Lösung 2.1:4c - Versionsgeschichte

Sekretariat1 um 11:02, 27. Aug. 2009

Sekretariat1 — Thu, 27 Aug 2009 11:02:52 GMT

← Nächstältere Version		Version vom 11:02, 27. Aug. 2009
Zeile 17:		Zeile 17:
	\end{align}\right.</math>}}		\end{align}\right.</math>}}

-	Eliminieren wir ''y'' und lösen die Gleichung für ''x'', erhalten wir die Schnittstellen	+	Eliminieren wir ''y'' und lösen die Gleichung für ''x'', erhalten wir die Schnittstellen.

-	{{Abgesetzte Formel\|\|<math>8-\tfrac{1}{8}x^2 = \tfrac{1}{4}x^2+2\,~~\textrm{.}~~</math>}}	+	{{Abgesetzte Formel\|\|<math>8-\tfrac{1}{8}x^2 = \tfrac{1}{4}x^2+2\,</math>}}

	oder		oder

-	{{Abgesetzte Formel\|\|<math>\tfrac{1}{4}x^2 + \tfrac{1}{8}x^2 = 8-2\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\tfrac{1}{4}x^2 + \tfrac{1}{8}x^2 = 8-2\,</math>,}}

	also ist		also ist

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
-	\bigl(\tfrac{1}{4}+\tfrac{1}{8}\bigr)x^2 &= 6\,,\\[5pt]	+	\bigl(\tfrac{1}{4}+\tfrac{1}{8}\bigr)x^2 &= 6\,\\[5pt]
-	\tfrac{3}{8}x^2 &= 6\,,\\[5pt]	+	\tfrac{3}{8}x^2 &= 6\,\\[5pt]
	x^2 &= 16\,\textrm{.}		x^2 &= 16\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}
Zeile 35:		Zeile 35:
	Also sind die Schnittstellen <math>x=-4</math> und <math>x=4\,</math>.		Also sind die Schnittstellen <math>x=-4</math> und <math>x=4\,</math>.

-	Die ~~Fälche~~ zwischen den Kurven ist also	+	Die Fläche zwischen den Kurven ist also

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

Moni um 22:30, 21. Aug. 2009

Moni — Fri, 21 Aug 2009 22:30:42 GMT

← Nächstältere Version		Version vom 22:30, 21. Aug. 2009
Zeile 1:		Zeile 1:
-	Zuerst ~~sehen~~ wir in der ~~Figur welches unseres~~ Gebiet ~~ist~~.	+	Zuerst betrachten wir den Graphen, der unser Gebiet einschließt.
-		+	Die Kurven <math>y=x^2/4+2</math> und <math>y=8-x^2/8</math> sind beide quadratische Funktionen. Die erste hat ihr Minimum <math>y=2</math> wenn <math>x=0</math> und die zweite das Maximum <math>y=8</math> wenn <math>x=0</math>. Die Kurven sehen ungefähr wie im Bild unten aus.
-	Die Kurven <math>y=x^2/4+2</math> und <math>y=8-x^2/8</math> sind beide quadratische Funktionen. Die erste hat ~~das~~ Minimum <math>y=2</math> wenn <math>x=0</math>, und die zweite das Maximum <math>y=8</math> wenn <math>x=0</math>. Die Kurven sehen ungefähr wie im Bild unten aus.	+

	[[Image:2_1_4_c.gif\|center]]		[[Image:2_1_4_c.gif\|center]]
Zeile 7:		Zeile 6:
	Die obere Grenze der Fläche ist die Funktion <math>y=8-x^2/8</math> und die untere Grenze die Funktion <math>y=x^2/4+2</math>. Bestimmen wir die Schnittstellen <math>x=a</math> und <math>x=b</math> der beiden Kurven, ist die Fläche		Die obere Grenze der Fläche ist die Funktion <math>y=8-x^2/8</math> und die untere Grenze die Funktion <math>y=x^2/4+2</math>. Bestimmen wir die Schnittstellen <math>x=a</math> und <math>x=b</math> der beiden Kurven, ist die Fläche

