5.3 The Fundamental Theorem of Calculus/23: Difference between revisions
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&=\frac{5\sqrt[5]{(1)^9}}{9}-\frac{5 \sqrt[5]{(0)^9}}{9} \\[2ex] | &=\frac{5\sqrt[5]{(1)^9}}{9}-\frac{5 \sqrt[5]{(0)^9}}{9} \\[2ex] | ||
&=\ | &=\frac{5}{9} | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 22:00, 7 September 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{0}^{1}x^{\frac {4}{5}}dx&={\frac {x^{{\frac {4}{5}}+1}}{{\frac {4}{5}}+1}}{\Bigg |}_{0}^{1}={\frac {x^{\frac {9}{5}}}{\frac {9}{5}}}{\Bigg |}_{0}^{1}\\[2ex]&={\frac {5{\sqrt[{5}]{(1)^{9}}}}{9}}-{\frac {5{\sqrt[{5}]{(0)^{9}}}}{9}}\\[2ex]&={\frac {5}{9}}\end{aligned}}}