Lösung 2.1:1d

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The modulus function,
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The modulus function, <math>|x|</math>, strips <math>x</math> of its sign, e.g.
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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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<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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{{Displayed math||<math>|-5|=5\,</math>, <math>\quad|3|=3\quad</math> and <math>\quad |-\pi|=\pi\,</math>.}}
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For positive values of <math>x</math>, the modulus function has no effect, since
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<math>|x|=x</math>, but for negative <math>x</math> the modulus function changes the sign of <math>x</math>, i.e. <math>|x|=-x</math> (remember that <math>x</math>
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is negative and therefore <math>-x</math> is positive).
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If we draw a graph of <math>y=|x|</math> it will consist of two parts. For
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<math>x\ge 0</math> we have <math>y=x</math>, and for <math>x\le 0</math> we have
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<math>y=-x\,</math>.
[[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 <math>y=|x|</math> and between <math>x=-1</math> and <math>x=2</math>.
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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
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>
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{{Displayed math||<math>\int\limits_{-1}^{2} |x|\,dx = \frac{1}{2}\cdot 1\cdot 1 + \frac{1}{2}\cdot 2\cdot 2 = \frac{5}{2}\,\textrm{.}</math>}}

Version vom 12:25, 21. Okt. 2008

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

\displaystyle |-5|=5\,, \displaystyle \quad|3|=3\quad and \displaystyle \quad |-\pi|=\pi\,.

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

If we draw a graph of \displaystyle y=|x| 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=|x| and between \displaystyle x=-1 and \displaystyle x=2.

This region consists of two triangles and we therefore obtain

\displaystyle \int\limits_{-1}^{2} |x|\,dx = \frac{1}{2}\cdot 1\cdot 1 + \frac{1}{2}\cdot 2\cdot 2 = \frac{5}{2}\,\textrm{.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:1d“