-	{{Abgesetzte Formel\|\|<math>\text{~~Area~~} = \int\limits_{a}^{b} \bigl(\bigl(8-\tfrac{1}{8}x^2\bigr) - \bigl(\tfrac{1}{4}x^2+2\bigr)\bigr)\,dx\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\text{Fläche} = \int\limits_{a}^{b} \bigl(\bigl(8-\tfrac{1}{8}x^2\bigr) - \bigl(\tfrac{1}{4}x^2+2\bigr)\bigr)\,dx\,\textrm{.}</math>}}

-	Die Fläche ist ~~der~~ Integral ~~des Unterschiedes~~ zwischen den beiden Funktionen.	+	Die Fläche ist das Integral der Differenz zwischen den beiden Funktionen.

-	Wir erhalten die Schnittstellen durch folgende Gleichung;	+	Wir erhalten die Schnittstellen durch folgende Gleichung:

	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}		{{Abgesetzte Formel\|\|<math>\left\{\begin{align}
-	y &= 8-\tfrac{1}{8}x^2\,,\\[5pt]	+	y &= 8-\tfrac{1}{8}x^2\,,\\[5pt]
-	y &= \tfrac{1}{4}x^2+2\,\textrm{.}	+	y &= \tfrac{1}{4}x^2+2\,\textrm{.}
	\end{align}\right.</math>}}		\end{align}\right.</math>}}

-	Eliminieren ~~Wir~~ ''y'', und lösen die Gleichung für ''x'', erhalten wir die Schnittstellen	+	Eliminieren wir ''y'' und lösen die Gleichung für ''x'', erhalten wir die Schnittstellen

	{{Abgesetzte Formel\|\|<math>8-\tfrac{1}{8}x^2 = \tfrac{1}{4}x^2+2\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>8-\tfrac{1}{8}x^2 = \tfrac{1}{4}x^2+2\,\textrm{.}</math>}}
Zeile 36:		Zeile 35:
	Also sind die Schnittstellen <math>x=-4</math> und <math>x=4\,</math>.		Also sind die Schnittstellen <math>x=-4</math> und <math>x=4\,</math>.

-	Die ~~Flche~~ zwischen den Kurven ist also	+	Die Fälche zwischen den Kurven ist also

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
-	\text{~~Area~~} &= \int\limits_{-4}^{4} \Bigl(\Bigl(8-\frac{1}{8}x^2\Bigr)-\Bigl( \frac{1}{4}x^2+2\Bigr) \Bigr)\,dx\\[5pt]	+	\text{Fläche} &= \int\limits_{-4}^{4} \Bigl(\Bigl(8-\frac{1}{8}x^2\Bigr)-\Bigl( \frac{1}{4}x^2+2\Bigr) \Bigr)\,dx\\[5pt]
	&= \int\limits_{-4}^{4}\Bigl(8-\frac{1}{8}x^2-\frac{1}{4}x^2-2\Bigr)\,dx\\[5pt]		&= \int\limits_{-4}^{4}\Bigl(8-\frac{1}{8}x^2-\frac{1}{4}x^2-2\Bigr)\,dx\\[5pt]
	&= \int\limits_{-4}^{4}\Bigl(6-\Bigl(\frac{1}{8}+\frac{1}{4}\Bigr)x^2\Bigr)\,dx\\[5pt]		&= \int\limits_{-4}^{4}\Bigl(6-\Bigl(\frac{1}{8}+\frac{1}{4}\Bigr)x^2\Bigr)\,dx\\[5pt]

Martin um 18:49, 28. Apr. 2009

Martin — Tue, 28 Apr 2009 18:49:02 GMT

← Nächstältere Version		Version vom 18:49, 28. Apr. 2009
Zeile 1:		Zeile 1:
-	~~First, we need a picture of what the region looks like~~.	+	Zuerst sehen wir in der Figur welches unseres Gebiet ist.

