5.3 The Fundamental Theorem of Calculus/10: Difference between revisions
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&\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] | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 19:24, 25 August 2022
Failed to parse (unknown function "\g"): {\displaystyle \begin{align} &\g(r)=\int_{0}^{r}\sqrt{x^2+4}dx \\[2ex] &\frac{d}{dx}\int_{0}^{r}\sqrt{x^2+4}dx \\[2ex] \end{align} }