Lösung 2.1:4e

Aus Online Mathematik Brückenkurs 2

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The double inequality means that we look for the area of the region which is bounded above in the y-direction by the straight line
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<center> [[Image:2_1_4e-1(3).gif]] </center>
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<math>y=x+\text{2}</math>
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and from below by the parabola
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<math>y=x^{2}</math>.
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<center> [[Image:2_1_4e-2(3).gif]] </center>
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If we sketch the line and the parabola, the region is given by the region shaded in the figure below.
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<center> [[Image:2_1_4e-3(3).gif]] </center>
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[[Image:2_1_4_e.gif|center]]
[[Image:2_1_4_e.gif|center]]
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As soon as we have determined the
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<math>x</math>
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-coordinates of the points of intersection,
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<math>x=a</math>
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and
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<math>x=b</math>, between the line and the parabola, we can calculate the area as the integral of the difference between the curves'
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<math>y</math>
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-values:
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Area=
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<math>\int\limits_{a}^{b}{\left( x+2-x^{2} \right)}\,dx</math>
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The curves' points of intersection are those points which lie on both curves, i.e. which satisfy both curves' equations:
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<math>\left\{ \begin{matrix}
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y=x+\text{2} \\
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y=x^{2} \\
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\end{matrix} \right.</math>
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By eliminating
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<math>y</math>, we obtain an equation for
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<math>x</math>,
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<math>x^{2}=x+2</math>
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If we move all x-terms to the left-hand side,
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<math>x^{2}-x=2</math>
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and complete the square, we obtain
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<math>\begin{align}
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& \left( x-\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}=2 \\
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& \left( x-\frac{1}{2} \right)^{2}=\frac{9}{4} \\
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\end{align}</math>
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Taking the root then gives that
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<math>x=\frac{1}{2}\pm \frac{3}{2}</math>. In other words,
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<math>x=-\text{1}</math>
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and
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<math>x=\text{2}</math>.
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The area of the region is now given by
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<math>\begin{align}
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& \text{Area}=\int\limits_{-1}^{2}{\left( x+2-x^{2} \right)}\,dx=\left[ \frac{x^{2}}{2}+2x-\frac{x^{3}}{3} \right]_{-1}^{2} \\
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& =\frac{2^{2}}{2}+2\centerdot 2-\frac{2^{3}}{3}-\left( \frac{\left( -1 \right)^{2}}{2}+2\centerdot \left( -1 \right)-\frac{\left( -1 \right)^{3}}{3} \right) \\
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& =2+4-\frac{8}{3}-\frac{1}{2}+2-\frac{1}{3} \\
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& =\frac{9}{2} \\
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\end{align}</math>

Version vom 13:05, 18. Okt. 2008

The double inequality means that we look for the area of the region which is bounded above in the y-direction by the straight line \displaystyle y=x+\text{2} and from below by the parabola \displaystyle y=x^{2}.

If we sketch the line and the parabola, the region is given by the region shaded in the figure below.



As soon as we have determined the \displaystyle x -coordinates of the points of intersection, \displaystyle x=a and \displaystyle x=b, between the line and the parabola, we can calculate the area as the integral of the difference between the curves' \displaystyle y -values:

Area= \displaystyle \int\limits_{a}^{b}{\left( x+2-x^{2} \right)}\,dx


The curves' points of intersection are those points which lie on both curves, i.e. which satisfy both curves' equations:


\displaystyle \left\{ \begin{matrix} y=x+\text{2} \\ y=x^{2} \\ \end{matrix} \right.


By eliminating \displaystyle y, we obtain an equation for \displaystyle x,


\displaystyle x^{2}=x+2


If we move all x-terms to the left-hand side,


\displaystyle x^{2}-x=2


and complete the square, we obtain


\displaystyle \begin{align} & \left( x-\frac{1}{2} \right)^{2}-\left( \frac{1}{2} \right)^{2}=2 \\ & \left( x-\frac{1}{2} \right)^{2}=\frac{9}{4} \\ \end{align}

Taking the root then gives that \displaystyle x=\frac{1}{2}\pm \frac{3}{2}. In other words, \displaystyle x=-\text{1} and \displaystyle x=\text{2}.

The area of the region is now given by


\displaystyle \begin{align} & \text{Area}=\int\limits_{-1}^{2}{\left( x+2-x^{2} \right)}\,dx=\left[ \frac{x^{2}}{2}+2x-\frac{x^{3}}{3} \right]_{-1}^{2} \\ & =\frac{2^{2}}{2}+2\centerdot 2-\frac{2^{3}}{3}-\left( \frac{\left( -1 \right)^{2}}{2}+2\centerdot \left( -1 \right)-\frac{\left( -1 \right)^{3}}{3} \right) \\ & =2+4-\frac{8}{3}-\frac{1}{2}+2-\frac{1}{3} \\ & =\frac{9}{2} \\ \end{align}