Processing Math: Done

Aus Online Mathematik Brückenkurs 2

Version vom 09:04, 28. Okt. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 12:59, 10. Mär. 2009 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
Zeile 13:		Zeile 13:
	The area of each part is given by the integrals		The area of each part is given by the integrals
-	{{Displayed math||<math>\begin{align}	+	{{Abgesetzte Formel||<math>\begin{align}
	\text{Left area} &= \int\limits_a^b (x+2-1)\,dx\,,\\[5pt]		\text{Left area} &= \int\limits_a^b (x+2-1)\,dx\,,\\[5pt]
	\text{Right area} &= \int\limits_b^c \Bigl(\frac{1}{x}-1\Bigr)\,dx\,,		\text{Right area} &= \int\limits_b^c \Bigl(\frac{1}{x}-1\Bigr)\,dx\,,
Zeile 26:		Zeile 26:
	*<math>x=a</math>: The point of intersection between <math>y=1</math> and <math>y=x+2</math> must satisfy both equations of the lines,		*<math>x=a</math>: The point of intersection between <math>y=1</math> and <math>y=x+2</math> must satisfy both equations of the lines,
-	{{Displayed math||<math>\left\{\begin{align}	+	{{Abgesetzte Formel||<math>\left\{\begin{align}
	y &= 1\,,\\[5pt]		y &= 1\,,\\[5pt]
	y &= x+2\,\textrm{.}		y &= x+2\,\textrm{.}
Zeile 36:		Zeile 36:
	*<math>x=b</math>: At the point where the curves <math>y=x+2</math> and <math>y=1/x</math> meet, we have that		*<math>x=b</math>: At the point where the curves <math>y=x+2</math> and <math>y=1/x</math> meet, we have that
-	{{Displayed math||<math>\left\{\begin{align}	+	{{Abgesetzte Formel||<math>\left\{\begin{align}
	y &= x+2\,,\\[5pt]		y &= x+2\,,\\[5pt]
	y &= 1/x\,\textrm{.}		y &= 1/x\,\textrm{.}
Zeile 43:		Zeile 43:
	:If we eliminate <math>y</math>, we obtain an equation for <math>x</math>,		:If we eliminate <math>y</math>, we obtain an equation for <math>x</math>,
-	{{Displayed math||<math>x+2=\frac{1}{x}\,,</math>}}	+	{{Abgesetzte Formel||<math>x+2=\frac{1}{x}\,,</math>}}
	:which we multiply by <math>x</math>,		:which we multiply by <math>x</math>,
-	{{Displayed math||<math>x^{2}+2x=1\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>x^{2}+2x=1\,\textrm{.}</math>}}
	:Completing the square of the left-hand side,		:Completing the square of the left-hand side,
-	{{Displayed math||<math>\begin{align}	+	{{Abgesetzte Formel||<math>\begin{align}
	(x+1)^2 - 1^2 &= 1\,,\\[5pt]		(x+1)^2 - 1^2 &= 1\,,\\[5pt]
	(x+1)^2 &= 2\,,		(x+1)^2 &= 2\,,
Zeile 66:		Zeile 66:
	The sub-areas are		The sub-areas are
-	{{Displayed math||<math>\begin{align}	+	{{Abgesetzte Formel||<math>\begin{align}
	\text{Left area}		\text{Left area}
	&= \int\limits_{-1}^{\sqrt{2}-1} (x+2-1)\,dx\\[5pt]		&= \int\limits_{-1}^{\sqrt{2}-1} (x+2-1)\,dx\\[5pt]
Zeile 86:		Zeile 86:
	The total area is		The total area is
-	{{Displayed math||<math>\begin{align}	+	{{Abgesetzte Formel||<math>\begin{align}
	\text{Area}		\text{Area}
	&= \text{(left area)} + \text{(right area)}\\[5pt]		&= \text{(left area)} + \text{(right area)}\\[5pt]

---

Version vom 12:59, 10. Mär. 2009

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=1x.

If we denote the x-coordinates of the intersection points between the curves 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 the curve y=1x is the upper limit.

The area of each part is given by the integrals

Left areaRight area=ba(x+2−1)dx=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 satisfy both equations of the lines,

yy=1=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

yy=x+2=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+1)2−12(x+1)2=1=2

and taking the square root gives that x=−12 , leading to

b=−1+2 . (The alternative b=−1−2  lies to the left of x=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

Left areaRight area=2−1−1(x+2−1)dx=2−1−1(x+1)dx= 2x2+x −12−1=22−12+2−1−2(−1)2+(−1)=222−22+1+2−1−21+1=22−22+1+2−1−21+1=1−2+21+2−1−21+1=1=12−1x1−1dx= lnx−x 12−1=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.