Lösung 2.2:2a

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The integral is a standard integral, with
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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,
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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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<math>u=\text{5}x</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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u &= 5x\\[5pt]
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du &= (5x)'\,dx = 5\,dx
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\end{align}\right\} = \frac{1}{5}\int\limits_0^{5\pi} \cos u\,du\,\textrm{.}</math>}}
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<math>\begin{align}
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As can be seen, the variable change replaced <math>dx</math> by <math>\tfrac{1}{5}\,du</math> and the new limits of integration become <math>u=5\cdot 0=0</math>
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& \int\limits_{0}^{\pi }{\cos 5x\,dx=\left\{ \begin{array}{*{35}l}
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and <math>u=5\cdot \pi = 5\pi\,</math>.
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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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Now, we have a standard integral which we can easily compute,
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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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{{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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<math>\begin{align}
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Note: If we draw the graph of <math>y=\cos 5x</math>, 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.
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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.
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[[Image:2_2_2_a.gif|center]]
[[Image:2_2_2_a.gif|center]]

Version vom 12:57, 28. Okt. 2008

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

\displaystyle \int\limits_0^{\pi} \cos 5x\,dx = \left\{\begin{align}

u &= 5x\\[5pt] du &= (5x)'\,dx = 5\,dx \end{align}\right\} = \frac{1}{5}\int\limits_0^{5\pi} \cos u\,du\,\textrm{.}

As can be seen, the variable change replaced \displaystyle dx by \displaystyle \tfrac{1}{5}\,du and the new limits of integration become \displaystyle u=5\cdot 0=0 and \displaystyle u=5\cdot \pi = 5\pi\,.

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

\displaystyle \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{.}


Note: If we draw the graph of \displaystyle y=\cos 5x, 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.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:2a“