. Okt. 2008
Zeile 1:		Zeile 1:
-	The double inequality means that we look for the area of the region which is bounded above in the y-direction by the straight line	+	The double inequality means that we look for the area of the region which is bounded above in the y-direction by the straight line <math>y=x+2</math> and from below by the parabola <math>y=x^2</math>.
-	<math>y=x+~~\text{~~2}</math>	+
-	and from below by the parabola	+
-	<math>y=x^{2}</math>.	+

	If we sketch the line and the parabola, the region is given by the region shaded in the figure below.		If we sketch the line and the parabola, the region is given by the region shaded in the figure below.
-
-

	[[Image:2_1_4_e.gif\|center]]		[[Image:2_1_4_e.gif\|center]]

		+	As soon as we have determined the ''x''-coordinates of the points of intersection,
		+	<math>x=a</math> and <math>x=b</math>, between the line and the parabola, we can calculate the area as the integral of the difference between the curves' ''y''-values,

-	~~As soon as we have determined the~~	+	{{Displayed math\|\|<math>\text{Area} = \int\limits_a^b \bigl(x+2-x^2\bigr)\,dx\,\textrm{.}</math>}}
-	<math>x~~</math>~~	+
-	-~~coordinates of the points of intersection,~~	+
-	~~<math>~~x~~=a</math>~~	+
-	~~and~~	+
-	~~<math>x=b</math>~~, ~~between the line and the parabola~~, ~~we can calculate the area as the integral of the difference between the curves'~~	+
-	~~<math>y~~</math>	+
-	~~-values:~~	+

-	~~Area=~~	+	The curves' points of intersection are those points which lie on both curves, i.e. which satisfy both curves' equations
-	~~<math>\int\limits_{a}^{b}{\left( x+2-x^{2} \right)}\~~,~~dx</math>~~	+

		+	{{Displayed math\|\|<math>\left\{\begin{align}
		+	y &= x+2\,,\\[5pt]
		+	y &= x^2\,\textrm{.}
		+	\end{align} \right.</math>}}

-	~~The curves' points of intersection are those points which lie on both curves, i.e. which satisfy both curves' equations:~~	+	By eliminating <math>y</math>, we obtain an equation for <math>x</math>,
-		+
-		+
-	~~<math>\left\{ \begin{matrix}~~	+
-	~~y=x+\text{2} \\~~	+
-	~~y=x^{2} \\~~	+
-	~~\end{matrix} \right.</math>~~	+
-		+
-		+
-	By eliminating	+
-	<math>y</math>, we obtain an equation for	+
-	<math>x</math>,	+

		+	{{Displayed math\|\|<math>x^{2}=x+2\,\textrm{.}</math>}}

-	~~<math>x^{2}=x+2</math>~~	+	If we move all ''x''-terms to the left-hand side,
-		+
-		+
-	If we move all x-terms to the left-hand side,	+
-		+
-		+
-	~~<math>x^{2}-x=2</math>~~	+

		+	{{Displayed math\|\|<math>x^2-x=2\,,</math>}}

	and complete the square, we obtain		and complete the square, we obtain

		+	{{Displayed math\|\|<math>\begin{align}
		+	\Bigl(x-\frac{1}{2}\Bigr)^2 - \Bigl(\frac{1}{2}\Bigr)^2 &= 2\\[5pt]
		+	\Bigl(x-\frac{1}{2}\Bigr)^2 &= \frac{9}{4}\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	Taking the root then gives that <math>x=\tfrac{1}{2}\pm \tfrac{3}{2}</math>. In other words, <math>x=-1</math> and <math>x=2\,</math>.
-	~~& \left( x-\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}=2 \\~~	+
-	~~& \left( x-\frac{1}{2} \right)^{2}=\frac{9}{4} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-	Taking the root then gives that	+
-	<math>x=\~~frac~~{1}{2}\pm \~~frac~~{3}{2}</math>. In other words,	+
-	<math>x=-~~\text{~~1}</math>	+
-	and	+
-	<math>x=\~~text{2}~~</math>.	+

	The area of the region is now given by		The area of the region is now given by

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\text{Area}
-	& \text{Area}=\int\limits_{-1}^{2}{\~~left~~( x+2-x^{2} \~~right~~)}\,dx=\~~left~~[ \frac{x^{2}}{2}+2x-\frac{x^{3}}{3} \~~right~~]_{-1}^{2} \\	+	&= \int\limits_{-1}^2 \bigl(x+2-x^2\bigr)\,dx\\[5pt]
-	& =\frac{2^{2}}{2}+2\~~centerdot~~ 2-\frac{2^{3}}{3}-\~~left~~( \frac{~~\left~~( -1 ~~\right~~)^{2}}{2}+2\~~centerdot \left~~( -1 ~~\right~~)-\frac{~~\left~~( -1 ~~\right~~)^{3}}{3} \~~right~~) \\	+	&= \Bigl[\ \frac{x^2}{2} + 2x - \frac{x^3}{3}\ \Bigr]_{-1}^2\\[5pt]
-	& =2+4-\frac{8}{3}-\frac{1}{2}+2-\frac{1}{3} \\	+	&= \frac{2^2}{2} + 2\cdot 2 - \frac{2^3}{3} - \Bigl( \frac{(-1)^2}{2} + 2\cdot (-1) - \frac{(-1)^3}{3}\Bigr)\\[5pt]
-	& =\frac{9}{2} \\	+	&= 2 + 4 - \frac{8}{3} - \frac{1}{2} + 2 - \frac{1}{3}\\[5pt]
-	\end{align}</math>	+	&= \frac{9}{2}\,\textrm{.}
		+	\end{align}</math>}}

