Lösung 2.2:2d

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What makes the integral not entirely simple is the expression
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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>,
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<math>\text{1}-x</math>
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under the root sign, so we try the substitution
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<math>u=\text{1}-x</math>,
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{{Displayed math||<math>\int\limits_0^1 \sqrt[3]{1-x}\,dx = \left\{ \begin{align}
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u &= 1-x\\[5pt]
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du &= (1-x)'\,dx = -\,dx
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\end{align} \right\} = -\int\limits_1^0 \sqrt[3]{u}\,du\,\textrm{.}</math>}}
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<math>\int\limits_{0}^{1}{\sqrt[3]{\text{1}-x}}\,dx=\left\{ \begin{matrix}
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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.
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u=\text{1}-x \\
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du=\left( \text{1}-x \right)^{\prime }\,dx=-\,dx \\
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\end{matrix} \right\}=-\int\limits_{1}^{0}{\sqrt[3]{u}}\,du</math>
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{{Displayed math||<math>-\int\limits_1^0 \sqrt[3]{u}\,du = +\int\limits_0^1 \sqrt[3]{u}\,du\,\textrm{.}</math>}}
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Note how the new limits of integration go from
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All that is now left is routine calculations,
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<math>\text{1}</math>
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to
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<math>0</math>
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(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.
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{{Displayed math||<math>\begin{align}
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<math>-\int\limits_{1}^{0}{\sqrt[3]{u}}\,du=+\int\limits_{0}^{1}{\sqrt[3]{u}}\,du</math>
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\int\limits_0^1 \sqrt[3]{u}\,du
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&= \int\limits_0^1 u^{1/3}\,du
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= \biggl[\ \frac{u^{1/3+1}}{\tfrac{1}{3}+1}\ \biggr]_0^1\\[5pt]
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All that is now left is routine calculations:
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&= \frac{3}{4}\Bigl[\ u^{4/3}\ \Bigr]_0^1
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= \frac{3}{4}\bigl( 1^{4/3}-0^{4/3} \bigr) = \frac{3}{4}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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& \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} \\
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& =\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} \\
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\end{align}</math>
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Version vom 13:56, 28. Okt. 2008

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“