5.4 Indefinite Integrals and the Net Change Theorem/31: Difference between revisions

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\int_{0}^{1}x\left(\sqrt[3]{x}+\sqrt[4]{x}\right)dx &=\int_{0}^{1}x\left(x^{\frac{1}{3}}+x^{\frac{1}{4}}\right)dx=\int_{0}^{1}\left(x^{\frac{4}{3}}+x^{\frac{5}{4}}\right)dx \\[2ex]
\int_{0}^{1}x\left(\sqrt[3]{x}+\sqrt[4]{x}\right)dx &=\int_{0}^{1}x\left(x^{\frac{1}{3}}+x^{\frac{1}{4}}\right)dx=\int_{0}^{1}\left(x^{\frac{4}{3}}+x^{\frac{5}{4}}\right)dx \\[2ex]


&= \left(\frac{3x^{\frac{7}{3}}}{7}+\frac{4x^{\frac{9}{4}}}{9}\right)\bigg|_{0}^{1} \\[2ex]
&= \left(\frac{3x^{\frac{7}{3}}}{7}+\frac{4x^{\frac{9}{4}}}{9}\right)\Bigg|_{0}^{1} \\[2ex]


&= \frac{3}{7}+\frac{4}{9} = \frac{27+28}{7*9} = \frac{55}{63}
&= \frac{3}{7}+\frac{4}{9} = \frac{27+28}{7*9} = \frac{55}{63}

Revision as of 19:07, 22 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\left({\sqrt[{3}]{x}}+{\sqrt[{4}]{x}}\right)dx&=\int _{0}^{1}x\left(x^{\frac {1}{3}}+x^{\frac {1}{4}}\right)dx=\int _{0}^{1}\left(x^{\frac {4}{3}}+x^{\frac {5}{4}}\right)dx\\[2ex]&=\left({\frac {3x^{\frac {7}{3}}}{7}}+{\frac {4x^{\frac {9}{4}}}{9}}\right){\Bigg |}_{0}^{1}\\[2ex]&={\frac {3}{7}}+{\frac {4}{9}}={\frac {27+28}{7*9}}={\frac {55}{63}}\end{aligned}}}