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

Revision as of 19:15, 25 August 2022

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

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

@@ Line 1: / Line 1: @@
 <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}dx</math>
-<math>1*\sqrt{x^2+4}
+<math>\sqrt{x^2+4}