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] | &=\left[\frac{5\sqrt[5]{(1)^9}}{9}\right]-\left[\frac{5 \sqrt[5]{(0)^9}}{9}\right] \\[2ex] | ||
&=\ | &=\frac{5}{9} | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Latest revision as of 22:01, 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]&=\left[{\frac {5{\sqrt[{5}]{(1)^{9}}}}{9}}\right]-\left[{\frac {5{\sqrt[{5}]{(0)^{9}}}}{9}}\right]\\[2ex]&={\frac {5}{9}}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a8fe31528d21b3d8d15367237fdbee8bc9a97541)