5.3 The Fundamental Theorem of Calculus/23: Difference between revisions
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\begin{align} | \begin{align} | ||
\int_{0}^{1}x^{\frac{4}{5}}dx &=\frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} \ | \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\sqrt[5]{(1)^9}}{9}-\frac{5 \sqrt[5]{(0)^9}}{9} \\[2ex] | ||
Revision as of 22:00, 7 September 2022
![{\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]&={\cfrac {5}{9}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/643ce20ae70e145b2cfe1b61654b6b0965481626)