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Aus Online Mathematik Brückenkurs 2

Version vom 12:46, 18. Okt. 2008 (bearbeiten) Ian (Diskussion | Beiträge) ← Zum vorherigen Versionsunterschied		Version vom 08:45, 19. Okt. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) K Zum nächsten Versionsunterschied →
Zeile 60:		Zeile 60:
-	<math>\left\{ \begin{matrix}	+	<math>\left\{ \begin{array}{*{35}l}
-	y=\text{1 } \\	+	y=\text{1} \\
	y=x+\text{2} \\		y=x+\text{2} \\
-	\end{matrix} \right.</math>	+	\end{array} \right.</math>
Zeile 82:		Zeile 82:
-	<math>\left\{ \begin{matrix}	+	<math>\left\{ \begin{array}{*{35}l}
	y=x+\text{2} \\		y=x+\text{2} \\
	y={1}/{x}\; \\		y={1}/{x}\; \\
-	\end{matrix} \right.</math>	+	\end{array} \right.</math>
Zeile 113:		Zeile 113:
	<math>x=-1\pm \sqrt{2}</math>, leading to		<math>x=-1\pm \sqrt{2}</math>, leading to
	<math>b=-1+\sqrt{2}</math>		<math>b=-1+\sqrt{2}</math>
-
	(the alternative		(the alternative
	<math>b=-1-\sqrt{2}</math>		<math>b=-1-\sqrt{2}</math>
Zeile 147:		Zeile 146:
	& =0-1-\ln \left( \sqrt{2}-1 \right)+\sqrt{2}-1 \\		& =0-1-\ln \left( \sqrt{2}-1 \right)+\sqrt{2}-1 \\
	& =\sqrt{2}-2-\ln \left( \sqrt{2}-1 \right) \\		& =\sqrt{2}-2-\ln \left( \sqrt{2}-1 \right) \\
		+	\end{align}</math>
		+
		+
		+	The total area is
		+
		+	Area = (left area +right area)
		+
		+	<math>\begin{align}
		+	& =1+\sqrt{2}-2-\ln \left( \sqrt{2}-1 \right) \\
		+	& =\sqrt{2}-1-\ln \left( \sqrt{2}-1 \right) \\
	\end{align}</math>		\end{align}</math>

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Version vom 08:45, 19. Okt. 2008

We start by drawing the three curves:

When we draw the curves on the same diagram, we see that the region is bounded from below in the y -direction by the horizontal line y=1 and above partly by y=x+2 and partly by y=x1.

If we denote the x -coordinates of the intersection points by x=a, x=b and x=c, as shown in the figure, we see that the region can be divided up into two parts. In the left part between x=a and x=b, the upper limit is y=x+2, whilst in the right part between x=b and x=c

y=1x is the upper limit.

The area of each part is given by the integrals

Left area =bax+2−1dx 

Right area =cbx1−1dx 

and the total area is the sum of these areas.

If we just manage to determine the curves' points of intersection, the rest is just a matter of integration.

To determine the points of intersection:

x=a: the point of intersection between y=1 and y=x+2 must both satisfy the equations of the lines

y=1y=x+2 

This gives that x must satisfy x+2=1, i.e. x=−1 Thus, a=−1.

x=b: At the point where the curves y=x+2 and y=1x meet, we have that

y=x+2y=1x 

If we eliminate y, we obtain an equation for x,

x+2=x1,

which we multiply by x,

x2+2x=1

Completing the square of the left-hand side,

x+12−12=1x+12=2,

and taking the root gives that x=−12 , leading to b=−1+2  (the alternative b=−1−2  lies to the left of b=a).

x=c: the final point of intersection is given by the condition that the equation to both curves, y=1 and y=1x, are satisfied simultaneously. We see almost immediately that this gives x=1, i.e. c= 1.

The sub-areas are

Leftarea = 2−1−1x+2−1dx=2−1−1x+1dx=2x2+x−12−1=22−12+2−1−2−12+−1=222−22+1+2−1−21+1=22−22+1+2−1−21+1=1−2+21+2−1−21+1=1

Rightarea =12−1x1−1dx=lnx−x12−1=ln1−1−ln2−1−2−1=0−1−ln2−1+2−1=2−2−ln2−1

The total area is

Area = (left area +right area)

=1+2−2−ln2−1=2−1−ln2−1