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

Latest revision as of 20:15, 6 September 2022

$g(r)=\int _{0}^{r}{\sqrt {x^{2}+4}}\,dx$

${\frac {d}{dr}}(g(r))={\frac {d}{dr}}\left[\int _{0}^{r}{\sqrt {x^{2}+4}}\,dx\right]=(1)\cdot {\sqrt {(r)^{2}+4}}-(0)\cdot {\sqrt {(0)^{2}+4}}={\sqrt {r^{2}+4}}$

${\text{Therefore, }}g'(r)={\sqrt {r^{2}+4}}$

@@ Line 1: / Line 1: @@
 <math>
-\begin{align}
+g(r)=\int_{0}^{r}\sqrt{x^2+4}\,dx
+</math>
-&\g(r)=\int_{0}^{r}\sqrt{x^2+4}dx \\[2ex]
-&\frac{d}{dx}\int_{0}^{r}\sqrt{x^2+4}dx \\[2ex]
+<math>
+\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}
+</math>
-\end{align}
+<math>
+\text{Therefore, } g'(r) =\sqrt{r^2 + 4}
 </math>

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

Latest revision as of 20:15, 6 September 2022

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