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Lösung 2.2:2a

Aus Online Mathematik Brückenkurs 2

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K (Lösning 2.2:2a moved to Solution 2.2:2a: Robot: moved page)
Zeile 1: Zeile 1:
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The integral is a standard integral, with
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<center> [[Image:2_2_2a-1(2).gif]] </center>
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<math>\text{5}x</math>
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as the argument of the cosine function. If we therefore substitute
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{{NAVCONTENT_START}}
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<math>u=\text{5}x</math>, we obtain the “correct” argument of the cosine,
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<center> [[Image:2_2_2a-2(2).gif]] </center>
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{{NAVCONTENT_STOP}}
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<math>\begin{align}
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& \int\limits_{0}^{\pi }{\cos 5x\,dx=\left\{ \begin{array}{*{35}l}
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u=\text{5}x \\
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du=\left( 5x \right)^{\prime }\,dx=5\,dx \\
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\end{array} \right\}} \\
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& =\frac{1}{5}\int\limits_{0}^{5\pi }{\cos u\,du} \\
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\end{align}</math>
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As can be seen, the variable change replaced
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<math>dx</math>
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by
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<math>\frac{1}{5}\,du</math>
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and the new limits of integration become
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<math>u=5\centerdot 0=0</math>
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and
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<math>u=5\centerdot \pi =5\pi </math>.
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Now, we have a standard integral which can easily compute:
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<math>\begin{align}
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& \frac{1}{5}\int\limits_{0}^{5\pi }{\cos u\,du}=\frac{1}{5}\left[ \sin u \right]_{0}^{5\pi } \\
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& =\frac{1}{5}\left( \sin 5\pi -\sin 0 \right)=\frac{1}{5}\left( 0-0 \right)=0 \\
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\end{align}</math>
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NOTE: if we draw the graph for
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<math>y=\cos 5x</math>, we see also that the area between the curve and
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<math>x</math>
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-axis above the
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<math>x</math>
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-axis is the same as the area under the
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<math>x</math>
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-axis.
[[Image:2_2_2_a.gif|center]]
[[Image:2_2_2_a.gif|center]]

Version vom 10:07, 19. Okt. 2008

The integral is a standard integral, with 5x as the argument of the cosine function. If we therefore substitute u=5x, we obtain the “correct” argument of the cosine,


0cos5xdx=u=5xdu=5xdx=5dx=5150cosudu

As can be seen, the variable change replaced dx by 51du and the new limits of integration become u=50=0 and u=5=5.

Now, we have a standard integral which can easily compute:


5150cosudu=51sinu05=51sin5−sin0=510−0=0


NOTE: if we draw the graph for y=cos5x, we see also that the area between the curve and x -axis above the x -axis is the same as the area under the x -axis.


Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:2a“