5.4 Indefinite Integrals and the Net Change Theorem/33 - Revision history

Dvaezazizi@laalliance.org: Protected "5.4 Indefinite Integrals and the Net Change Theorem/33" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-21T19:41:09Z

Protected "5.4 Indefinite Integrals and the Net Change Theorem/33" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision

Revision as of 19:41, 21 September 2022

2022-09-21T15:30:41Z

← Older revision		Revision as of 15:30, 21 September 2022
Line 3:		Line 3:


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

	&= 2\sqrt{5}x^{\frac{1}{2}}\bigg\|_{1}^{4} \\[2ex]		&= 2\sqrt{5}x^{\frac{1}{2}}\bigg\|_{1}^{4} \\[2ex]

2022-09-21T15:30:14Z

← Older revision		Revision as of 15:30, 21 September 2022
Line 7:		Line 7:
	&= 2\sqrt{5}x^{\frac{1}{2}}\bigg\|_{1}^{4} \\[2ex]		&= 2\sqrt{5}x^{\frac{1}{2}}\bigg\|_{1}^{4} \\[2ex]

	&= 2\sqrt{5}\sqrt{4}-2\sqrt{5}{\sqrt{1}} = 4\sqrt{5}-2\sqrt{5} \\[2ex]		&= [2\sqrt{5}\sqrt{4}]-[2\sqrt{5}{\sqrt{1}}] = 4\sqrt{5}-2\sqrt{5} \\[2ex]

	&= 2\sqrt{5}		&= 2\sqrt{5}
	\end{align}		\end{align}
	</math>		</math>

2022-09-21T15:29:58Z

← Older revision		Revision as of 15:29, 21 September 2022
Line 5:		Line 5:
	\int_{1}^{4}\sqrt{\frac{5}{x}}dy &= \int_{1}^{4}\frac{\sqrt{5}}{\sqrt{x}}dx = 5^\frac{1}{2}\int_{1}^{4}x^{-\frac{1}{2}}dx\\[2ex]		\int_{1}^{4}\sqrt{\frac{5}{x}}dy &= \int_{1}^{4}\frac{\sqrt{5}}{\sqrt{x}}dx = 5^\frac{1}{2}\int_{1}^{4}x^{-\frac{1}{2}}dx\\[2ex]

	&= 2\sqrt{5}x^{\frac{1}{2}}\bigg\|_{1}^{4} = 2\sqrt{5}\sqrt{4}-2\sqrt{5}{\sqrt{1}} ~~\\[2ex]~~		&= 2\sqrt{5}x^{\frac{1}{2}}\bigg\|_{1}^{4} \\[2ex]

	&= 4\sqrt{5}-2\sqrt{5} \\[2ex]		&= 2\sqrt{5}\sqrt{4}-2\sqrt{5}{\sqrt{1}} = 4\sqrt{5}-2\sqrt{5} \\[2ex]

	&= 2\sqrt{5}		&= 2\sqrt{5}
	\end{align}		\end{align}
	</math>		</math>

2022-09-21T15:29:29Z

← Older revision		Revision as of 15:29, 21 September 2022
Line 5:		Line 5:
	\int_{1}^{4}\sqrt{\frac{5}{x}}dy &= \int_{1}^{4}\frac{\sqrt{5}}{\sqrt{x}}dx = 5^\frac{1}{2}\int_{1}^{4}x^{-\frac{1}{2}}dx\\[2ex]		\int_{1}^{4}\sqrt{\frac{5}{x}}dy &= \int_{1}^{4}\frac{\sqrt{5}}{\sqrt{x}}dx = 5^\frac{1}{2}\int_{1}^{4}x^{-\frac{1}{2}}dx\\[2ex]

	&= 2\sqrt{5}x^{\frac{1}{2}}\bigg\|_{1}^{4} ~~\\[2ex]~~		&= 2\sqrt{5}x^{\frac{1}{2}}\bigg\|_{1}^{4} = 2\sqrt{5}\sqrt{4}-2\sqrt{5}{\sqrt{1}} \\[2ex]

	&= 2\sqrt{5}\sqrt{4}-2\sqrt{5}{\sqrt{1}} \\[2ex]

	&= 4\sqrt{5}-2\sqrt{5} \\[2ex]		&= 4\sqrt{5}-2\sqrt{5} \\[2ex]

2022-09-21T15:29:16Z

2022-09-21T15:28:33Z

← Older revision		Revision as of 15:28, 21 September 2022
Line 5:		Line 5:
	\int_{1}^{4}\sqrt{\frac{5}{x}}dy &= \int_{1}^{4}\frac{\sqrt{5}}{\sqrt{x}}dx = 5^\frac{1}{2}\int_{1}^{4}x^{-\frac{1}{2}}dx\\[2ex]		\int_{1}^{4}\sqrt{\frac{5}{x}}dy &= \int_{1}^{4}\frac{\sqrt{5}}{\sqrt{x}}dx = 5^\frac{1}{2}\int_{1}^{4}x^{-\frac{1}{2}}dx\\[2ex]

