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

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


1\cdot\sqrt{r^2+4}dx-\sqrt{0^2+4} - 0\cdot\sqrt{0^2+4}
1\cdot\sqrt{r^2+4} - 0\cdot\sqrt{0^2+4}




\end{align}
\end{align}
</math>
</math>

Revision as of 19:37, 25 August 2022

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \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] 1\cdot\sqrt{r^2+4} - 0\cdot\sqrt{0^2+4} \end{align} }