5.4 Indefinite Integrals and the Net Change Theorem/39: Difference between revisions

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&= 2(x)^\frac{1}{2} + \frac{6}{5}(x)^\frac{5}{6}\bigg|_{1}^{64} \\[2ex]
&= 2(x)^\frac{1}{2} + \frac{6}{5}(x)^\frac{5}{6}\bigg|_{1}^{64} \\[2ex]


&= \left[2(64)^\frac{1}{2} + \frac{6}{5}(64)^\frac{5}{6}\right] - \left[(2(1)^\frac{1}{2} + \frac{6}{5}(1)^\frac{5}{6})\right]
&= \left[2(64)^\frac{1}{2} + \frac{6}{5}(64)^\frac{5}{6}\right] - \left[(2(1)^\frac{1}{2} + \frac{6}{5}(1)^\frac{5}{6})\right] \\[2ex]
 
&= \frac{256}{5}


\end{align}
\end{align}
</math>
</math>
= <math>2(64)^\frac{1}{2} + \frac{6}{5}(64)^\frac{5}{6} - (2(1)^\frac{1}{2} + \frac{6}{5}(1)^\frac{5}{6})</math>
= <math>16+38.4 - (2+1.2)</math>
= <math>54.4 - 3.2</math>
= <math>51.2</math>
= <math>\frac{256}{5}</math>

Revision as of 16:24, 21 September 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{1}^{64}{\frac {1+{\sqrt[{3}]{x}}}{\sqrt {x}}}dx&=\int _{1}^{64}\left({\frac {1}{x^{1/2}}}+{\frac {x^{1/3}}{x^{1/2}}}\right)dx=\int _{1}^{64}\left(x^{-1/2}+x^{{\frac {1}{3}}-{\frac {1}{2}}}\right)dx=\int _{1}^{64}\left(x^{-{\frac {1}{2}}}+x^{-{\frac {1}{6}}}\right)dx\\[2ex]&=\left[{\frac {x^{\frac {1}{2}}}{\frac {1}{2}}}+{\frac {x^{\frac {5}{6}}}{\frac {5}{6}}}\right]_{1}^{64}=\left[2x^{\frac {1}{2}}+{\frac {6}{5}}x^{\frac {5}{6}}\right]_{1}^{64}\\[2ex]&=2(x)^{\frac {1}{2}}+{\frac {6}{5}}(x)^{\frac {5}{6}}{\bigg |}_{1}^{64}\\[2ex]&=\left[2(64)^{\frac {1}{2}}+{\frac {6}{5}}(64)^{\frac {5}{6}}\right]-\left[(2(1)^{\frac {1}{2}}+{\frac {6}{5}}(1)^{\frac {5}{6}})\right]\\[2ex]&={\frac {256}{5}}\end{aligned}}}