Lösung 2.2:2d
Aus Online Mathematik Brückenkurs 2
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Version vom 10:22, 11. 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} |
