5.3 The Fundamental Theorem of Calculus/28: Difference between revisions

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&= \left[3(1)+\frac{2(1)^{5/2}}{5}\right]-\left[3(0)+\frac{2(0)^{5/2}}{5}\right] \\[2ex]
&= \left[3(1)+\frac{2(1)^{5/2}}{5}\right]-\left[3(0)+\frac{2(0)^{5/2}}{5}\right] \\[2ex]


&= 3+\frac{2}{5} = \frac{15}{5}+\frac{2}{5} = \frac{17}{5}   
&= 3+\frac{2}{5} = \frac{15}{5}+\frac{2}{5} = \frac{17}{6}   


\end{align}
\end{align}
</math>
</math>

Revision as of 04:20, 26 August 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}\left(3+x{\sqrt {x}}\right)dx&=\int _{0}^{1}\left(3+x^{1}{x}^{\frac {1}{2}}\right)dx=\int _{0}^{1}\left(3+x^{1+{\frac {1}{2}}}\right)dx=\int _{0}^{1}\left(3+x^{\frac {3}{2}}\right)dx\\[2ex]&=3x+{\frac {x^{{\frac {3}{2}}+1}}{{\frac {3}{2}}+1}}{\bigg |}_{0}^{1}=3x+{\frac {x^{\tfrac {5}{2}}}{\frac {5}{2}}}{\bigg |}_{0}^{1}=3x+{\frac {2x^{\frac {5}{2}}}{5}}{\bigg |}_{0}^{1}\\[2ex]&=\left[3(1)+{\frac {2(1)^{5/2}}{5}}\right]-\left[3(0)+{\frac {2(0)^{5/2}}{5}}\right]\\[2ex]&=3+{\frac {2}{5}}={\frac {15}{5}}+{\frac {2}{5}}={\frac {17}{6}}\end{aligned}}}