Lösung 2.1:4b - Versionsgeschichte

Sekretariat1 um 10:54, 27. Aug. 2009

2009-08-27T10:54:00Z

← Nächstältere Version		Version vom 10:54, 27. Aug. 2009
Zeile 5:		Zeile 5:
	&= -\bigl(x^2 - 2x- 2\bigr)\\[5pt]		&= -\bigl(x^2 - 2x- 2\bigr)\\[5pt]
	&= -\bigl((x-1)^2 - 1^2 - 2\bigr)\\[5pt]		&= -\bigl((x-1)^2 - 1^2 - 2\bigr)\\[5pt]
-	&= -(x-1)^2 + 3	+	&= -(x-1)^2 + 3 \textrm{.}
	\end{align}</math>}}		\end{align}</math>}}

Zeile 12:		Zeile 12:
	[[Image:2_1_4_b.gif\|center]]		[[Image:2_1_4_b.gif\|center]]

-	~~die~~ Fläche die wir bestimmen sollen, ist im Bild ~~schrafiert~~.	+	Die Fläche, die wir bestimmen sollen, ist im Bild schraffiert.

	Diese Fläche bestimmen wir mit dem Integral		Diese Fläche bestimmen wir mit dem Integral
Zeile 18:		Zeile 18:
	{{Abgesetzte Formel\|\|<math>\text{Fläche} = \int\limits_a^b \bigl(-x^2+2x+2\bigr)\,dx\,,</math>}}		{{Abgesetzte Formel\|\|<math>\text{Fläche} = \int\limits_a^b \bigl(-x^2+2x+2\bigr)\,dx\,,</math>}}

-	Wo ''a'' und ''b'' die Schnittstellen ~~von~~ der Parabel und der ''x''-Achse sind, also die Wurzeln von	+	wobei ''a'' und ''b'' die Schnittstellen der Parabel und der ''x''-Achse sind, also die Wurzeln von

	{{Abgesetzte Formel\|\|<math>0=-x^{2}+2x+2</math>}}		{{Abgesetzte Formel\|\|<math>0=-x^{2}+2x+2</math>}}

-	oder, durch quadratische Ergänzung (siehe oben),	+	oder, durch quadratische Ergänzung (siehe oben)

-	{{Abgesetzte Formel\|\|<math>0=-(x-1)^2+3</math>}}	+	{{Abgesetzte Formel\|\|<math>0=-(x-1)^2+3 \textrm{,}</math>}}

	also		also
Zeile 42:		Zeile 42:
	So erhalten wir die Stammfunktion		So erhalten wir die Stammfunktion

-	{{Abgesetzte Formel\|\|<math>\text{Fläche} = \Bigl[\ -\frac{(x-1)^3}{3} + 3x\ \Bigr]_{1-\sqrt{3}}^{1+\sqrt{3}}\,~~\textrm{.}~~</math>}}	+	{{Abgesetzte Formel\|\|<math>\text{Fläche} = \Bigl[\ -\frac{(x-1)^3}{3} + 3x\ \Bigr]_{1-\sqrt{3}}^{1+\sqrt{3}}\,</math>.}}

-	~~und daher~~	+	Daraus folgt

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

Moni um 22:23, 21. Aug. 2009

2009-08-21T22:23:15Z

← Nächstältere Version		Version vom 22:23, 21. Aug. 2009
Zeile 8:		Zeile 8:
	\end{align}</math>}}		\end{align}</math>}}

-	Wir ~~sehan~~ dass die Funktion ~~ein~~ Parabel mit ~~den Maxima~~ <math>y=3</math> ~~wenn~~ <math>x=1</math> ist.	+	Wir sehen, dass die Funktion eine Parabel mit dem Maximum <math>y=3</math> bei <math>x=1</math> ist.

	[[Image:2_1_4_b.gif\|center]]		[[Image:2_1_4_b.gif\|center]]

-	die Fläche die wir bestimmen ~~soll~~ ist im Bild ~~geschattet~~.	+	die Fläche die wir bestimmen sollen, ist im Bild schrafiert.