-	~~Both curves,~~ <math>y=x^2/4+2</math> ~~and~~ <math>y=8-x^2/8</math>~~, are parabolas, the first with a minimum value~~ <math>y=2</math> ~~when~~ <math>x=0</math>, ~~and the second with a maximum value of~~ <math>y=8</math> ~~when~~ <math>x=0</math>. ~~Roughly speaking, the curves have the appearance shown in the figure below, where the shaded region whose area we are trying to find lies between the curves~~.	+	Die Kurven <math>y=x^2/4+2</math> und <math>y=8-x^2/8</math> sind beide quadratische Funktionen. Die erste hat das Minimum <math>y=2</math> wenn <math>x=0</math>, und die zweite das Maximum <math>y=8</math> wenn <math>x=0</math>. Die Kurven sehen ungefähr wie im Bild unten aus.

	[[Image:2_1_4_c.gif\|center]]		[[Image:2_1_4_c.gif\|center]]

-	~~The region is bounded above by the parabola~~ <math>y=8-x^2/8</math> ~~and below by the parabola~~ <math>y=x^2/4+2</math>. ~~If we can determine the ''x''-coordinates,~~ <math>x=a</math> ~~and~~ <math>x=b</math>, ~~for the points of intersection between the curves, the area we are looking for will be given by~~	+	Die obere Grenze der Fläche ist die Funktion <math>y=8-x^2/8</math> und die untere Grenze die Funktion <math>y=x^2/4+2</math>. Bestimmen wir die Schnittstellen <math>x=a</math> und <math>x=b</math> der beiden Kurven, ist die Fläche

	{{Abgesetzte Formel\|\|<math>\text{Area} = \int\limits_{a}^{b} \bigl(\bigl(8-\tfrac{1}{8}x^2\bigr) - \bigl(\tfrac{1}{4}x^2+2\bigr)\bigr)\,dx\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\text{Area} = \int\limits_{a}^{b} \bigl(\bigl(8-\tfrac{1}{8}x^2\bigr) - \bigl(\tfrac{1}{4}x^2+2\bigr)\bigr)\,dx\,\textrm{.}</math>}}

-	~~The integrand is the ''y''-value for the upper parabola minus the corresponding ''y''-value for the lower parabola~~.	+	Die Fläche ist der Integral des Unterschiedes zwischen den beiden Funktionen.

-	~~At the points where the curves intersect each other, the ''x''- and ''y''-coordinates are equal, which gives the equation system,~~	+	Wir erhalten die Schnittstellen durch folgende Gleichung;

	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}		{{Abgesetzte Formel\|\|<math>\left\{\begin{align}
Zeile 18:		Zeile 18:
	\end{align}\right.</math>}}		\end{align}\right.</math>}}

-	~~If we eliminate~~ ''y'' ~~from this system~~, ~~we get the following equation for~~ ''x'',	+	Eliminieren Wir ''y'', und lösen die Gleichung für ''x'', erhalten wir die Schnittstellen

	{{Abgesetzte Formel\|\|<math>8-\tfrac{1}{8}x^2 = \tfrac{1}{4}x^2+2\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>8-\tfrac{1}{8}x^2 = \tfrac{1}{4}x^2+2\,\textrm{.}</math>}}

-	~~If we move the <math>x^2</math>-terms onto one side and the constants onto the other, we obtain~~	+	oder

	{{Abgesetzte Formel\|\|<math>\tfrac{1}{4}x^2 + \tfrac{1}{8}x^2 = 8-2\,,</math>}}		{{Abgesetzte Formel\|\|<math>\tfrac{1}{4}x^2 + \tfrac{1}{8}x^2 = 8-2\,,</math>}}

-	~~i.e.~~	+	also ist

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 34:		Zeile 34:
	\end{align}</math>}}		\end{align}</math>}}

-	~~The ''x''-coordinates of the points of intersection are therefore equal to~~ <math>x=-4</math> ~~and~~ <math>x=4\,</math>.	+	Also sind die Schnittstellen <math>x=-4</math> und <math>x=4\,</math>.

