Lösung 2.1:1d

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The modulus function, <math>|x|</math>, strips <math>x</math> of its sign, e.g.
The modulus function, <math>|x|</math>, strips <math>x</math> of its sign, e.g.
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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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{{Abgesetzte Formel||<math>|-5|=5\,</math>, <math>\quad|3|=3\quad</math> and <math>\quad |-\pi|=\pi\,</math>.}}
For positive values of <math>x</math>, the modulus function has no effect, since
For positive values of <math>x</math>, the modulus function has no effect, since
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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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{{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>}}
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{{Abgesetzte Formel||<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 12:57, 10. Mär. 2009

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“