-	Diese Fläche bestimmen wir mit ~~den~~ Integral	+	Diese Fläche bestimmen wir mit dem Integral

	{{Abgesetzte Formel\|\|<math>\text{Fläche} = \int\limits_a^b \bigl(-x^2+2x+2\bigr)\,dx\,,</math>}}		{{Abgesetzte Formel\|\|<math>\text{Fläche} = \int\limits_a^b \bigl(-x^2+2x+2\bigr)\,dx\,,</math>}}
Zeile 26:		Zeile 26:
	{{Abgesetzte Formel\|\|<math>0=-(x-1)^2+3</math>}}		{{Abgesetzte Formel\|\|<math>0=-(x-1)^2+3</math>}}

-	~~oder auch~~	+	also

	{{Abgesetzte Formel\|\|<math>(x-1)^2=3\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>(x-1)^2=3\,\textrm{.}</math>}}

-	Die Gleichung hat also die Wurzeln <math>x = 1\pm \sqrt{3}\,</math>, ~~und also~~ <math>x=1-\sqrt{3}</math> und <math>x=1+\sqrt{3}\,</math>.	+	Die Gleichung hat also die Wurzeln <math>x = 1\pm \sqrt{3}\</math> , <math>x=1-\sqrt{3}</math> und <math>x=1+\sqrt{3}\</math>.

	Die Fläche ist also		Die Fläche ist also

-	{{Abgesetzte Formel\|\|<math>\text{~~Area~~} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl(-x^2+2x+2\bigr)\,dx\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\text{Fläche} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl(-x^2+2x+2\bigr)\,dx\,\textrm{.}</math>}}

-	Wir schreiben hier den Integranden in ~~der~~ quadratisch ~~ergänzten~~ Form.	+	Wir schreiben hier den Integranden in die quadratisch ergänzte Form.

-	{{Abgesetzte Formel\|\|<math>\text{~~Area~~} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl( -(x-1)^2 + 3\bigr)\,dx\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\text{Fläche} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl( -(x-1)^2 + 3\bigr)\,dx\,,</math>}}

-	~~Wir~~ erhalten ~~die~~ die Stammfunktion,	+	So erhalten wir die Stammfunktion

-	{{Abgesetzte Formel\|\|<math>\text{~~Area~~} = \Bigl[\ -\frac{(x-1)^3}{3} + 3x\ \Bigr]_{1-\sqrt{3}}^{1+\sqrt{3}}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\text{Fläche} = \Bigl[\ -\frac{(x-1)^3}{3} + 3x\ \Bigr]_{1-\sqrt{3}}^{1+\sqrt{3}}\,\textrm{.}</math>}}

-	und daher ~~erhalten wir~~	+	und daher

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
-	\text{~~Area~~} &= -\frac{(1+\sqrt{3}-1)^3}{3}+3(1+\sqrt{3}\,)-\Bigl(-\frac{(1-\sqrt{3}-1)^3}{3}+3(1-\sqrt{3}\,)\Bigr)\\[5pt]	+	\text{Fläche} &= -\frac{(1+\sqrt{3}-1)^3}{3}+3(1+\sqrt{3}\,)-\Bigl(-\frac{(1-\sqrt{3}-1)^3}{3}+3(1-\sqrt{3}\,)\Bigr)\\[5pt]
	&= -\frac{(\sqrt{3}\,)^3}{3} + 3 + 3\sqrt{3} + \frac{(-\sqrt{3}\,)^3}{3} - 3 + 3\sqrt{3}\\[5pt]		&= -\frac{(\sqrt{3}\,)^3}{3} + 3 + 3\sqrt{3} + \frac{(-\sqrt{3}\,)^3}{3} - 3 + 3\sqrt{3}\\[5pt]
	&= -\frac{\sqrt{3}\sqrt{3}\sqrt{3}}{3} + 3\sqrt{3} + \frac{(-\sqrt{3}\,)(-\sqrt{3}\,)(-\sqrt{3}\,)}{3} + 3\sqrt{3}\\[5pt]		&= -\frac{\sqrt{3}\sqrt{3}\sqrt{3}}{3} + 3\sqrt{3} + \frac{(-\sqrt{3}\,)(-\sqrt{3}\,)(-\sqrt{3}\,)}{3} + 3\sqrt{3}\\[5pt]
Zeile 56:		Zeile 56:
	\end{align}</math>}}		\end{align}</math>}}

-	Hinweis: Die Rechnungen werden ~~unständiger~~ wenn wir mit ~~den~~ Ausdruck	+	Hinweis: Die Rechnungen werden umständlich, wenn wir mit dem Ausdruck

	{{Abgesetzte Formel\|\|<math>\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\bigl(-x^2+2x+2 \bigr)}\,dx = \cdots</math>}}		{{Abgesetzte Formel\|\|<math>\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\bigl(-x^2+2x+2 \bigr)}\,dx = \cdots</math>}}

	rechnen.		rechnen.

