5.3 The Fundamental Theorem of Calculus/23 - Revision history

Dvaezazizi@laalliance.org at 22:01, 7 September 2022

2022-09-07T22:01:23Z

← Older revision		Revision as of 22:01, 7 September 2022
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	\int_{0}^{1}x^{\frac{4}{5}}dx &=\frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} \Bigg\|_{0}^{1} =\frac{x^{\frac{9}{5}}}{\frac{9}{5}} \Bigg\|_{0}^{1} \\[2ex]		\int_{0}^{1}x^{\frac{4}{5}}dx &=\frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} \Bigg\|_{0}^{1} =\frac{x^{\frac{9}{5}}}{\frac{9}{5}} \Bigg\|_{0}^{1} \\[2ex]

	&=\frac{5\sqrt[5]{(1)^9}}{9}-\frac{5 \sqrt[5]{(0)^9}}{9} \\[2ex]		&=\left[\frac{5\sqrt[5]{(1)^9}}{9}\right]-\left[\frac{5 \sqrt[5]{(0)^9}}{9}\right] \\[2ex]

	&=\frac{5}{9}		&=\frac{5}{9}

2022-09-07T22:00:54Z

2022-09-07T22:00:47Z

← Older revision		Revision as of 22:00, 7 September 2022
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	\begin{align}		\begin{align}

	\int_{0}^{1}x^{\frac{4}{5}}dx &=\frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} \Bigg\|_{0}^{1} =\frac{x^{\frac{9}{5}}}{\frac{9}{5}} \~~bigg~~\|_{0}^{1} \\[2ex]		\int_{0}^{1}x^{\frac{4}{5}}dx &=\frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} \Bigg\|_{0}^{1} =\frac{x^{\frac{9}{5}}}{\frac{9}{5}} \Bigg\|_{0}^{1} \\[2ex]

	&=\frac{5\sqrt[5]{(1)^9}}{9}-\frac{5 \sqrt[5]{(0)^9}}{9} \\[2ex]		&=\frac{5\sqrt[5]{(1)^9}}{9}-\frac{5 \sqrt[5]{(0)^9}}{9} \\[2ex]

2022-09-07T22:00:37Z

← Older revision		Revision as of 22:00, 7 September 2022
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	\begin{align}		\begin{align}

	\int_{0}^{1}x^{\frac{4}{5}}dx &=\frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} \~~bigg~~\|_{0}^{1} =\frac{x^{\frac{9}{5}}}{\frac{9}{5}} \bigg\|_{0}^{1} \\[2ex]		\int_{0}^{1}x^{\frac{4}{5}}dx &=\frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} \Bigg\|_{0}^{1} =\frac{x^{\frac{9}{5}}}{\frac{9}{5}} \bigg\|_{0}^{1} \\[2ex]

	&=\frac{5\sqrt[5]{(1)^9}}{9}-\frac{5 \sqrt[5]{(0)^9}}{9} \\[2ex]		&=\frac{5\sqrt[5]{(1)^9}}{9}-\frac{5 \sqrt[5]{(0)^9}}{9} \\[2ex]

2022-09-06T20:51:56Z

Protected "5.3 The Fundamental Theorem of Calculus/23" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 20:51, 6 September 2022
(No difference)

2022-09-06T20:50:31Z

← Older revision		Revision as of 20:50, 6 September 2022
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	\int_{0}^{1}x^{\frac{4}{5}}dx &=\frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} \bigg\|_{0}^{1} =\frac{x^{\frac{9}{5}}}{\frac{9}{5}} \bigg\|_{0}^{1} \\[2ex]		\int_{0}^{1}x^{\frac{4}{5}}dx &=\frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} \bigg\|_{0}^{1} =\frac{x^{\frac{9}{5}}}{\frac{9}{5}} \bigg\|_{0}^{1} \\[2ex]

	&=\frac{5\sqrt[5]{1^9}}{9}-\frac{5 \sqrt[5]{0^9}}{9} \\[2ex]		&=\frac{5\sqrt[5]{(1)^9}}{9}-\frac{5 \sqrt[5]{(0)^9}}{9} \\[2ex]

	&=\cfrac{5}{9}		&=\cfrac{5}{9}

2022-09-06T20:50:15Z

← Older revision		Revision as of 20:50, 6 September 2022
Line 4:		Line 4:
	\int_{0}^{1}x^{\frac{4}{5}}dx &=\frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} \bigg\|_{0}^{1} =\frac{x^{\frac{9}{5}}}{\frac{9}{5}} \bigg\|_{0}^{1} \\[2ex]		\int_{0}^{1}x^{\frac{4}{5}}dx &=\frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} \bigg\|_{0}^{1} =\frac{x^{\frac{9}{5}}}{\frac{9}{5}} \bigg\|_{0}^{1} \\[2ex]

	&=\frac{5~~\cdot~~ \sqrt[5]{1^9}}{9}-\frac{5 \sqrt[5]{0^9}}{9} \\[2ex]		&=\frac{5\sqrt[5]{1^9}}{9}-\frac{5 \sqrt[5]{0^9}}{9} \\[2ex]

	&=\cfrac{5}{9}		&=\cfrac{5}{9}

2022-09-06T20:49:36Z

← Older revision		Revision as of 20:49, 6 September 2022
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	\begin{align}		\begin{align}

			\int_{0}^{1}x^{\frac{4}{5}}dx &=\frac{x^{\frac{4}{5}+1}}{\frac{4}{5}+1} \bigg\|_{0}^{1} =\frac{x^{\frac{9}{5}}}{\frac{9}{5}} \bigg\|_{0}^{1} \\[2ex]

	&~~\int_{0}^{1}x^{~~\frac~~{4}{5}}dx \\[2ex]~~		&=\frac{5\cdot \sqrt[5]{1^9}}{9}-\frac{5 \sqrt[5]{0^9}}{9} \\[2ex]


	~~&\cfrac~~{5\cdot \sqrt[5]{x^~~4}}{~~9~~}\bigg\|_{0}^{1} \\[2ex]~~

	~~&\cfrac{5\cdot \sqrt{1^4~~}}{9}-\~~cfrac~~{5 \sqrt[5]{0^4}}{9} \\[2ex]

	&=\cfrac{5}{9}		&=\cfrac{5}{9}


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

2022-09-06T20:41:13Z

2022-08-25T20:03:44Z