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

From Burton Tech. Points Wiki
Jump to navigation Jump to search
No edit summary
No edit summary
Line 2: Line 2:
\begin{align}
\begin{align}


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


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


&\sqrt{x^2+4}
&\sqrt{x^2+4}

Revision as of 19:19, 25 August 2022

<math> \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]

&\sqrt{x^2+4}

\end{align} <math>