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Lösung 2.2:2d

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
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K (Lösning 2.2:2d moved to Solution 2.2:2d: Robot: moved page)
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What makes the integral not entirely simple is the expression
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<center> [[Image:2_2_2d.gif]] </center>
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<math>\text{1}-x</math>
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under the root sign, so we try the substitution
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<math>u=\text{1}-x</math>,
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<math>\int\limits_{0}^{1}{\sqrt[3]{\text{1}-x}}\,dx=\left\{ \begin{matrix}
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u=\text{1}-x \\
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du=\left( \text{1}-x \right)^{\prime }\,dx=-\,dx \\
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\end{matrix} \right\}=-\int\limits_{1}^{0}{\sqrt[3]{u}}\,du</math>
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Note how the new limits of integration go from
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<math>\text{1}</math>
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to
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<math>0</math>
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(and not the other way around!). It is possible to change the order of the limits we change sign at the same time, i.e.
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<math>-\int\limits_{1}^{0}{\sqrt[3]{u}}\,du=+\int\limits_{0}^{1}{\sqrt[3]{u}}\,du</math>
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All that is now left is routine calculations:
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<math>\begin{align}
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& \int\limits_{0}^{1}{\sqrt[3]{u}}\,du=\int\limits_{0}^{1}{u^{\frac{1}{3}}}\,du=\left[ \frac{u^{\frac{1}{3}+1}}{\frac{1}{3}+1} \right]_{0}^{1} \\
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& =\frac{3}{4}\left[ u^{\frac{4}{3}} \right]_{0}^{1}=\frac{3}{4}\left( 1^{\frac{4}{3}}-0^{\frac{4}{3}} \right)=\frac{3}{4} \\
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\end{align}</math>

Version vom 12:34, 20. Okt. 2008

What makes the integral not entirely simple is the expression 1−x under the root sign, so we try the substitution u=1−x,


1031−xdx=u=1−xdu=1−xdx=−dx=−013udu 


Note how the new limits of integration go from 1 to 0 (and not the other way around!). It is possible to change the order of the limits we change sign at the same time, i.e.


−013udu=+103udu 


All that is now left is routine calculations:


103udu=10u31du=u31+131+110=43u3410=43134−034=43

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