Martin um 18:40, 28. Apr. 2009

2009-04-28T18:40:13Z

← Nächstältere Version		Version vom 18:40, 28. Apr. 2009
Zeile 1:		Zeile 1:
-	~~By completing the square of the equation of the curve~~	+	Durch quadratische Ergänzung erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 8:		Zeile 8:
	\end{align}</math>}}		\end{align}</math>}}

-	~~we can read off that the curve is a downward parabola with maximum value~~ <math>y=3</math> ~~when~~ <math>x=1</math>.	+	Wir sehan dass die Funktion ein Parabel mit den Maxima <math>y=3</math> wenn <math>x=1</math> ist.

	[[Image:2_1_4_b.gif\|center]]		[[Image:2_1_4_b.gif\|center]]

-	~~The region whose area we shall determine is the one shaded in the figure~~.	+	die Fläche die wir bestimmen soll ist im Bild geschattet.

-	~~We can express this area using the integral~~	+	Diese Fläche bestimmen wir mit den Integral

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

-	~~where~~ ''a'' ~~and~~ ''b'' ~~are the~~ ''x''-~~coordinates for the points of intersection between the parabola and the ''x''-axis.~~	+	Wo ''a'' und ''b'' die Schnittstellen von der Parabel und der ''x''-Achse sind, also die Wurzeln von
-		+
-	~~A solution plan is to first determine the intersection points, <math>x=a</math>~~	+
-	~~and <math>x=b</math>, and then calculate the area using the integral formula above.~~	+
-		+
-	~~The parabola cuts the ''x''-axis when its ''y''-coordinate is zero~~, ~~i.e.~~	+

	{{Abgesetzte Formel\|\|<math>0=-x^{2}+2x+2</math>}}		{{Abgesetzte Formel\|\|<math>0=-x^{2}+2x+2</math>}}

-	~~and because we have already completed the square of the right-hand side once~~, ~~the equation can be written as~~	+	oder, durch quadratische Ergänzung (siehe oben),

	{{Abgesetzte Formel\|\|<math>0=-(x-1)^2+3</math>}}		{{Abgesetzte Formel\|\|<math>0=-(x-1)^2+3</math>}}

-	or	+	oder auch

	{{Abgesetzte Formel\|\|<math>(x-1)^2=3\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>(x-1)^2=3\,\textrm{.}</math>}}

-	~~Taking the square root gives~~ <math>x = 1\pm \sqrt{3}\,</math>~~. The points of intersection are~~ <math>x=1-\sqrt{3}</math> ~~and~~ <math>x=1+\sqrt{3}\,</math>.	+	Die Gleichung hat also die Wurzeln <math>x = 1\pm \sqrt{3}\,</math>, und also <math>x=1-\sqrt{3}</math> und <math>x=1+\sqrt{3}\,</math>.

-	~~The area we are looking for is therefore given by~~	+	Die Fläche ist also

	{{Abgesetzte Formel\|\|<math>\text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl(-x^2+2x+2\bigr)\,dx\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl(-x^2+2x+2\bigr)\,dx\,\textrm{.}</math>}}

-	~~Instead of directly starting to calculate, we can start from the integrand~~ in ~~the form we obtain after completing its square,~~	+	Wir schreiben hier den Integranden in der quadratisch ergänzten Form.

