5.3 The Fundamental Theorem of Calculus/10: Difference between revisions
Jump to navigation
Jump to search
Andy Arevalo (talk | contribs) No edit summary |
Andy Arevalo (talk | contribs) No edit summary |
||
| Line 7: | Line 7: | ||
\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)]\cdotF(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 20:01, 25 August 2022
Failed to parse (unknown function "\cdotF"): {\displaystyle \begin{align} 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)]\cdotF(b(x))\\[2ex] 1\cdot\sqrt{r^2+4} - 0\cdot\sqrt{0^2+4} \\[2ex] =\sqrt{r^2 + 4} \end{align} }