Lösung 2.2:2d

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 2.2:2d moved to Solution 2.2:2d: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
What makes the integral not entirely simple is the expression
-
<center> [[Image:2_2_2d.gif]] </center>
+
<math>\text{1}-x</math>
-
{{NAVCONTENT_STOP}}
+
under the root sign, so we try the substitution
 +
<math>u=\text{1}-x</math>,
 +
 
 +
 
 +
<math>\int\limits_{0}^{1}{\sqrt[3]{\text{1}-x}}\,dx=\left\{ \begin{matrix}
 +
u=\text{1}-x \\
 +
du=\left( \text{1}-x \right)^{\prime }\,dx=-\,dx \\
 +
\end{matrix} \right\}=-\int\limits_{1}^{0}{\sqrt[3]{u}}\,du</math>
 +
 
 +
 
 +
Note how the new limits of integration go from
 +
<math>\text{1}</math>
 +
to
 +
<math>0</math>
 +
(and not the other way around!). It is possible to change the order of the limits we change sign at the same time, i.e.
 +
 
 +
 
 +
<math>-\int\limits_{1}^{0}{\sqrt[3]{u}}\,du=+\int\limits_{0}^{1}{\sqrt[3]{u}}\,du</math>
 +
 
 +
 
 +
All that is now left is routine calculations:
 +
 
 +
 
 +
<math>\begin{align}
 +
& \int\limits_{0}^{1}{\sqrt[3]{u}}\,du=\int\limits_{0}^{1}{u^{\frac{1}{3}}}\,du=\left[ \frac{u^{\frac{1}{3}+1}}{\frac{1}{3}+1} \right]_{0}^{1} \\
 +
& =\frac{3}{4}\left[ u^{\frac{4}{3}} \right]_{0}^{1}=\frac{3}{4}\left( 1^{\frac{4}{3}}-0^{\frac{4}{3}} \right)=\frac{3}{4} \\
 +
\end{align}</math>

Version vom 12:34, 20. Okt. 2008

What makes the integral not entirely simple is the expression \displaystyle \text{1}-x under the root sign, so we try the substitution \displaystyle u=\text{1}-x,


\displaystyle \int\limits_{0}^{1}{\sqrt[3]{\text{1}-x}}\,dx=\left\{ \begin{matrix} u=\text{1}-x \\ du=\left( \text{1}-x \right)^{\prime }\,dx=-\,dx \\ \end{matrix} \right\}=-\int\limits_{1}^{0}{\sqrt[3]{u}}\,du


Note how the new limits of integration go from \displaystyle \text{1} to \displaystyle 0 (and not the other way around!). It is possible to change the order of the limits we change sign at the same time, i.e.


\displaystyle -\int\limits_{1}^{0}{\sqrt[3]{u}}\,du=+\int\limits_{0}^{1}{\sqrt[3]{u}}\,du


All that is now left is routine calculations:


\displaystyle \begin{align} & \int\limits_{0}^{1}{\sqrt[3]{u}}\,du=\int\limits_{0}^{1}{u^{\frac{1}{3}}}\,du=\left[ \frac{u^{\frac{1}{3}+1}}{\frac{1}{3}+1} \right]_{0}^{1} \\ & =\frac{3}{4}\left[ u^{\frac{4}{3}} \right]_{0}^{1}=\frac{3}{4}\left( 1^{\frac{4}{3}}-0^{\frac{4}{3}} \right)=\frac{3}{4} \\ \end{align}

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