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Lösung 2.2:2b

Aus Online Mathematik Brückenkurs 2

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If we set
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If we set <math>u=2x+3</math>, the integral simplifies to <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 <math>dx</math> and <math>du</math>, which can give undesired effects. In this case, however, we have
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<math>u=\text{2}x+\text{3}</math>
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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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{{Displayed math||<math>du = (2x+3)'\,dx = 2\,dx</math>}}
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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 <math>u = 2x+3</math> seems to work, in spite of everything,
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{{Displayed math||<math>\begin{align}
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\int\limits_0^{1/2} e^{2x+3}\,dx &= \left\{\begin{align}
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u &= 2x+3\\[5pt]
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du &= 2\,dx
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\end{align}\right\} = \frac{1}{2}\int\limits_3^4 e^u\,du\\[5pt]
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&= \frac{1}{2}\Bigl[\ e^u\ \Bigr]_3^4 = \frac{1}{2}\bigl(e^4-e^3\bigr)\,\textrm{.}
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\end{align}</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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Note: Another possible substitution is <math>u=e^{2x+3}</math> which also happens to work (usually, such an extensive substitution almost always fails).
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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).
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Version vom 13:28, 28. Okt. 2008

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

du=(2x+3)dx=2dx

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

120e2x+3dx=udu=2x+3=2dx=2143eudu=21 eu 43=21e4−e3.


Note: Another possible substitution is u=e2x+3 which also happens to work (usually, such an extensive substitution almost always fails).

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