5.3 The Fundamental Theorem of Calculus/10: Difference between revisions
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\text{Therefore, } g'(r) | \text{Therefore, } g'(r) =\sqrt{r^2 + 4} | ||
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Revision as of 20:04, 6 September 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 g(r)=\int_{0}^{r}\sqrt{x^2+4}\,dx }
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 \frac{d}{dr}(g(r)) = \frac{d}{dr}\left[\int_{0}^{r}\sqrt{x^2+4}\,dx\right] = (1)\cdot\sqrt{(r)^2+4} - (0)\cdot\sqrt{(0)^2+4} =\sqrt{r^2 + 4} }
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 \text{Therefore, } g'(r) =\sqrt{r^2 + 4} }