Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Lösung 2.2:2b

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 2.2:2b moved to Solution 2.2:2b: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
If we set
-
<center> [[Image:2_2_2b.gif]] </center>
+
<math>u=\text{2}x+\text{3}</math>
-
{{NAVCONTENT_STOP}}
+
, 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\text{ }</math>
 +
and
 +
<math>du</math>, which can give undesired effects. In this case, however, we have
 +
 
 +
 
 +
<math>du=\left( \text{2}x+\text{3} \right)^{\prime }\,dx=2\,dx</math>
 +
 
 +
 
 +
which only affects by a constant factor, so the substitution
 +
<math>u=\text{2}x+\text{3}</math>
 +
seems to work, in spite of everything:
 +
 
 +
 
 +
<math>\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}</math>
 +
 
 +
 
 +
NOTE: Another possible substitution is
 +
<math>u=e^{2x+3}</math>
 +
which also happens to work (usually, such an extensive substitution almost always fails).

Version vom 12:36, 20. 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+3dx=2dx 


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


120e2x+3dx=u=2x+3du=2dx=2143eudu=21eu43=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“