Aus Online Mathematik Brückenkurs 2

Version vom 11:32, 18. Okt. 2008 (bearbeiten) Ian (Diskussion | Beiträge) ← Zum vorherigen Versionsunterschied		Version vom 14:03, 21. Okt. 2008 (bearbeiten) (rückgängig) Tek (Diskussion | Beiträge) K Zum nächsten Versionsunterschied →
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>}}

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Version vom 14:03, 21. Okt. 2008

By completing the square of the equation of the curve

\displaystyle \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}

we can read off that the curve is a downward parabola with maximum value \displaystyle y=3 when \displaystyle x=1.

The region whose area we shall determine is the one shaded in the figure.

We can express this area using the integral

\displaystyle \text{Area} = \int\limits_a^b \bigl(-x^2+2x+2\bigr)\,dx\,,

where a and b are the x-coordinates for the points of intersection between the parabola and the x-axis.

A solution plan is to first determine the intersection points, \displaystyle x=a and \displaystyle x=b, and then calculate the area using the integral formula above.

The parabola cuts the x-axis when its y-coordinate is zero, i.e.

\displaystyle 0=-x^{2}+2x+2

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

\displaystyle 0=-(x-1)^2+3

or

\displaystyle (x-1)^2=3\,\textrm{.}

Taking the square root gives \displaystyle x = 1\pm \sqrt{3}\,. The points of intersection are \displaystyle x=1-\sqrt{3} and \displaystyle x=1+\sqrt{3}\,.

The area we are looking for is therefore given by

\displaystyle \text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl(-x^2+2x+2\bigr)\,dx\,\textrm{.}

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

\displaystyle \text{Area} = \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}} \bigl( -(x-1)^2 + 3\bigr)\,dx\,,

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

\displaystyle \text{Area} = \Bigl[\ -\frac{(x-1)^3}{3} + 3x\ \Bigr]_{1-\sqrt{3}}^{1+\sqrt{3}}\,\textrm{.}

(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,

\displaystyle \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}

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

\displaystyle \int\limits_{1-\sqrt{3}}^{1+\sqrt{3}}{\bigl(-x^2+2x+2 \bigr)}\,dx = \cdots