Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
No jsMath TeX fonts found -- using image fonts instead.
These may be slow and might not print well.
Use the jsMath control panel to get additional information.
jsMath Control PanelHide this Message

jsMath

Lösung 2.2:2c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 2.2:2c moved to Solution 2.2:2c: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
If we focus on the integrand, then the substitution
-
<center> [[Image:2_2_2c.gif]] </center>
+
<math>u=\text{3}x+\text{1}</math>
-
{{NAVCONTENT_STOP}}
+
seems suitable, since we then get
 +
<math>\sqrt{u}</math>
 +
which we can integrate. There is also no risk involved in using a linear substitution such as
 +
<math>u=\text{3}x+\text{1}</math>, because the relation between
 +
<math>dx\text{ }</math>
 +
and
 +
<math>du</math>
 +
will be a constant factor,
 +
 
 +
 
 +
<math>du=\left( \text{3}x+\text{1} \right)^{1}\,dx=3\,dx</math>
 +
 
 +
 
 +
which does not cause any problems.
 +
 
 +
We obtain
 +
 
 +
 
 +
<math>\begin{align}
 +
& \int\limits_{0}^{5}{\sqrt{3x+1}}\,dx=\left\{ \begin{matrix}
 +
u=\text{3}x+\text{1} \\
 +
du=3\,dx \\
 +
\end{matrix} \right\}=\frac{1}{3}\int\limits_{1}^{16}{\sqrt{u}}\,du \\
 +
& =\frac{1}{3}\int\limits_{1}^{16}{u^{{1}/{2}\;}}\,du=\frac{1}{3}\left[ \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1} \right]_{1}^{16} \\
 +
& =\frac{1}{3}\left[ \frac{2}{3}u\sqrt{u} \right]_{1}^{16}=\frac{2}{9}\left( 16\sqrt{16}-1\sqrt{1} \right) \\
 +
& =\frac{2}{9}\left( 16\centerdot 4-1 \right)=\frac{2\centerdot 63}{9}=14 \\
 +
\end{align}</math>

Version vom 12:10, 20. Okt. 2008

If we focus on the integrand, then the substitution u=3x+1 seems suitable, since we then get u  which we can integrate. There is also no risk involved in using a linear substitution such as u=3x+1, because the relation between dx and du will be a constant factor,


du=3x+11dx=3dx 


which does not cause any problems.

We obtain


503x+1dx=u=3x+1du=3dx=31161udu=31161u12du=31u21+121+1116=3132uu116=921616−11=92164−1=9263=14

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