-	~~The area of the area between the curves is given by~~	+	Die Flche zwischen den Kurven ist also

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

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

Tekbot — Wed, 11 Mar 2009 10:18:35 GMT

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

← Nächstältere Version

Version vom 10:18, 11. Mär. 2009

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

Tekbot — Tue, 10 Mar 2009 12:59:44 GMT

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

Tek um 14:25, 21. Okt. 2008

Tek — Tue, 21 Oct 2008 14:25:23 GMT

← Nächstältere Version		Version vom 14:25, 21. Okt. 2008
Zeile 1:		Zeile 1:
	First, we need a picture of what the region looks like.		First, we need a picture of what the region looks like.

-	Both curves,	+	Both curves, <math>y=x^2/4+2</math> and <math>y=8-x^2/8</math>, are parabolas, the first with a minimum value <math>y=2</math> when <math>x=0</math>, and the second with a maximum value of <math>y=8</math> when <math>x=0</math>. Roughly speaking, the curves have the appearance shown in the figure below, where the shaded region whose area we are trying to find lies between the curves.
-	<math>y=~~\frac{1}{4}~~x^{2}+2</math>	+
-	and	+
-	<math>y=8-~~\frac{1}{8}~~x^{2}</math>, are parabolas, the first with a minimum value	+
-	<math>y=~~\text{~~2}</math>	+
-	when	+
-	<math>x=0</math>, and the second with a maximum value of	+
-	<math>y=~~\text{~~8}</math>	+
-	when	+
-	<math>x=0</math>. Roughly speaking, the curves have the appearance shown in the figure below, where the shaded region whose area we are trying to find lies between the curves.	+
-		+

	[[Image:2_1_4_c.gif\|center]]		[[Image:2_1_4_c.gif\|center]]

		+	The region is bounded above by the parabola <math>y=8-x^2/8</math> and below by the parabola <math>y=x^2/4+2</math>. If we can determine the ''x''-coordinates, <math>x=a</math> and <math>x=b</math>, for the points of intersection between the curves, the area we are looking for will be given by

-	~~The region is bounded above by the parabola~~	+	{{Displayed math\|\|<math>\text{Area} = \int\limits_{a}^{b} \bigl(\bigl(8-\tfrac{1}{8}x^2\bigr) - \bigl(\tfrac{1}{4}x^2+2\bigr)\bigr)\,dx\,\textrm{.}</math>}}
-	<math>y=8-\~~frac~~{1}{8}x^{2~~}</math>~~	+
-	~~and below by the parabola~~	+
-	~~<math>y=~~\~~frac~~{1}{4}x^{2}+2~~</math>. If we can determine the~~	+
-	~~<math>x</math>~~	+
-	~~-coordinates~~,	+
-	~~<math>x=a~~\~~text~~{ }</math>	+
-	~~and~~	+
-	~~<math>x=b</math>, for the points of intersection between the curves, the area we are looking for will be given by~~	+

-	~~Area =~~	+	The integrand is the ''y''-value for the upper parabola minus the corresponding ''y''-value for the lower parabola.
-	~~<math>\int\limits_{a}^{b}{\left( \left( 8~~-~~\frac{1}{8}x^{2} \right)~~-~~\left( \frac{1}{4}x^{2}+2 \right) \right)}\,dx</math>~~	+

		+	At the points where the curves intersect each other, the ''x''- and ''y''-coordinates are equal, which gives the equation system,

-	~~The integrand is the~~	+	{{Displayed math\|\|<math>\left\{\begin{align}
-	<math>y</math>	+	y &= 8-\tfrac{1}{8}x^2\,,\\[5pt]
-	-~~value for the upper parabola minus the corresponding~~	+	y &= \tfrac{1}{4}x^2+2\,\textrm{.}
-	~~<math>~~y</math>	+	\end{align}\right.</math>}}
-	~~-value for the lower parabola.~~	+

