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

Lösung 2.2:4a

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 2.2:4a moved to Solution 2.2:4a: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
What makes our integral differ from that in the exercise´s text is that there is a term
-
<center> [[Image:2_2_4a.gif]] </center>
+
<math>x^{2}+4</math>
-
{{NAVCONTENT_STOP}}
+
instead of
 +
<math>x^{2}+1</math>, but if we factor out the 4 from the denominator,
 +
 
 +
 
 +
<math>\int{\frac{\,dx}{x^{2}+4}}=\int{\frac{\,dx}{4\left( \frac{1}{4}x^{2}+1 \right)}}=\frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}x^{2}+1}}</math>,
 +
 
 +
we obtain the correct second term in the denominator. On the other hand, there is no longer
 +
<math>x^{2}</math>
 +
but
 +
<math>\frac{1}{4}x^{2}</math>, although we can get around this by substituting
 +
<math>u=\frac{1}{2}x</math>,
 +
 
 +
 
 +
<math>\begin{align}
 +
& \frac{1}{4}\int{\frac{\,dx}{\frac{1}{4}x^{2}+1}}=\frac{1}{4}\int{\frac{\,dx}{\left( \frac{x}{2} \right)^{2}+1}}=\left\{ \begin{matrix}
 +
u=\frac{1}{2}x \\
 +
du=\frac{1}{2}\,dx \\
 +
\end{matrix} \right\} \\
 +
& =\frac{1}{4}\int{\frac{2\,du}{u^{2}+1}}=\frac{1}{2}\int{\frac{\,du}{u^{2}+1}} \\
 +
& =\frac{1}{2}\arctan u+C=\frac{1}{2}\arctan \frac{x}{2}+C \\
 +
\end{align}</math>

Version vom 12:22, 21. Okt. 2008

What makes our integral differ from that in the exercise´s text is that there is a term x2+4 instead of x2+1, but if we factor out the 4 from the denominator,


dxx2+4=dx441x2+1=41dx41x2+1 ,

we obtain the correct second term in the denominator. On the other hand, there is no longer x2 but 41x2, although we can get around this by substituting u=21x,


41dx41x2+1=41dx2x2+1=u=21xdu=21dx=412duu2+1=21duu2+1=21arctanu+C=21arctanx2+C

Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:4a“