Ian um 13:05, 18. Okt. 2008

Ian — Sat, 18 Oct 2008 13:05:02 GMT

← Nächstältere Version		Version vom 13:05, 18. Okt. 2008
Zeile 1:		Zeile 1:
-	~~{{NAVCONTENT_START}}~~	+	The double inequality means that we look for the area of the region which is bounded above in the y-direction by the straight line
-	<~~center~~> ~~[[Image:2_1_4e-1(3).gif]]~~ </~~center~~>	+	<math>y=x+\text{2}</math>
-	~~{{NAVCONTENT_STOP}}~~	+	and from below by the parabola
-	~~{{NAVCONTENT_START}}~~	+	<math>y=x^{2}</math>.
-	<~~center~~> ~~[[Image:2_1_4e-~~2~~(3).gif]]~~ </~~center~~>	+
-	~~{{NAVCONTENT_STOP}}~~	+	If we sketch the line and the parabola, the region is given by the region shaded in the figure below.
-	~~{{NAVCONTENT_START}}~~	+
-	~~<center> [[Image:2_1_4e-3(3)~~.~~gif]] </center>~~	+
-	~~{{NAVCONTENT_STOP}}~~	+


	[[Image:2_1_4_e.gif\|center]]		[[Image:2_1_4_e.gif\|center]]
		+
		+
		+	As soon as we have determined the
		+	<math>x</math>
		+	-coordinates of the points of intersection,
		+	<math>x=a</math>
		+	and
		+	<math>x=b</math>, between the line and the parabola, we can calculate the area as the integral of the difference between the curves'
		+	<math>y</math>
		+	-values:
		+
		+	Area=
		+	<math>\int\limits_{a}^{b}{\left( x+2-x^{2} \right)}\,dx</math>
		+
		+
		+	The curves' points of intersection are those points which lie on both curves, i.e. which satisfy both curves' equations:
		+
		+
		+	<math>\left\{ \begin{matrix}
		+	y=x+\text{2} \\
		+	y=x^{2} \\
		+	\end{matrix} \right.</math>
		+
		+
		+	By eliminating
		+	<math>y</math>, we obtain an equation for
		+	<math>x</math>,
		+
		+
		+	<math>x^{2}=x+2</math>
		+
		+
		+	If we move all x-terms to the left-hand side,
		+
		+
		+	<math>x^{2}-x=2</math>
		+
		+
		+	and complete the square, we obtain
		+
		+
		+	<math>\begin{align}
		+	& \left( x-\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}=2 \\
		+	& \left( x-\frac{1}{2} \right)^{2}=\frac{9}{4} \\
		+	\end{align}</math>
		+
		+	Taking the root then gives that
		+	<math>x=\frac{1}{2}\pm \frac{3}{2}</math>. In other words,
		+	<math>x=-\text{1}</math>
		+	and
		+	<math>x=\text{2}</math>.
		+
		+	The area of the region is now given by
		+
		+
		+	<math>\begin{align}
		+	& \text{Area}=\int\limits_{-1}^{2}{\left( x+2-x^{2} \right)}\,dx=\left[ \frac{x^{2}}{2}+2x-\frac{x^{3}}{3} \right]_{-1}^{2} \\
		+	& =\frac{2^{2}}{2}+2\centerdot 2-\frac{2^{3}}{3}-\left( \frac{\left( -1 \right)^{2}}{2}+2\centerdot \left( -1 \right)-\frac{\left( -1 \right)^{3}}{3} \right) \\
		+	& =2+4-\frac{8}{3}-\frac{1}{2}+2-\frac{1}{3} \\
		+	& =\frac{9}{2} \\
		+	\end{align}</math>

Tekbot: Lösning 2.1:4e moved to Solution 2.1:4e: Robot: moved page

Tekbot — Tue, 16 Sep 2008 14:39:41 GMT

Lösning 2.1:4e moved to Solution 2.1:4e: Robot: moved page

← Nächstältere Version

Version vom 14:39, 16. Sep. 2008

Madde um 12:06, 21. Aug. 2008

Madde — Thu, 21 Aug 2008 12:06:08 GMT

← Nächstältere Version		Version vom 12:06, 21. Aug. 2008
Zeile 8:		Zeile 8:
	<center> [[Image:2_1_4e-3(3).gif]] </center>		<center> [[Image:2_1_4e-3(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+
		+
		+	[[Image:2_1_4_e.gif\|center]]

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

Tekbot — Thu, 21 Aug 2008 07:59:34 GMT

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

← Nächstältere Version		Version vom 07:59, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_1_4e-1(3).gif]] </center>	+	<center> [[Image:2_1_4e-1(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_1_4e-2(3).gif]] </center>	+	<center> [[Image:2_1_4e-2(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_1_4e-3(3).gif]] </center>	+	<center> [[Image:2_1_4e-3(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}