-	~~At the points where the curves intersect each other, the~~	+	If we eliminate ''y'' from this system, we get the following equation for ''x'',
-	~~<math>x</math>~~	+
-	~~- and~~	+
-	~~<math>~~y~~</math>~~	+
-	~~-coordinates are equal~~, ~~which gives~~ the equation ~~system~~,	+

-		+	{{Displayed math\|\|<math>8-\tfrac{1}{8}x^2 = \tfrac{1}{4}x^2+2\,\textrm{.}</math>}}
-	<math>~~\left\{ \begin{matrix}~~	+
-	y=8-\~~frac~~{1}{8}x^{2~~} \\~~	+
-	y=\~~frac~~{1}{4}x^{2}+2 \\	+
-	~~\end~~{~~matrix~~} ~~\right.~~</math>	+
-		+
-		+
-	~~If we eliminate~~	+
-	~~<math>y</math>~~	+
-	~~from this system, we get the following equation for~~	+
-	~~<math>x</math>,~~	+
-		+
-		+
-	~~<math>8-\frac{1~~}{8}~~x^{2}=\frac{1}{4}x^{2}+2</math>~~	+

		+	If we move the <math>x^2</math>-terms onto one side and the constants onto the other, we obtain

-	~~If we move the~~	+	{{Displayed math\|\|<math>\tfrac{1}{4}x^2 + \tfrac{1}{8}x^2 = 8-2\,,</math>}}
-	~~<math>x^~~{~~2}</~~math>	+
-	~~terms onto one side and the constants onto the other, we obtain~~	+
-		+
-		+
-	<math>\~~frac~~{1}{4}x^{2}+\~~frac~~{1}{8}x^{2}=8-2</math>,	+

	i.e.		i.e.

		+	{{Displayed math\|\|<math>\begin{align}
		+	\bigl(\tfrac{1}{4}+\tfrac{1}{8}\bigr)x^2 &= 6\,,\\[5pt]
		+	\tfrac{3}{8}x^2 &= 6\,,\\[5pt]
		+	x^2 &= 16\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	The ''x''-coordinates of the points of intersection are therefore equal to <math>x=-4</math> and <math>x=4\,</math>.
-	~~& \left( \frac{1}{4}+\frac{1}{8} \right)x^{2}=6 \\~~	+
-	~~& \frac{3}{8}x^{2}=6 \\~~	+
-	~~& x^{2}=16 \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	The	+
-	~~<math>~~x~~</math>~~	+
-	-coordinates of the points of intersection are therefore equal to	+
-	<math>x=-~~\text{~~4 }</math>	+
-	and	+
-	<math>x=~~\text{~~4}</math>.	+

	The area of the area between the curves is given by		The area of the area between the curves is given by