	{{Abgesetzte Formel\|\|<math>\text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl( -(x-1)^2 + 3\bigr)\,dx\,,</math>}}		{{Abgesetzte Formel\|\|<math>\text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl( -(x-1)^2 + 3\bigr)\,dx\,,</math>}}

-	~~which seems easier. Because the expression <math>x-1</math> inside the square is a linear expression, we can write down a primitive function “in the usual way”~~,	+	Wir erhalten die die Stammfunktion,

	{{Abgesetzte Formel\|\|<math>\text{Area} = \Bigl[\ -\frac{(x-1)^3}{3} + 3x\ \Bigr]_{1-\sqrt{3}}^{1+\sqrt{3}}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\text{Area} = \Bigl[\ -\frac{(x-1)^3}{3} + 3x\ \Bigr]_{1-\sqrt{3}}^{1+\sqrt{3}}\,\textrm{.}</math>}}

-	~~(If one is uncertain of this step, it is possible to differentiate the primitive function and see that one really does get the integrand back). Hence,~~	+	und daher erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 61:		Zeile 56:
	\end{align}</math>}}		\end{align}</math>}}

-		+	Hinweis: Die Rechnungen werden unständiger wenn wir mit den Ausdruck
-	~~Note~~: ~~The calculations become a lot more complicated if one starts from~~	+

	{{Abgesetzte Formel\|\|<math>\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\bigl(-x^2+2x+2 \bigr)}\,dx = \cdots</math>}}		{{Abgesetzte Formel\|\|<math>\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\bigl(-x^2+2x+2 \bigr)}\,dx = \cdots</math>}}
		+
		+	rechnen.

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

2009-03-11T10:18:15Z

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

← Nächstältere Version

Version vom 10:18, 11. Mär. 2009

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

2009-03-10T12:59:34Z

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

Tek um 14:03, 21. Okt. 2008

2008-10-21T14:03:20Z

← Nächstältere Version		Version vom 14:03, 21. Okt. 2008
Zeile 1:		Zeile 1:
	By completing the square of the equation of the curve		By completing the square of the equation of the curve

		+	{{Displayed math\|\|<math>\begin{align}
		+	y &= -x^2 + 2x + 2\\[5pt]
		+	&= -\bigl(x^2 - 2x- 2\bigr)\\[5pt]
		+	&= -\bigl((x-1)^2 - 1^2 - 2\bigr)\\[5pt]
		+	&= -(x-1)^2 + 3
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	we can read off that the curve is a downward parabola with maximum value <math>y=3</math> when <math>x=1</math>.
-	~~& y=-x^{2}+2x+2=-\left( x^{2}-2x-2 \right) \\~~	+
-	~~& =-\left( \left( x-1 \right)^{2}-1^{2}-2 \right)=-\left( x-1 \right)^{2}+3 \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	we can read off that the curve is a downward parabola with maximum value	+
-	<math>y=~~\text{~~3 }</math>	+
-	when	+
-	<math>x=~~\text{~~1}</math>	+
-		+
-		+

	[[Image:2_1_4_b.gif\|center]]		[[Image:2_1_4_b.gif\|center]]
-

	The region whose area we shall determine is the one shaded in the figure.		The region whose area we shall determine is the one shaded in the figure.
Zeile 22:		Zeile 16:
	We can express this area using the integral		We can express this area using the integral

-	~~Area=~~	+	{{Displayed math\|\|<math>\text{Area} = \int\limits_a^b \bigl(-x^2+2x+2\bigr)\,dx\,,</math>}}
-	<math>\~~int\limits_~~{a}^{b}{\~~left~~( -x^{2}+2x+2 \~~right~~)}\,dx</math>	+

-	where	+	where ''a'' and ''b'' are the ''x''-coordinates for the points of intersection between the parabola and the ''x''-axis.
-	~~<math>~~a~~</math>~~	+
-	and	+
-	~~<math>~~b~~</math>~~	+
-	are the	+
-	~~<math>~~x~~</math>~~	+
-	-coordinates for the points of intersection between the parabola and the	+
-	~~<math>~~x~~</math>~~	+
-	-axis.	+

-	A solution plan is to first determine the intersection points,	+	A solution plan is to first determine the intersection points, <math>x=a</math>
-	<math>x=a</math>	+	and <math>x=b</math>, and then calculate the area using the integral formula above.
-	and	+
-	<math>x=b</math>, and then calculate the area using the integral formula above.	+

-	The parabola cuts the	+	The parabola cuts the ''x''-axis when its ''y''-coordinate is zero, i.e.
-	~~<math>~~x~~</math>~~	+
-	-axis when its	+
-	~~<math>~~y~~</math>~~	+
-	-coordinate is zero, i.e.	+
-		+
-		+
-	~~<math>0=-x^{2}+2x+2</math>~~	+

