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]


\frac{d}{dx}\int_{b(x)}^{a(x)}F(t)dt=\frac{d}{dx}[b(x)]\cdotF(b(x))\\[2ex]
\frac{d}{dx}\int_{b(x)}^{a(x)}F(t)dt=\frac{d}{dx}[b(x)] \cdot F(b(x))\\[2ex]


1\cdot\sqrt{r^2+4} - 0\cdot\sqrt{0^2+4} \\[2ex]
1\cdot\sqrt{r^2+4} - 0\cdot\sqrt{0^2+4} \\[2ex]

Revision as of 20:03, 25 August 2022

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}g(r)=\int _{0}^{r}{\sqrt {x^{2}+4}}dx\\[2ex]{\frac {d}{dx}}\int _{0}^{r}{\sqrt {x^{2}+4}}dx\\[2ex]{\frac {d}{dx}}\int _{b(x)}^{a(x)}F(t)dt={\frac {d}{dx}}[b(x)]\cdot F(b(x))\\[2ex]1\cdot {\sqrt {r^{2}+4}}-0\cdot {\sqrt {0^{2}+4}}\\[2ex]={\sqrt {r^{2}+4}}\end{aligned}}}