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

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\begin{align}
\begin{align}


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




\frac{d}{dx}\int_{0}^{r}\sqrt{x^2+4}dx \\[2ex]
\frac{d}{dr}(g(r)) = \frac{d}{dr}\left[\int_{0}^{r}\sqrt{x^2+4}\,dx\right] = 


\frac{d}{dx}\int_{a(x)}^{b(x)}F(t)dt=\frac{d}{dx}[b(x)] \cdot F(b(x))-\frac{d}{dx}[a(x)] \cdot F(a(x))\\[2ex]
1\cdot\sqrt{(r)^2+4} - 0\cdot\sqrt{(0)^2+4}
 
1\cdot\sqrt{r^2+4} - 0\cdot\sqrt{0^2+4} \\[2ex]


=\sqrt{r^2 + 4}
=\sqrt{r^2 + 4}

Revision as of 20:02, 6 September 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\\{\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}}\end{aligned}}}

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