		+	{{Displayed math\|\|<math>0=-x^{2}+2x+2</math>}}

	and because we have already completed the square of the right-hand side once, the equation can be written as		and because we have already completed the square of the right-hand side once, the equation can be written as

-		+	{{Displayed math\|\|<math>0=-(x-1)^2+3</math>}}
-	<math>0=-~~\left~~( x-1 ~~\right~~)^{2}+3</math>	+

	or		or

		+	{{Displayed math\|\|<math>(x-1)^2=3\,\textrm{.}</math>}}

-	~~<math>\left( x-1 \right)^{2}=3</math>.~~	+	Taking the square root gives <math>x = 1\pm \sqrt{3}\,</math>. The points of intersection are <math>x=1-\sqrt{3}</math> and <math>x=1+\sqrt{3}\,</math>.
-		+
-	Taking the root gives	+
-	<math>x=1\pm \sqrt{3}</math>. The points of intersection	+
-	<math>x=1-\sqrt{3}</math>	+
-	and	+
-	<math>x=1+\sqrt{3}</math>.	+

	The area we are looking for is therefore given by		The area we are looking for is therefore given by

-	~~Area~~	+	{{Displayed math\|\|<math>\text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl(-x^2+2x+2\bigr)\,dx\,\textrm{.}</math>}}
-	<math>=\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\~~left~~( -x^{2}+2x+2 \~~right~~)}\,dx</math>	+
-		+

	Instead of directly starting to calculate, we can start from the integrand in the form we obtain after completing its square,		Instead of directly starting to calculate, we can start from the integrand in the form we obtain after completing its square,

-	~~Area=~~	+	{{Displayed math\|\|<math>\text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl( -(x-1)^2 + 3\bigr)\,dx\,,</math>}}
-	<math>=\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\~~left~~( -~~\left~~( x-1 ~~\right~~)^{2}+3 \~~right~~)}\,dx~~</math>~~	+
-		+
-		+
-	~~which seems easier. Because the expression~~	+
-	~~<math>x-1</math>~~	+
-	~~inside the square is a linear expression~~, ~~we can write down a primitive function “in the usual way”~~,	+
-		+
-	~~Area~~	+
-	<math>~~=\left[ -\frac{\left( x-1 \right)^{3~~}}~~{3}+3x \right]_{1-\sqrt{3}}^{1+\sqrt{3}}</math>~~	+
-		+

-	~~(If one~~ is ~~uncertain of this step~~, ~~it is possible to differentiate the~~ primitive function ~~and see that one really does get~~ the ~~integral back). Hence~~,	+	which seems easier. Because the expression <math>x-1</math> inside the square is a linear expression, we can write down a primitive function “in the usual way”,

		+	{{Displayed math\|\|<math>\text{Area} = \Bigl[\ -\frac{(x-1)^3}{3} + 3x\ \Bigr]_{1-\sqrt{3}}^{1+\sqrt{3}}\,\textrm{.}</math>}}

