Lösung 2.2:2d

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What makes the integral not entirely simple is the expression <math>1-x</math> under the root sign, so we try the substitution <math>u=1-x</math>,
What makes the integral not entirely simple is the expression <math>1-x</math> under the root sign, so we try the substitution <math>u=1-x</math>,
-
{{Displayed math||<math>\int\limits_0^1 \sqrt[3]{1-x}\,dx = \left\{ \begin{align}
+
{{Abgesetzte Formel||<math>\int\limits_0^1 \sqrt[3]{1-x}\,dx = \left\{ \begin{align}
u &= 1-x\\[5pt]
u &= 1-x\\[5pt]
du &= (1-x)'\,dx = -\,dx
du &= (1-x)'\,dx = -\,dx
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Note how the new limits of integration go from 1 to 0 (and not the other way around!). It is possible to change the order of the limits if we change sign at the same time, i.e.
Note how the new limits of integration go from 1 to 0 (and not the other way around!). It is possible to change the order of the limits if we change sign at the same time, i.e.
-
{{Displayed math||<math>-\int\limits_1^0 \sqrt[3]{u}\,du = +\int\limits_0^1 \sqrt[3]{u}\,du\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>-\int\limits_1^0 \sqrt[3]{u}\,du = +\int\limits_0^1 \sqrt[3]{u}\,du\,\textrm{.}</math>}}
All that is now left is routine calculations,
All that is now left is routine calculations,
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\int\limits_0^1 \sqrt[3]{u}\,du
\int\limits_0^1 \sqrt[3]{u}\,du
&= \int\limits_0^1 u^{1/3}\,du
&= \int\limits_0^1 u^{1/3}\,du

Version vom 13:01, 10. Mär. 2009

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

\displaystyle \int\limits_0^1 \sqrt[3]{1-x}\,dx = \left\{ \begin{align}

u &= 1-x\\[5pt] du &= (1-x)'\,dx = -\,dx \end{align} \right\} = -\int\limits_1^0 \sqrt[3]{u}\,du\,\textrm{.}

Note how the new limits of integration go from 1 to 0 (and not the other way around!). It is possible to change the order of the limits if 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\,\textrm{.}

All that is now left is routine calculations,

\displaystyle \begin{align}

\int\limits_0^1 \sqrt[3]{u}\,du &= \int\limits_0^1 u^{1/3}\,du = \biggl[\ \frac{u^{1/3+1}}{\tfrac{1}{3}+1}\ \biggr]_0^1\\[5pt] &= \frac{3}{4}\Bigl[\ u^{4/3}\ \Bigr]_0^1 = \frac{3}{4}\bigl( 1^{4/3}-0^{4/3} \bigr) = \frac{3}{4}\,\textrm{.} \end{align}

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