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

Revision as of 19:13, 25 August 2022

$g(r)=\int _{0}^{r}{\sqrt {x^{2}+4}}dx$

${\frac {d}{dx}}\int _{0}^{r}{\sqrt {x^{2}+4}}dx$

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	<math> g(r)=\int_{0}^{r}\sqrt{x^2+4} dx </math> <br><br>		<math> g(r)=\int_{0}^{r}\sqrt{x^2+4} dx </math> <br><br>
	<math>\frac{d}{dx}\int_{0}^{r}\sqrt{x^2+4}</math>		<math>\frac{d}{dx}\int_{0}^{r}\sqrt{x^2+4}dx</math>