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

Lösung 2.2:2c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
If we focus on the integrand, then the substitution <math>u=3x+1</math> 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=3x+1</math>, because the relation between <math>dx</math> and <math>du</math> will be a constant factor,
If we focus on the integrand, then the substitution <math>u=3x+1</math> 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=3x+1</math>, because the relation between <math>dx</math> and <math>du</math> will be a constant factor,
-
{{Displayed math||<math>du = (3x+1)'\,dx = 3\,dx\,,</math>}}
+
{{Abgesetzte Formel||<math>du = (3x+1)'\,dx = 3\,dx\,,</math>}}
which does not cause any problems.
which does not cause any problems.
Zeile 7: Zeile 7:
We obtain
We obtain
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\int\limits_0^5 \sqrt{3x+1}\,dx
\int\limits_0^5 \sqrt{3x+1}\,dx
&= \left\{\begin{align}
&= \left\{\begin{align}

Version vom 13:01, 10. Mär. 2009

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+1)dx=3dx

which does not cause any problems.

We obtain

503x+1dx=udu=3x+1=3dx=31161udu=31161u12du=31 21+1u12+1 116=31 32uu 116=921616−11=92164−1=9263=14.
Von „http://wiki.math.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_2.2:2c“