Całka oznaczona
: 04 mar 2012, 16:54
\(\int_{0}^{2} \frac{dx}{ \sqrt[3]{(x-1)^2} }=\)
\(\int \frac{1}{(x-1)^{\frac{2}{3}}}dx=\begin{cases}k=x-1\therefore dk=dx\end{cases}=\int\frac{1}{k^{\frac{2}{3}}}dk=\int k^{-\frac{2}{3}}dk=3k^{\frac{1}{3}}+C=3\sqrt[3]{k}+C=3\sqrt[3]{x-1}+C\)
no i podstawic teraz granice
\(\[3\sqrt[3]{x-1}\]_{0}^{2}=3-3\sqrt[3]{i^{2}}\)