5.3 The Fundamental Theorem of Calculus/10 - Revision history

Dvaezazizi@laalliance.org: Protected "5.3 The Fundamental Theorem of Calculus/10" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-06T20:15:06Z

Protected "5.3 The Fundamental Theorem of Calculus/10" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision

Revision as of 20:15, 6 September 2022

2022-09-06T20:04:21Z

2022-09-06T20:04:12Z

2022-09-06T20:04:07Z

← Older revision		Revision as of 20:04, 6 September 2022
Line 7:		Line 7:
	\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} =\sqrt{r^2 + 4}
			</math>

			<math>
			\text{Therefore, } g'(r) = =\sqrt{r^2 + 4}
	</math>		</math>

2022-09-06T20:03:29Z

← Older revision		Revision as of 20:03, 6 September 2022
Line 6:		Line 6:
	<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} =\sqrt{r^2 + 4}
	</math>		</math>

2022-09-06T20:03:09Z

2022-09-06T20:03:04Z

← Older revision		Revision as of 20:03, 6 September 2022
Line 1:		Line 1:
	<math>		<math>
	~~\begin{align}~~

	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}

	=\sqrt{r^2 + 4}


	~~\end{align~~}
	</math>		</math>

2022-09-06T20:02:33Z

2022-09-06T20:02:25Z

← Older revision		Revision as of 20:02, 6 September 2022
Line 2:		Line 2:
	\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}

2022-08-25T20:07:43Z

← Older revision		Revision as of 20:07, 25 August 2022
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)] \cdot F(b(x))-\frac{d}{dx}[a(x)] \cdot F(a(x))\\[2ex]		\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} \\[2ex]		1\cdot\sqrt{r^2+4} - 0\cdot\sqrt{0^2+4} \\[2ex]