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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=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.