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Aus Online Mathematik Brückenkurs 2

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Version vom 10:18, 11. 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

\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}