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\text{Area} &= \int\limits_{-4}^{4} \Bigl(\Bigl(8-\frac{1}{8}x^2\Bigr)-\Bigl( \frac{1}{4}x^2+2\Bigr) \Bigr)\,dx\\[5pt]
-	& \text{Area}=\int\limits_{-4}^{4}{\~~left~~( \~~left~~( 8-\frac{1}{8}x^{2} \~~right~~)-\~~left~~( \frac{1}{4}x^{2}+2 \~~right~~) \~~right~~)}\,dx \\	+	&= \int\limits_{-4}^{4}\Bigl(8-\frac{1}{8}x^2-\frac{1}{4}x^2-2\Bigr)\,dx\\[5pt]
-	& =\int\limits_{-4}^{4}{\~~left~~( 8-\frac{1}{8}x^{2}-\frac{1}{4}x^{2}-2 \~~right~~)}\,dx \\	+	&= \int\limits_{-4}^{4}\Bigl(6-\Bigl(\frac{1}{8}+\frac{1}{4}\Bigr)x^2\Bigr)\,dx\\[5pt]
-	& =\int\limits_{-4}^{4}{\~~left~~( 6-\~~left~~( \frac{1}{8}+\frac{1}{4} \~~right~~)x^{2} \~~right~~)}\,dx \\	+	&= \int\limits_{-4}^{4}\Bigl(6-\frac{3}{8}x^2\Bigr)\,dx\\[5pt]
-	& =\int\limits_{-4}^{4}{\~~left~~( 6-\frac{3}{8}x^{2} \~~right~~)}\,dx \\	+	&= \Bigl[\ 6x-\frac{3}{8}\frac{x^3}{3}\ \Bigr]_{-4}^{4}\\[5pt]
-	& =\~~left~~[ 6x-\frac{3}{8}\frac{x^{3}}{3} \~~right~~]_{-4}^{4} \\	+	&= \Bigl[\ 6x-\frac{x^3}{8}\ \Bigr]_{-4}^{4}\\[5pt]
-	& =\~~left~~[ 6x-\frac{x^{3}}{8} \~~right~~]_{-4}^{4} \\	+	&= 6\cdot 4 - \frac{4^3}{8} - \Bigl(6\cdot (-4) - \frac{(-4)^3}{8}\Bigr)\\[5pt]
-	& =6\~~centerdot~~ 4-\frac{4^{3}}{8}-\~~left~~( 6\~~left~~( -4 ~~\right~~)-\frac{~~\left~~( -4 ~~\right~~)^{3}}{8} \~~right~~) \\	+	&= 24-8+24-8\\[5pt]
-	& =24-8+24-8=32 \\	+	&= 32\,\textrm{.}
-	\end{align}</math>	+	\end{align}</math>}}

Ian um 12:01, 18. Okt. 2008

Ian — Sat, 18 Oct 2008 12:01:02 GMT

← Nächstältere Version		Version vom 12:01, 18. Okt. 2008
Zeile 1:		Zeile 1:
-	~~{{NAVCONTENT_START}}~~	+	First, we need a picture of what the region looks like.
-	<~~center~~> ~~[[Image:2_1_4c-~~1(4~~).gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	Both curves,
-	{~~{NAVCONTENT_START}~~}	+	<math>y=\frac{1}{4}x^{2}+2</math>
-	<~~center~~> ~~[[Image:2_1_4c-~~2~~(4).gif]]~~ </~~center~~>	+	and
-	~~{{NAVCONTENT_STOP}}~~	+	<math>y=8-\frac{1}{8}x^{2}</math>, are parabolas, the first with a minimum value
-	~~{{NAVCONTENT_START}}~~	+	<math>y=\text{2}</math>
-	<~~center~~> ~~[[Image:2_1_4c-3(4).gif]]~~ </~~center~~>	+	when
-	{~~{NAVCONTENT_STOP}~~}	+	<math>x=0</math>, and the second with a maximum value of
-	~~{{NAVCONTENT_START}}~~	+	<math>y=\text{8}</math>
-	<~~center~~> ~~[[Image:2_1_4c-4(4).gif]]~~ </~~center~~>	+	when
-	~~{{NAVCONTENT_STOP}}~~	+	<math>x=0</math>. Roughly speaking, the curves have the appearance shown in the figure below, where the shaded region whose area we are trying to find lies between the curves.


