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Lösung 2.1:4c

Aus Online Mathematik Brückenkurs 2

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K (Lösning 2.1:4c moved to Solution 2.1:4c: Robot: moved page)
Zeile 1: Zeile 1:
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{{NAVCONTENT_START}}
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First, we need a picture of what the region looks like.
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<center> [[Image:2_1_4c-1(4).gif]] </center>
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{{NAVCONTENT_STOP}}
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Both curves,
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<math>y=\frac{1}{4}x^{2}+2</math>
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<center> [[Image:2_1_4c-2(4).gif]] </center>
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and
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{{NAVCONTENT_STOP}}
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<math>y=8-\frac{1}{8}x^{2}</math>, are parabolas, the first with a minimum value
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{{NAVCONTENT_START}}
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<math>y=\text{2}</math>
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<center> [[Image:2_1_4c-3(4).gif]] </center>
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when
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{{NAVCONTENT_STOP}}
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<math>x=0</math>, and the second with a maximum value of
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{{NAVCONTENT_START}}
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<math>y=\text{8}</math>
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<center> [[Image:2_1_4c-4(4).gif]] </center>
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when
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{{NAVCONTENT_STOP}}
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<math>x=0</math>. Roughly speaking, the curves have the appearance shown in the figure below, where the shaded region whose area we are trying to find lies between the curves.
[[Image:2_1_4_c.gif|center]]
[[Image:2_1_4_c.gif|center]]
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 +
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The region is bounded above by the parabola
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<math>y=8-\frac{1}{8}x^{2}</math>
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and below by the parabola
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<math>y=\frac{1}{4}x^{2}+2</math>. If we can determine the
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<math>x</math>
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-coordinates,
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<math>x=a\text{ }</math>
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and
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<math>x=b</math>, for the points of intersection between the curves, the area we are looking for will be given by
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Area =
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<math>\int\limits_{a}^{b}{\left( \left( 8-\frac{1}{8}x^{2} \right)-\left( \frac{1}{4}x^{2}+2 \right) \right)}\,dx</math>
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 +
 +
The integrand is the
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<math>y</math>
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-value for the upper parabola minus the corresponding
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<math>y</math>
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-value for the lower parabola.
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 +
At the points where the curves intersect each other, the
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<math>x</math>
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- and
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<math>y</math>
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-coordinates are equal, which gives the equation system,
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 +
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<math>\left\{ \begin{matrix}
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y=8-\frac{1}{8}x^{2} \\
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y=\frac{1}{4}x^{2}+2 \\
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\end{matrix} \right.</math>
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If we eliminate
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<math>y</math>
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from this system, we get the following equation for
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<math>x</math>,
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 +
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<math>8-\frac{1}{8}x^{2}=\frac{1}{4}x^{2}+2</math>
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If we move the
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<math>x^{2}</math>
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terms onto one side and the constants onto the other, we obtain
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<math>\frac{1}{4}x^{2}+\frac{1}{8}x^{2}=8-2</math>,
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i.e.
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<math>\begin{align}
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& \left( \frac{1}{4}+\frac{1}{8} \right)x^{2}=6 \\
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& \frac{3}{8}x^{2}=6 \\
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& x^{2}=16 \\
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\end{align}</math>
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 +
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The
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<math>x</math>
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-coordinates of the points of intersection are therefore equal to
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<math>x=-\text{4 }</math>
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and
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<math>x=\text{4}</math>.
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The area of the area between the curves is given by
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 +
 +
<math>\begin{align}
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& \text{Area}=\int\limits_{-4}^{4}{\left( \left( 8-\frac{1}{8}x^{2} \right)-\left( \frac{1}{4}x^{2}+2 \right) \right)}\,dx \\
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& =\int\limits_{-4}^{4}{\left( 8-\frac{1}{8}x^{2}-\frac{1}{4}x^{2}-2 \right)}\,dx \\
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& =\int\limits_{-4}^{4}{\left( 6-\left( \frac{1}{8}+\frac{1}{4} \right)x^{2} \right)}\,dx \\
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& =\int\limits_{-4}^{4}{\left( 6-\frac{3}{8}x^{2} \right)}\,dx \\
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& =\left[ 6x-\frac{3}{8}\frac{x^{3}}{3} \right]_{-4}^{4} \\
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& =\left[ 6x-\frac{x^{3}}{8} \right]_{-4}^{4} \\
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& =6\centerdot 4-\frac{4^{3}}{8}-\left( 6\left( -4 \right)-\frac{\left( -4 \right)^{3}}{8} \right) \\
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& =24-8+24-8=32 \\
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\end{align}</math>

Version vom 12:01, 18. Okt. 2008

First, we need a picture of what the region looks like.

Both curves, y=41x2+2 and y=8−81x2, are parabolas, the first with a minimum value y=2 when x=0, and the second with a maximum value of y=8 when x=0. Roughly speaking, the curves have the appearance shown in the figure below, where the shaded region whose area we are trying to find lies between the curves.



The region is bounded above by the parabola y=8−81x2 and below by the parabola y=41x2+2. If we can determine the x -coordinates, x=a and x=b, for the points of intersection between the curves, the area we are looking for will be given by

Area = ba8−81x2−41x2+2dx 


The integrand is the y -value for the upper parabola minus the corresponding y -value for the lower parabola.

At the points where the curves intersect each other, the x - and y -coordinates are equal, which gives the equation system,


y=8−81x2y=41x2+2 


If we eliminate y from this system, we get the following equation for x,


8−81x2=41x2+2


If we move the x2 terms onto one side and the constants onto the other, we obtain


41x2+81x2=8−2,

i.e.


41+81x2=683x2=6x2=16


The x -coordinates of the points of intersection are therefore equal to x=−4 and x=4.

The area of the area between the curves is given by


\displaystyle \begin{align} & \text{Area}=\int\limits_{-4}^{4}{\left( \left( 8-\frac{1}{8}x^{2} \right)-\left( \frac{1}{4}x^{2}+2 \right) \right)}\,dx \\ & =\int\limits_{-4}^{4}{\left( 8-\frac{1}{8}x^{2}-\frac{1}{4}x^{2}-2 \right)}\,dx \\ & =\int\limits_{-4}^{4}{\left( 6-\left( \frac{1}{8}+\frac{1}{4} \right)x^{2} \right)}\,dx \\ & =\int\limits_{-4}^{4}{\left( 6-\frac{3}{8}x^{2} \right)}\,dx \\ & =\left[ 6x-\frac{3}{8}\frac{x^{3}}{3} \right]_{-4}^{4} \\ & =\left[ 6x-\frac{x^{3}}{8} \right]_{-4}^{4} \\ & =6\centerdot 4-\frac{4^{3}}{8}-\left( 6\left( -4 \right)-\frac{\left( -4 \right)^{3}}{8} \right) \\ & =24-8+24-8=32 \\ \end{align}

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:4c“