5.3 The Fundamental Theorem of Calculus/23: Difference between revisions
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&\int_{0}^{1}x^\cfrac{4}{5}dx \\[2ex] | &\int_{0}^{1}x^\cfrac{4}{5}dx \\[2ex] | ||
&= \cfrac{5\sqrt[5]{x^4}}{9} | &= \cfrac{5\sqrt[5]{x^4}}{9} \\[2ex] | ||
&= \cfrac{5\sqrt[5]{x^4}}{9}\bigg|_{0}^{1} | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 19:04, 25 August 2022
![{\displaystyle {\begin{aligned}&\int _{0}^{1}x^{\cfrac {4}{5}}dx\\[2ex]&={\cfrac {5{\sqrt[{5}]{x^{4}}}}{9}}\\[2ex]&={\cfrac {5{\sqrt[{5}]{x^{4}}}}{9}}{\bigg |}_{0}^{1}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/7c075deeb762fa0d99a584aad7d928000e7e2c09)