	&= 2\sqrt{5}x^{\frac{1}{2}~~}\bigg\|_{1}^{4} = 2\sqrt{5}\sqrt{x}\bigg\|_{1}^{4} = 2\sqrt{5x~~}\bigg\|_{1}^{4} \\[2ex]		&= 2\sqrt{5}x^{\frac{1}{2}}\bigg\|_{1}^{4} \\[2ex]

	&= 2\sqrt{5\~~times4~~}-2\sqrt{5\~~times1~~} \\[2ex]		&= 2\sqrt{5}\sqrt{4}-2\sqrt{5}{\sqrt{1}} \\[2ex]

	&= 2\sqrt{20}-2\sqrt{5} = 4\sqrt{5}-2\sqrt{5} = 2\sqrt{5}		&= 2\sqrt{20}-2\sqrt{5} = 4\sqrt{5}-2\sqrt{5} = 2\sqrt{5}
	\end{align}		\end{align}
	</math>		</math>

2022-09-21T15:27:24Z

← Older revision		Revision as of 15:27, 21 September 2022
Line 5:		Line 5:
	\int_{1}^{4}\sqrt{\frac{5}{x}}dy &= \int_{1}^{4}\frac{\sqrt{5}}{\sqrt{x}}dx = 5^\frac{1}{2}\int_{1}^{4}x^{-\frac{1}{2}}dx\\[2ex]		\int_{1}^{4}\sqrt{\frac{5}{x}}dy &= \int_{1}^{4}\frac{\sqrt{5}}{\sqrt{x}}dx = 5^\frac{1}{2}\int_{1}^{4}x^{-\frac{1}{2}}dx\\[2ex]

	&= \sqrt{5}2x^{\frac{1}{2}}\bigg\|_{1}^{4} = \sqrt{5}~~\times2~~\sqrt{x}\bigg\|_{1}^{4} = 2\sqrt{5x}\bigg\|_{1}^{4} \\[2ex]		&= 2\sqrt{5}x^{\frac{1}{2}}\bigg\|_{1}^{4} = 2\sqrt{5}\sqrt{x}\bigg\|_{1}^{4} = 2\sqrt{5x}\bigg\|_{1}^{4} \\[2ex]

	&= 2\sqrt{5\times4}-2\sqrt{5\times1} \\[2ex]		&= 2\sqrt{5\times4}-2\sqrt{5\times1} \\[2ex]

2022-09-21T15:27:01Z

← Older revision		Revision as of 15:27, 21 September 2022
Line 5:		Line 5:
	\int_{1}^{4}\sqrt{\frac{5}{x}}dy &= \int_{1}^{4}\frac{\sqrt{5}}{\sqrt{x}}dx = 5^\frac{1}{2}\int_{1}^{4}x^{-\frac{1}{2}}dx\\[2ex]		\int_{1}^{4}\sqrt{\frac{5}{x}}dy &= \int_{1}^{4}\frac{\sqrt{5}}{\sqrt{x}}dx = 5^\frac{1}{2}\int_{1}^{4}x^{-\frac{1}{2}}dx\\[2ex]

	&= \sqrt{5}~~\times2x~~^\frac{1}{2}\bigg\|_{1}^{4} = \sqrt{5}\times2\sqrt{x}\bigg\|_{1}^{4} = 2\sqrt{5x}\bigg\|_{1}^{4} \\[2ex]		&= \sqrt{5}2x^{\frac{1}{2}}\bigg\|_{1}^{4} = \sqrt{5}\times2\sqrt{x}\bigg\|_{1}^{4} = 2\sqrt{5x}\bigg\|_{1}^{4} \\[2ex]

	&= 2\sqrt{5\times4}-2\sqrt{5\times1} \\[2ex]		&= 2\sqrt{5\times4}-2\sqrt{5\times1} \\[2ex]

2022-09-21T15:26:28Z

← Older revision		Revision as of 15:26, 21 September 2022
Line 3:		Line 3:


	\int_{1}^{4}\sqrt{\frac{5}{x}}dy &= \int_{1}^{4}\frac{\sqrt{5}}{\sqrt{x}}dx = 5^\frac{1}{2}\int_{1}^{4}x^\frac{-1}{2}dx\\[2ex]		\int_{1}^{4}\sqrt{\frac{5}{x}}dy &= \int_{1}^{4}\frac{\sqrt{5}}{\sqrt{x}}dx = 5^\frac{1}{2}\int_{1}^{4}x^{-\frac{1}{2}}dx\\[2ex]

	&= \sqrt{5}\times2x^\frac{1}{2}\bigg\|_{1}^{4} = \sqrt{5}\times2\sqrt{x}\bigg\|_{1}^{4} = 2\sqrt{5x}\bigg\|_{1}^{4} \\[2ex]		&= \sqrt{5}\times2x^\frac{1}{2}\bigg\|_{1}^{4} = \sqrt{5}\times2\sqrt{x}\bigg\|_{1}^{4} = 2\sqrt{5x}\bigg\|_{1}^{4} \\[2ex]