-	~~<math>\begin{align}~~	+	(If one is uncertain of this step, it is possible to differentiate the primitive function and see that one really does get the integrand back). Hence,
-	~~& \text{Area}=-\frac{\left~~( ~~1+\sqrt{3}-1 \right~~)~~^{3}}{3}+3\left( 1+\sqrt{3} \right)-\left( -\frac{\left( 1-\sqrt{3}-1 \right)^{3}}{3}+3\left( 1-\sqrt{3} \right) \right) \\~~	+
-	~~& =-\frac{\left( \sqrt{3} \right)^{3}}{3}+3+3\sqrt{3}+\frac{\left( -\sqrt{3} \right)^{3}}{3}-3+3\sqrt{3} \\~~	+
-	~~& =-\frac{\sqrt{3}\sqrt{3}\sqrt{3}}{3}+3\sqrt{3}+\frac{\left( -\sqrt{3} \right)\left( -\sqrt{3} \right)\left( -\sqrt{3} \right)}{3}+3\sqrt{3} \\~~	+
-	~~& =-\frac{3\sqrt{3}}{3}+3\sqrt{3}-\frac{3\sqrt{3}}{3}+3\sqrt{3} \\~~	+
-	~~& =-\sqrt{3}+3\sqrt{3}-\sqrt{3}+3\sqrt{3} \\~~	+
-	~~& =\left( -1+3-1+3 \right)\sqrt{3}=4\sqrt{3} \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\text{Area} &= -\frac{(1+\sqrt{3}-1)^3}{3}+3(1+\sqrt{3}\,)-\Bigl(-\frac{(1-\sqrt{3}-1)^3}{3}+3(1-\sqrt{3}\,)\Bigr)\\[5pt]
		+	&= -\frac{(\sqrt{3}\,)^3}{3} + 3 + 3\sqrt{3} + \frac{(-\sqrt{3}\,)^3}{3} - 3 + 3\sqrt{3}\\[5pt]
		+	&= -\frac{\sqrt{3}\sqrt{3}\sqrt{3}}{3} + 3\sqrt{3} + \frac{(-\sqrt{3}\,)(-\sqrt{3}\,)(-\sqrt{3}\,)}{3} + 3\sqrt{3}\\[5pt]
		+	&= -\frac{3\sqrt{3}}{3} + 3\sqrt{3} - \frac{3\sqrt{3}}{3} + 3\sqrt{3}\\[5pt]
		+	&= -\sqrt{3} + 3\sqrt{3} - \sqrt{3} + 3\sqrt{3}\\[5pt]
		+	&= (-1+3-1+3)\sqrt{3}\\[5pt]
		+	&= 4\sqrt{3}\,\textrm{.}
		+	\end{align}</math>}}

-	NOTE: The calculations become a lot more complicated if one starts from

		+	Note: The calculations become a lot more complicated if one starts from

-	<math>=\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\~~left~~( -x^{2}+2x+2 \~~right~~)}\,dx=~~....~~</math>	+	{{Displayed math\|\|<math>\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\bigl(-x^2+2x+2 \bigr)}\,dx = \cdots</math>}}

Ian um 11:32, 18. Okt. 2008

2008-10-18T11:32:50Z

← Nächstältere Version		Version vom 11:32, 18. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	By completing the square of the equation of the curve
-	~~<center> [[Image:2_1_4b~~-1(4)~~.gif]] </center>~~	+
-	{~~{NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>\begin{align}
-	~~<center> [[Image:2_1_4b~~-2(4)~~.gif]] </center>~~	+	& y=-x^{2}+2x+2=-\left( x^{2}-2x-2 \right) \\
-	{~~{NAVCONTENT_STOP}~~}	+	& =-\left( \left( x-1 \right)^{2}-1^{2}-2 \right)=-\left( x-1 \right)^{2}+3 \\
-	{~~{NAVCONTENT_START}~~}	+	\end{align}</math>
-	<~~center~~> ~~[[Image:2_1_4b-~~3~~(4).gif]]~~ </~~center~~>	+
-	~~{{NAVCONTENT_STOP}}~~	+
-	{~~{NAVCONTENT_START~~}}	+	we can read off that the curve is a downward parabola with maximum value
-	~~<center> [[Image:2_1_4b-4(4).gif]]~~ </~~center~~>	+	<math>y=\text{3 }</math>
-	~~{{NAVCONTENT_STOP}}~~	+	when
		+	<math>x=\text{1}</math>
		+