	[[Image:2_1_4_c.gif\|center]]		[[Image:2_1_4_c.gif\|center]]
		+
		+
		+	The region is bounded above by the parabola
		+	<math>y=8-\frac{1}{8}x^{2}</math>
		+	and below by the parabola
		+	<math>y=\frac{1}{4}x^{2}+2</math>. If we can determine the
		+	<math>x</math>
		+	-coordinates,
		+	<math>x=a\text{ }</math>
		+	and
		+	<math>x=b</math>, for the points of intersection between the curves, the area we are looking for will be given by
		+
		+	Area =
		+	<math>\int\limits_{a}^{b}{\left( \left( 8-\frac{1}{8}x^{2} \right)-\left( \frac{1}{4}x^{2}+2 \right) \right)}\,dx</math>
		+
		+
		+	The integrand is the
		+	<math>y</math>
		+	-value for the upper parabola minus the corresponding
		+	<math>y</math>
		+	-value for the lower parabola.
		+
		+	At the points where the curves intersect each other, the
		+	<math>x</math>
		+	- and
		+	<math>y</math>
		+	-coordinates are equal, which gives the equation system,
		+
		+
		+	<math>\left\{ \begin{matrix}
		+	y=8-\frac{1}{8}x^{2} \\
		+	y=\frac{1}{4}x^{2}+2 \\
		+	\end{matrix} \right.</math>
		+
		+
		+	If we eliminate
		+	<math>y</math>
		+	from this system, we get the following equation for
		+	<math>x</math>,
		+
		+
		+	<math>8-\frac{1}{8}x^{2}=\frac{1}{4}x^{2}+2</math>
		+
		+
		+	If we move the
		+	<math>x^{2}</math>
		+	terms onto one side and the constants onto the other, we obtain
		+
		+
		+	<math>\frac{1}{4}x^{2}+\frac{1}{8}x^{2}=8-2</math>,
		+
		+	i.e.
		+
		+
		+	<math>\begin{align}
		+	& \left( \frac{1}{4}+\frac{1}{8} \right)x^{2}=6 \\
		+	& \frac{3}{8}x^{2}=6 \\
		+	& x^{2}=16 \\
		+	\end{align}</math>
		+
		+
		+	The
		+	<math>x</math>
		+	-coordinates of the points of intersection are therefore equal to
		+	<math>x=-\text{4 }</math>
		+	and
		+	<math>x=\text{4}</math>.
		+
		+	The area of the area between the curves is given by
		+
		+
		+	<math>\begin{align}
		+	& \text{Area}=\int\limits_{-4}^{4}{\left( \left( 8-\frac{1}{8}x^{2} \right)-\left( \frac{1}{4}x^{2}+2 \right) \right)}\,dx \\
		+	& =\int\limits_{-4}^{4}{\left( 8-\frac{1}{8}x^{2}-\frac{1}{4}x^{2}-2 \right)}\,dx \\
		+	& =\int\limits_{-4}^{4}{\left( 6-\left( \frac{1}{8}+\frac{1}{4} \right)x^{2} \right)}\,dx \\
		+	& =\int\limits_{-4}^{4}{\left( 6-\frac{3}{8}x^{2} \right)}\,dx \\
		+	& =\left[ 6x-\frac{3}{8}\frac{x^{3}}{3} \right]_{-4}^{4} \\
		+	& =\left[ 6x-\frac{x^{3}}{8} \right]_{-4}^{4} \\
		+	& =6\centerdot 4-\frac{4^{3}}{8}-\left( 6\left( -4 \right)-\frac{\left( -4 \right)^{3}}{8} \right) \\
		+	& =24-8+24-8=32 \\
		+	\end{align}</math>

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

Tekbot — Tue, 16 Sep 2008 14:39:01 GMT

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

← Nächstältere Version

Version vom 14:39, 16. Sep. 2008

Madde um 12:03, 21. Aug. 2008

Madde — Thu, 21 Aug 2008 12:03:55 GMT

← Nächstältere Version		Version vom 12:03, 21. Aug. 2008
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		+
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Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

Tekbot — Thu, 21 Aug 2008 07:59:14 GMT

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

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← Nächstältere Version		Version vom 12:59, 10. Mär. 2009
Zeile 7:		Zeile 7:
	The region is bounded above by the parabola <math>y=8-x^2/8</math> and below by the parabola <math>y=x^2/4+2</math>. If we can determine the ''x''-coordinates, <math>x=a</math> and <math>x=b</math>, for the points of intersection between the curves, the area we are looking for will be given by		The region is bounded above by the parabola <math>y=8-x^2/8</math> and below by the parabola <math>y=x^2/4+2</math>. If we can determine the ''x''-coordinates, <math>x=a</math> and <math>x=b</math>, for the points of intersection between the curves, the area we are looking for will be given by

