5.3 The Fundamental Theorem of Calculus/10: Difference between revisions
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<math> | <math> | ||
g(r)=\int_{0}^{r}\sqrt{x^2+4}\,dx | g(r)=\int_{0}^{r}\sqrt{x^2+4}\,dx | ||
</math> | |||
<math> | |||
\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} | </math> | ||
=\sqrt{r^2 + 4} | |||
\ | <math> | ||
\text{Therefore, } g'(r) =\sqrt{r^2 + 4} | |||
</math> | </math> | ||
Latest revision as of 20:15, 6 September 2022
![{\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/19a27b2aeb03a9b4e43b9eb01e532c47576d3cf5)
