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

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\begin{align}
\begin{align}


&\g(r)=\int_{0}^{r}\sqrt{x^2+4}dx \\[2ex]
g(r)=\int_{0}^{r}\sqrt{x^2+4}dx \\[2ex]




&\frac{d}{dx}\int_{0}^{r}\sqrt{x^2+4}dx \\[2ex]
\frac{d}{dx}\int_{0}^{r}\sqrt{x^2+4}dx \\[2ex]




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

Revision as of 19:30, 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]\end{aligned}}}

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