Lösung 2.1:1d

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The modulus function,
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<center> [[Image:2_1_1d-1(2).gif]] </center>
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<math>\left| x \right|</math>, strips
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<math>x</math>
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of its sign, e.g.
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<center> [[Image:2_1_1d-2(2).gif]] </center>
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<math>\left| -5 \right|=5</math>,
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<math>\left| 3 \right|=3</math>
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and
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<math>\left| -\pi \right|=\pi </math>
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For positive values of
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<math>x</math>, the modulus function has no effect, since
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<math>\left| x \right|=x</math>, but for negative
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<math>x</math>
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the modulus function changes the sign of
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<math>x</math>, i.e.
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<math>\left| -x \right|=x</math>
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(remember that
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<math>x</math>
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is negative and therefore
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<math>-x</math>
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is positive).
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If we draw a graph of
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<math>y=\left| x \right|</math>
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it will consist of two parts. For
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<math>x\ge 0</math>
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we have
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<math>y=x</math>, and for
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<math>x\le 0</math>
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we have
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<math>y=-x</math>
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[[Image:2_1_1_d1.gif|center]]
[[Image:2_1_1_d1.gif|center]]
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The value of the integral is the area of the region under the graph
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<math>y=\left| x \right|</math>
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and between
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<math>x=-1</math>
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and
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<math>x=2</math>.
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[[Image:2_1_1_d2.gif|center]]
[[Image:2_1_1_d2.gif|center]]
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This region consists of two triangles and we therefore obtain
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<math>\int\limits_{-1}^{2}{\left| x \right|}\,dx=\frac{1}{2}\centerdot 1\centerdot 1+\frac{1}{2}\centerdot 2\centerdot 2=\frac{5}{2}</math>

Version vom 12:24, 17. Okt. 2008

The modulus function, \displaystyle \left| x \right|, strips \displaystyle x of its sign, e.g.


\displaystyle \left| -5 \right|=5, \displaystyle \left| 3 \right|=3 and \displaystyle \left| -\pi \right|=\pi


For positive values of \displaystyle x, the modulus function has no effect, since \displaystyle \left| x \right|=x, but for negative \displaystyle x the modulus function changes the sign of \displaystyle x, i.e. \displaystyle \left| -x \right|=x (remember that \displaystyle x is negative and therefore \displaystyle -x is positive).

If we draw a graph of \displaystyle y=\left| x \right| it will consist of two parts. For \displaystyle x\ge 0 we have \displaystyle y=x, and for \displaystyle x\le 0 we have \displaystyle y=-x




The value of the integral is the area of the region under the graph \displaystyle y=\left| x \right| and between \displaystyle x=-1 and \displaystyle x=2.



This region consists of two triangles and we therefore obtain

\displaystyle \int\limits_{-1}^{2}{\left| x \right|}\,dx=\frac{1}{2}\centerdot 1\centerdot 1+\frac{1}{2}\centerdot 2\centerdot 2=\frac{5}{2}

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