5.3 The Fundamental Theorem of Calculus/35 - Revision history

Dvaezazizi@laalliance.org: Protected "5.3 The Fundamental Theorem of Calculus/35" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-06T21:34:31Z

Protected "5.3 The Fundamental Theorem of Calculus/35" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 21:34, 6 September 2022
(No difference)

2022-08-25T19:44:19Z

2022-08-25T19:43:45Z

← Older revision		Revision as of 19:43, 25 August 2022
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	&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9} = \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|} \\[2ex]		&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9} \\[2ex]
			&= \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|} \\[2ex]

	&= \ln{\|9^{\frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3} \\[2ex]		&= \ln{\|9^{\frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3} \\[2ex]

2022-08-25T19:41:16Z

← Older revision		Revision as of 19:41, 25 August 2022
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	\begin{align}		\begin{align}

	\int_{1}^{9}\frac{1}{2x}dx &= \frac{1}{2}\int_{1}^{9}\frac{1}{x}dx		\int_{1}^{9}\frac{1}{2x}dx &= \frac{1}{2}\int_{1}^{9}\frac{1}{x}dx \\[2ex]

2022-08-25T19:40:18Z

← Older revision		Revision as of 19:40, 25 August 2022
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	&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9}		&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9} = \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|} \\[2ex]

	&= \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|} \\[2ex]

	&= \ln{\|9^{\frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3} \\[2ex]		&= \ln{\|9^{\frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3} \\[2ex]

2022-08-25T19:40:01Z

2022-08-25T19:39:51Z

← Older revision		Revision as of 19:39, 25 August 2022
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	\int_{1}^{9}\frac{1}{2x}dx &= \frac{1}{2}\int_{1}^{9}\frac{1}{x}dx		\int_{1}^{9}\frac{1}{2x}dx &= \frac{1}{2}\int_{1}^{9}\frac{1}{x}dx
			\

	&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9} = \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|} \\[2ex]		&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9}

			&= \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|} \\[2ex]

	&= \ln{\|9^{\frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3} \\[2ex]		&= \ln{\|9^{\frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3} \\[2ex]

2022-08-25T19:37:30Z

← Older revision		Revision as of 19:37, 25 August 2022
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	\int_{1}^{9}\frac{1}{2x}dx &= \frac{1}{2}\int_{1}^{9}\frac{1}{x}dx		\int_{1}^{9}\frac{1}{2x}dx &= \frac{1}{2}\int_{1}^{9}\frac{1}{x}dx

	&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9} = \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|}		&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9} = \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|} \\[2ex]

	&= \~~ln{\|9^{~~\~~frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3}~~

			&= \ln{\|9^{\frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3} \\[2ex]

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

2022-08-25T19:36:04Z

← Older revision		Revision as of 19:36, 25 August 2022
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	\begin{align}		\begin{align}

	\int_{1}^{9}\frac{1}{2x}dx = \frac{1}{2}\int_{1}^{9}\frac{1}{x}dx		\int_{1}^{9}\frac{1}{2x}dx &= \frac{1}{2}\int_{1}^{9}\frac{1}{x}dx

	&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9} = \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|} = \ln{\|9^{\frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3}		&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9} = \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|}

			&= \ln{\|9^{\frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3}


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

2022-08-25T19:32:56Z

← Older revision		Revision as of 19:32, 25 August 2022
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	\int_{1}^{9}\frac{1}{2x}dx = \frac{1}{2}\int_{1}^{9}\frac{1}{x}dx		\int_{1}^{9}\frac{1}{2x}dx = \frac{1}{2}\int_{1}^{9}\frac{1}{x}dx

	&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9} = \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|} \ln{\|9^{\frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3}		&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9} = \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|} = \ln{\|9^{\frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3}


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

← Older revision		Revision as of 19:43, 25 August 2022
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	&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9} = \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|} \\[2ex]		&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9} \\[2ex]
			&= \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|} \\[2ex]

	&= \ln{\|9^{\frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3} \\[2ex]		&= \ln{\|9^{\frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3} \\[2ex]

← Older revision		Revision as of 19:40, 25 August 2022
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	&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9}		&= \frac{1}{2}\ln{\|x\|}\bigg\|_{1}^{9} = \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|} \\[2ex]

	&= \frac{1}{2}\ln{\|9\|}-\frac{1}{2}\ln{\|1\|} \\[2ex]

	&= \ln{\|9^{\frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3} \\[2ex]		&= \ln{\|9^{\frac{1}{2}}\|} - \ln{\|1^{\frac{1}{2}}\|} = \ln{3}-0 = \ln{3} \\[2ex]