Aus Online Mathematik Brückenkurs 2

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}