	[[Image:2_1_4_b.gif\|center]]		[[Image:2_1_4_b.gif\|center]]
		+
		+
		+	The region whose area we shall determine is the one shaded in the figure.
		+
		+	We can express this area using the integral
		+
		+	Area=
		+	<math>\int\limits_{a}^{b}{\left( -x^{2}+2x+2 \right)}\,dx</math>
		+
		+	where
		+	<math>a</math>
		+	and
		+	<math>b</math>
		+	are the
		+	<math>x</math>
		+	-coordinates for the points of intersection between the parabola and the
		+	<math>x</math>
		+	-axis.
		+
		+	A solution plan is to first determine the intersection points,
		+	<math>x=a</math>
		+	and
		+	<math>x=b</math>, and then calculate the area using the integral formula above.
		+
		+	The parabola cuts the
		+	<math>x</math>
		+	-axis when its
		+	<math>y</math>
		+	-coordinate is zero, i.e.
		+
		+
		+	<math>0=-x^{2}+2x+2</math>
		+
		+
		+	and because we have already completed the square of the right-hand side once, the equation can be written as
		+
		+
		+	<math>0=-\left( x-1 \right)^{2}+3</math>
		+
		+	or
		+
		+
		+	<math>\left( x-1 \right)^{2}=3</math>.
		+
		+	Taking the root gives
		+	<math>x=1\pm \sqrt{3}</math>. The points of intersection
		+	<math>x=1-\sqrt{3}</math>
		+	and
		+	<math>x=1+\sqrt{3}</math>.
		+
		+	The area we are looking for is therefore given by
		+
		+	Area
		+	<math>=\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\left( -x^{2}+2x+2 \right)}\,dx</math>
		+
		+
		+	Instead of directly starting to calculate, we can start from the integrand in the form we obtain after completing its square,
		+
		+	Area=
		+	<math>=\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\left( -\left( x-1 \right)^{2}+3 \right)}\,dx</math>
		+
		+
		+	which seems easier. Because the expression
		+	<math>x-1</math>
		+	inside the square is a linear expression, we can write down a primitive function “in the usual way”,
		+
		+	Area
		+	<math>=\left[ -\frac{\left( x-1 \right)^{3}}{3}+3x \right]_{1-\sqrt{3}}^{1+\sqrt{3}}</math>
		+
		+
		+	(If one is uncertain of this step, it is possible to differentiate the primitive function and see that one really does get the integral back). Hence,
		+
		+
		+	<math>\begin{align}
		+	& \text{Area}=-\frac{\left( 1+\sqrt{3}-1 \right)^{3}}{3}+3\left( 1+\sqrt{3} \right)-\left( -\frac{\left( 1-\sqrt{3}-1 \right)^{3}}{3}+3\left( 1-\sqrt{3} \right) \right) \\
		+	& =-\frac{\left( \sqrt{3} \right)^{3}}{3}+3+3\sqrt{3}+\frac{\left( -\sqrt{3} \right)^{3}}{3}-3+3\sqrt{3} \\
		+	& =-\frac{\sqrt{3}\sqrt{3}\sqrt{3}}{3}+3\sqrt{3}+\frac{\left( -\sqrt{3} \right)\left( -\sqrt{3} \right)\left( -\sqrt{3} \right)}{3}+3\sqrt{3} \\
		+	& =-\frac{3\sqrt{3}}{3}+3\sqrt{3}-\frac{3\sqrt{3}}{3}+3\sqrt{3} \\
		+	& =-\sqrt{3}+3\sqrt{3}-\sqrt{3}+3\sqrt{3} \\
		+	& =\left( -1+3-1+3 \right)\sqrt{3}=4\sqrt{3} \\
		+	\end{align}</math>
		+
		+
		+	NOTE: The calculations become a lot more complicated if one starts from
		+
		+
		+	<math>=\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\left( -x^{2}+2x+2 \right)}\,dx=....</math>

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

2008-09-16T14:38:41Z

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

← Nächstältere Version

Version vom 14:38, 16. Sep. 2008

Madde um 12:02, 21. Aug. 2008

2008-08-21T12:02:42Z

← Nächstältere Version		Version vom 12:02, 21. Aug. 2008
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		+
		+
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Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

2008-08-21T07:59:04Z

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

← Nächstältere Version		Version vom 07:59, 21. Aug. 2008
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← Nächstältere Version		Version vom 12:59, 10. Mär. 2009
Zeile 1:		Zeile 1:
	By completing the square of the equation of the curve		By completing the square of the equation of the curve

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	y &= -x^2 + 2x + 2\\[5pt]		y &= -x^2 + 2x + 2\\[5pt]
	&= -\bigl(x^2 - 2x- 2\bigr)\\[5pt]		&= -\bigl(x^2 - 2x- 2\bigr)\\[5pt]
Zeile 16:		Zeile 16:
	We can express this area using the integral		We can express this area using the integral

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

	where ''a'' and ''b'' are the ''x''-coordinates for the points of intersection between the parabola and the ''x''-axis.		where ''a'' and ''b'' are the ''x''-coordinates for the points of intersection between the parabola and the ''x''-axis.
Zeile 25:		Zeile 25:
	The parabola cuts the ''x''-axis when its ''y''-coordinate is zero, i.e.		The parabola cuts the ''x''-axis when its ''y''-coordinate is zero, i.e.

