5.3 The Fundamental Theorem of Calculus/10: Difference between revisions

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1\cdot\sqrt{r^2+4} - 0\cdot\sqrt{0^2+4} \\[2ex]
1\cdot\sqrt{r^2+4} - 0\cdot\sqrt{0^2+4} \\[2ex]


&= \sqrt{r^2} + {4}
=\sqrt{r^2} + {4}




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

Revision as of 19:38, 25 August 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}g(r)=\int _{0}^{r}{\sqrt {x^{2}+4}}dx\\[2ex]{\frac {d}{dx}}\int _{0}^{r}{\sqrt {x^{2}+4}}dx\\[2ex]1\cdot {\sqrt {r^{2}+4}}-0\cdot {\sqrt {0^{2}+4}}\\[2ex]={\sqrt {r^{2}}}+{4}\end{aligned}}}