5.3 The Fundamental Theorem of Calculus/10: Difference between revisions
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\frac{d}{dr}(g(r)) = \frac{d}{dr}\left[\int_{0}^{r}\sqrt{x^2+4}\,dx\right] = | \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} | (1)\cdot\sqrt{(r)^2+4} - (0)\cdot\sqrt{(0)^2+4} =\sqrt{r^2 + 4} | ||
</math> | |||
<math> | |||
\text{Therefore, } g'(r) = =\sqrt{r^2 + 4} | |||
</math> | </math> | ||
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 (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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} }