-	{{~~Displayed math~~\|\|<math>0=-x^{2}+2x+2</math>}}	+	{{Abgesetzte Formel\|\|<math>0=-x^{2}+2x+2</math>}}

	and because we have already completed the square of the right-hand side once, the equation can be written as		and because we have already completed the square of the right-hand side once, the equation can be written as

-	{{~~Displayed math~~\|\|<math>0=-(x-1)^2+3</math>}}	+	{{Abgesetzte Formel\|\|<math>0=-(x-1)^2+3</math>}}

	or		or

-	{{~~Displayed math~~\|\|<math>(x-1)^2=3\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>(x-1)^2=3\,\textrm{.}</math>}}

	Taking the square root gives <math>x = 1\pm \sqrt{3}\,</math>. The points of intersection are <math>x=1-\sqrt{3}</math> and <math>x=1+\sqrt{3}\,</math>.		Taking the square root gives <math>x = 1\pm \sqrt{3}\,</math>. The points of intersection are <math>x=1-\sqrt{3}</math> and <math>x=1+\sqrt{3}\,</math>.
Zeile 39:		Zeile 39:
	The area we are looking for is therefore given by		The area we are looking for is therefore given by

-	{{~~Displayed math~~\|\|<math>\text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl(-x^2+2x+2\bigr)\,dx\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl(-x^2+2x+2\bigr)\,dx\,\textrm{.}</math>}}

	Instead of directly starting to calculate, we can start from the integrand in the form we obtain after completing its square,		Instead of directly starting to calculate, we can start from the integrand in the form we obtain after completing its square,

-	{{~~Displayed math~~\|\|<math>\text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl( -(x-1)^2 + 3\bigr)\,dx\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>\text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl( -(x-1)^2 + 3\bigr)\,dx\,,</math>}}

	which seems easier. Because the expression <math>x-1</math> inside the square is a linear expression, we can write down a primitive function “in the usual way”,		which seems easier. Because the expression <math>x-1</math> inside the square is a linear expression, we can write down a primitive function “in the usual way”,

-	{{~~Displayed math~~\|\|<math>\text{Area} = \Bigl[\ -\frac{(x-1)^3}{3} + 3x\ \Bigr]_{1-\sqrt{3}}^{1+\sqrt{3}}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\text{Area} = \Bigl[\ -\frac{(x-1)^3}{3} + 3x\ \Bigr]_{1-\sqrt{3}}^{1+\sqrt{3}}\,\textrm{.}</math>}}

	(If one is uncertain of this step, it is possible to differentiate the primitive function and see that one really does get the integrand back). Hence,		(If one is uncertain of this step, it is possible to differentiate the primitive function and see that one really does get the integrand back). Hence,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\text{Area} &= -\frac{(1+\sqrt{3}-1)^3}{3}+3(1+\sqrt{3}\,)-\Bigl(-\frac{(1-\sqrt{3}-1)^3}{3}+3(1-\sqrt{3}\,)\Bigr)\\[5pt]		\text{Area} &= -\frac{(1+\sqrt{3}-1)^3}{3}+3(1+\sqrt{3}\,)-\Bigl(-\frac{(1-\sqrt{3}-1)^3}{3}+3(1-\sqrt{3}\,)\Bigr)\\[5pt]
	&= -\frac{(\sqrt{3}\,)^3}{3} + 3 + 3\sqrt{3} + \frac{(-\sqrt{3}\,)^3}{3} - 3 + 3\sqrt{3}\\[5pt]		&= -\frac{(\sqrt{3}\,)^3}{3} + 3 + 3\sqrt{3} + \frac{(-\sqrt{3}\,)^3}{3} - 3 + 3\sqrt{3}\\[5pt]
Zeile 64:		Zeile 64:
	Note: The calculations become a lot more complicated if one starts from		Note: The calculations become a lot more complicated if one starts from

-	{{~~Displayed math~~\|\|<math>\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\bigl(-x^2+2x+2 \bigr)}\,dx = \cdots</math>}}	+	{{Abgesetzte Formel\|\|<math>\int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\bigl(-x^2+2x+2 \bigr)}\,dx = \cdots</math>}}