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Lösung 2.2:2a

Aus Online Mathematik Brückenkurs 2

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
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The integral is a standard integral, with <math>5x</math> as the argument of the cosine function. If we therefore substitute <math>u=5x</math>, we obtain the “correct” argument of the cosine,
The integral is a standard integral, with <math>5x</math> as the argument of the cosine function. If we therefore substitute <math>u=5x</math>, we obtain the “correct” argument of the cosine,
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{{Displayed math||<math>\int\limits_0^{\pi} \cos 5x\,dx = \left\{\begin{align}
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{{Abgesetzte Formel||<math>\int\limits_0^{\pi} \cos 5x\,dx = \left\{\begin{align}
u &= 5x\\[5pt]
u &= 5x\\[5pt]
du &= (5x)'\,dx = 5\,dx
du &= (5x)'\,dx = 5\,dx
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Now, we have a standard integral which we can easily compute,
Now, we have a standard integral which we can easily compute,
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{{Displayed math||<math>\frac{1}{5}\int\limits_0^{5\pi} \cos u\,du = \frac{1}{5}\Bigl[\ \sin u\ \Bigr]_0^{5\pi} = \frac{1}{5}( \sin 5\pi -\sin 0) = \frac{1}{5}(0-0) = 0\,\textrm{.}</math>}}
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{{Abgesetzte Formel||<math>\frac{1}{5}\int\limits_0^{5\pi} \cos u\,du = \frac{1}{5}\Bigl[\ \sin u\ \Bigr]_0^{5\pi} = \frac{1}{5}( \sin 5\pi -\sin 0) = \frac{1}{5}(0-0) = 0\,\textrm{.}</math>}}

Version vom 13:01, 10. Mär. 2009

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=udu=5x=(5x)dx=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 we can easily compute,

5150cosudu=51 sinu 05=51(sin5−sin0)=51(0−0)=0. 


Note: If we draw the graph of 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“