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