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]

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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 \displaystyle y=1 and above partly by \displaystyle y=x+2 and partly by \displaystyle y=1/x\,.

If we denote the x-coordinates of the intersection points between the curves by \displaystyle x=a, \displaystyle x=b and \displaystyle x=c, as shown in the figure, we see that the region can be divided up into two parts. In the left part between \displaystyle x=a and \displaystyle x=b, the upper limit is \displaystyle y=x+2, whilst in the right part between \displaystyle x=b and \displaystyle x=c the curve \displaystyle y=1/x is the upper limit.

The area of each part is given by the integrals

\displaystyle \begin{align} \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\,, \end{align}

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:

• \displaystyle x=a: The point of intersection between \displaystyle y=1 and \displaystyle y=x+2 must satisfy both equations of the lines,

\displaystyle \left\{\begin{align} y &= 1\,,\\[5pt] y &= x+2\,\textrm{.} \end{align}\right.

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

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

\displaystyle \left\{\begin{align} y &= x+2\,,\\[5pt] y &= 1/x\,\textrm{.} \end{align}\right.

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

\displaystyle x+2=\frac{1}{x}\,,

which we multiply by \displaystyle x,

\displaystyle x^{2}+2x=1\,\textrm{.}

Completing the square of the left-hand side,

\displaystyle \begin{align} (x+1)^2 - 1^2 &= 1\,,\\[5pt] (x+1)^2 &= 2\,, \end{align}

and taking the square root gives that \displaystyle x=-1\pm \sqrt{2}, leading to

\displaystyle b=-1+\sqrt{2}. (The alternative \displaystyle b=-1-\sqrt{2} lies to the left of \displaystyle x=a\,.)

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

The sub-areas are

\displaystyle \begin{align} \text{Left area} &= \int\limits_{-1}^{\sqrt{2}-1} (x+2-1)\,dx\\[5pt] &= \int\limits_{-1}^{\sqrt{2}-1} (x+1)\,dx\\[5pt] &= \Bigl[\ \frac{x^2}{2} + x\ \Bigr]_{-1}^{\sqrt{2}-1}\\[5pt] &= \frac{\bigl(\sqrt{2}-1\bigr)^2}{2} + \sqrt{2} - 1 - \Bigl(\frac{(-1)^2}{2} + (-1) \Bigr)\\[5pt] &= \frac{\bigl(\sqrt{2}\bigr)^2-2\sqrt{2}+1}{2} + \sqrt{2} - 1 - \frac{1}{2} + 1\\[5pt] &= \frac{2-2\sqrt{2}+1}{2} + \sqrt{2} - 1 - \frac{1}{2} + 1\\[5pt] &= 1 - \sqrt{2} + \frac{1}{2} + \sqrt{2} - 1 - \frac{1}{2} + 1\\[5pt] &= 1\,,\\[10pt] \text{Right area} &= \int\limits_{\sqrt{2}-1}^1 \Bigl(\frac{1}{x}-1\Bigr)\,dx\\[5pt] &= \Bigl[\ \ln |x| - x\ \Bigr]_{\sqrt{2}-1}^1\\[5pt] &= \ln 1 - 1 - \Bigl( \ln \bigl(\sqrt{2}-1\bigr)-\bigl(\sqrt{2}-1\bigr)\Bigr)\\[5pt] &= 0 - 1 - \ln \bigl(\sqrt{2}-1\bigr) + \sqrt{2} - 1\\[5pt] &= \sqrt{2} - 2 - \ln\bigl(\sqrt{2}-1\bigr)\,\textrm{.} \end{align}

The total area is

\displaystyle \begin{align} \text{Area} &= \text{(left area)} + \text{(right area)}\\[5pt] &= 1 + \sqrt{2} - 2 - \ln\bigl(\sqrt{2}-1\bigr)\\[5pt] &= \sqrt{2} - 1 - \ln\bigl(\sqrt{2}-1\bigr)\,\textrm{.} \end{align}