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

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

Revision as of 19:18, 25 August 2022

<math> \begin{align}

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

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

&\sqrt{x^2+4}

\end{align} <math>