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Lösung 2.1:1d

Aus Online Mathematik Brückenkurs 2

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K (Lösning 2.1:1d moved to Solution 2.1:1d: Robot: moved page)
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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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{{NAVCONTENT_STOP}}
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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
 +
<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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 +
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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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 +
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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, x, strips x of its sign, e.g.


−5=5, 3=3 and −=


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

If we draw a graph of y=x it will consist of two parts. For x0 we have y=x, and for x0 we have y=−x




The value of the integral is the area of the region under the graph y=x and between x=−1 and 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.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.1:1d“