-	{{~~Displayed math~~\|\|<math>\text{Area} = \int\limits_{a}^{b} \bigl(\bigl(8-\tfrac{1}{8}x^2\bigr) - \bigl(\tfrac{1}{4}x^2+2\bigr)\bigr)\,dx\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\text{Area} = \int\limits_{a}^{b} \bigl(\bigl(8-\tfrac{1}{8}x^2\bigr) - \bigl(\tfrac{1}{4}x^2+2\bigr)\bigr)\,dx\,\textrm{.}</math>}}

	The integrand is the ''y''-value for the upper parabola minus the corresponding ''y''-value for the lower parabola.		The integrand is the ''y''-value for the upper parabola minus the corresponding ''y''-value for the lower parabola.
Zeile 13:		Zeile 13:
	At the points where the curves intersect each other, the ''x''- and ''y''-coordinates are equal, which gives the equation system,		At the points where the curves intersect each other, the ''x''- and ''y''-coordinates are equal, which gives the equation system,

-	{{~~Displayed math~~\|\|<math>\left\{\begin{align}	+	{{Abgesetzte Formel\|\|<math>\left\{\begin{align}
	y &= 8-\tfrac{1}{8}x^2\,,\\[5pt]		y &= 8-\tfrac{1}{8}x^2\,,\\[5pt]
	y &= \tfrac{1}{4}x^2+2\,\textrm{.}		y &= \tfrac{1}{4}x^2+2\,\textrm{.}
Zeile 20:		Zeile 20:
	If we eliminate ''y'' from this system, we get the following equation for ''x'',		If we eliminate ''y'' from this system, we get the following equation for ''x'',

-	{{~~Displayed math~~\|\|<math>8-\tfrac{1}{8}x^2 = \tfrac{1}{4}x^2+2\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>8-\tfrac{1}{8}x^2 = \tfrac{1}{4}x^2+2\,\textrm{.}</math>}}

	If we move the <math>x^2</math>-terms onto one side and the constants onto the other, we obtain		If we move the <math>x^2</math>-terms onto one side and the constants onto the other, we obtain

-	{{~~Displayed math~~\|\|<math>\tfrac{1}{4}x^2 + \tfrac{1}{8}x^2 = 8-2\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\tfrac{1}{4}x^2 + \tfrac{1}{8}x^2 = 8-2\,,</math>}}

	i.e.		i.e.

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\bigl(\tfrac{1}{4}+\tfrac{1}{8}\bigr)x^2 &= 6\,,\\[5pt]		\bigl(\tfrac{1}{4}+\tfrac{1}{8}\bigr)x^2 &= 6\,,\\[5pt]
	\tfrac{3}{8}x^2 &= 6\,,\\[5pt]		\tfrac{3}{8}x^2 &= 6\,,\\[5pt]
Zeile 38:		Zeile 38:
	The area of the area between the curves is given by		The area of the area between the curves is given by

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\text{Area} &= \int\limits_{-4}^{4} \Bigl(\Bigl(8-\frac{1}{8}x^2\Bigr)-\Bigl( \frac{1}{4}x^2+2\Bigr) \Bigr)\,dx\\[5pt]		\text{Area} &= \int\limits_{-4}^{4} \Bigl(\Bigl(8-\frac{1}{8}x^2\Bigr)-\Bigl( \frac{1}{4}x^2+2\Bigr) \Bigr)\,dx\\[5pt]
	&= \int\limits_{-4}^{4}\Bigl(8-\frac{1}{8}x^2-\frac{1}{4}x^2-2\Bigr)\,dx\\[5pt]		&= \int\limits_{-4}^{4}\Bigl(8-\frac{1}{8}x^2-\frac{1}{4}x^2-2\Bigr)\,dx\\[5pt]