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

Revision as of 19:30, 25 August 2022

${\begin{aligned}g(r)=\int _{0}^{r}{\sqrt {x^{2}+4}}dx\\[2ex]{\frac {d}{dx}}\int _{0}^{r}{\sqrt {x^{2}+4}}dx\\[2ex]\end{aligned}}$

@@ Line 2: / Line 2: @@
 \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}
 </math>