Lösung 2.1:4d

Aus Online Mathematik Brückenkurs 2

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=1 y=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,

\displaystyle x^{2}+2x=1

Completing the square of the left-hand side,

\displaystyle \begin{align} & \left( x+1 \right)^{2}-1^{2}=1 \\ & \left( x+1 \right)^{2}=2 \\ \end{align},

and taking the 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 b=a).

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

The sub-areas are

\displaystyle \begin{align} & \text{Leftarea }=\text{ }\int\limits_{-1}^{\sqrt{2}-1}{\left( x+2-1 \right)}\,dx=\int\limits_{-1}^{\sqrt{2}-1}{\left( x+1 \right)}\,dx \\ & =\left[ \frac{x^{2}}{2}+x \right]_{-1}^{\sqrt{2}-1} \\ & =\frac{\left( \sqrt{2}-1 \right)^{2}}{2}+\sqrt{2}-1-\left( \frac{\left( -1 \right)^{2}}{2}+\left( -1 \right) \right) \\ & =\frac{\left( \sqrt{2} \right)^{2}-2\sqrt{2}+1}{2}+\sqrt{2}-1-\frac{1}{2}+1 \\ & =\frac{2-2\sqrt{2}+1}{2}+\sqrt{2}-1-\frac{1}{2}+1 \\ & =1-\sqrt{2}+\frac{1}{2}+\sqrt{2}-1-\frac{1}{2}+1 \\ & =1 \\ \end{align}

\displaystyle \begin{align} & \text{Rightarea }=\int\limits_{\sqrt{2}-1}^{1}{\left( \frac{1}{x}-1 \right)}\,dx=\left[ \ln \left| x \right|-x \right]_{\sqrt{2}-1}^{1} \\ & =\ln 1-1-\left( \ln \left( \sqrt{2}-1 \right)-\left( \sqrt{2}-1 \right) \right) \\ & =0-1-\ln \left( \sqrt{2}-1 \right)+\sqrt{2}-1 \\ & =\sqrt{2}-2-\ln \left( \sqrt{2}-1 \right) \\ \end{align}