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 \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}