Lösung 2.2:2b

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{{NAVCONTENT_START}}
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If we set
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<center> [[Image:2_2_2b.gif]] </center>
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<math>u=\text{2}x+\text{3}</math>
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{{NAVCONTENT_STOP}}
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, the integral simplifies to
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<math>e^{u}</math>. However, this is only part of the truth. We must in addition take account of the relation between the integration element
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<math>dx\text{ }</math>
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and
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<math>du</math>, which can give undesired effects. In this case, however, we have
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<math>du=\left( \text{2}x+\text{3} \right)^{\prime }\,dx=2\,dx</math>
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which only affects by a constant factor, so the substitution
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<math>u=\text{2}x+\text{3}</math>
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seems to work, in spite of everything:
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<math>\begin{align}
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& \int\limits_{0}^{{1}/{2}\;}{e^{\text{2}x+\text{3}}}\,dx=\left\{ \begin{matrix}
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u=\text{2}x+\text{3} \\
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du=2\,dx \\
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\end{matrix} \right\}=\frac{1}{2}\int\limits_{3}^{4}{e^{u}\,du} \\
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& =\frac{1}{2}\left[ e^{u} \right]_{3}^{4}=\frac{1}{2}\left( e^{4}-e^{3} \right) \\
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\end{align}</math>
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NOTE: Another possible substitution is
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<math>u=e^{2x+3}</math>
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which also happens to work (usually, such an extensive substitution almost always fails).

Version vom 12:36, 20. Okt. 2008

If we set \displaystyle u=\text{2}x+\text{3} , the integral simplifies to \displaystyle e^{u}. However, this is only part of the truth. We must in addition take account of the relation between the integration element \displaystyle dx\text{ } and \displaystyle du, which can give undesired effects. In this case, however, we have


\displaystyle du=\left( \text{2}x+\text{3} \right)^{\prime }\,dx=2\,dx


which only affects by a constant factor, so the substitution \displaystyle u=\text{2}x+\text{3} seems to work, in spite of everything:


\displaystyle \begin{align} & \int\limits_{0}^{{1}/{2}\;}{e^{\text{2}x+\text{3}}}\,dx=\left\{ \begin{matrix} u=\text{2}x+\text{3} \\ du=2\,dx \\ \end{matrix} \right\}=\frac{1}{2}\int\limits_{3}^{4}{e^{u}\,du} \\ & =\frac{1}{2}\left[ e^{u} \right]_{3}^{4}=\frac{1}{2}\left( e^{4}-e^{3} \right) \\ \end{align}


NOTE: Another possible substitution is \displaystyle u=e^{2x+3} which also happens to work (usually, such an extensive substitution almost always fails).

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