5.4 Indefinite Integrals and the Net Change Theorem/29 - Revision history

Dvaezazizi@laalliance.org: Protected "5.4 Indefinite Integrals and the Net Change Theorem/29" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-21T19:40:53Z

Protected "5.4 Indefinite Integrals and the Net Change Theorem/29" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 19:40, 21 September 2022
(No difference)

2022-09-21T15:20:24Z

2022-09-21T15:19:25Z

← Older revision		Revision as of 15:19, 21 September 2022
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	\int_{-2}^{-1}\left(4y^3+\frac{2}{y^3}\right)dy &= \int_{-2}^{-1}\left(4y^3+2y^{-3}\right)dy\\[2ex]		\int_{-2}^{-1}\left(4y^3+\frac{2}{y^3}\right)dy &= \int_{-2}^{-1}\left(4y^3+2y^{-3}\right)dy\\[2ex]
	&= \left[y^4-y^-2\right]_{-2}^{-1} \\[2ex]		&= \left[y^{4}-y^{-2}\right]_{-2}^{-1} \\[2ex]
	&= (1-1)-\left(16-\frac{1}{4}\right) \\[2ex]		&= (1-1)-\left(16-\frac{1}{4}\right) \\[2ex]
	&= \frac{-63}{4}		&= \frac{-63}{4}

2022-09-21T15:19:00Z

← Older revision		Revision as of 15:19, 21 September 2022
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	\begin{align}		\begin{align}

	\int_{-2}^{-1}\left(4y^3+\frac{2}{y^3}\right)dy \\[2ex]		\int_{-2}^{-1}\left(4y^3+\frac{2}{y^3}\right)dy &= \int_{-2}^{-1}\left(4y^3+2y^{-3}\right)dy\\[2ex]
	&= \left[y^4-y^-2\right]_{-2}^{-1} \\[2ex]		&= \left[y^4-y^-2\right]_{-2}^{-1} \\[2ex]
	&= (1-1)-\left(16-\frac{1}{4}\right) \\[2ex]		&= (1-1)-\left(16-\frac{1}{4}\right) \\[2ex]

2022-09-21T15:17:49Z

2022-09-21T15:15:56Z

← Older revision		Revision as of 15:15, 21 September 2022
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	<math>\~~int\limits_~~{2}^{-1}\left(4y^3+\frac{2}{y^3}\right)dy~~</math><br>~~		<math>
	~~<math>~~= y^4-y^-2\bigg\|_{-2}^{-1}~~</math><br>~~		\int_{2}^{-1}\left(4y^3+\frac{2}{y^3}\right)dy
	~~<math>~~= (1-1)-\left(16-\frac{1}{4}\right)~~</math><br>~~		= y^4-y^-2\bigg\|_{-2}^{-1}
	~~<math>~~= \frac{-63}{4}		= (1-1)-\left(16-\frac{1}{4}\right)
			= \frac{-63}{4}
	</math>		</math>

2022-08-26T18:49:07Z

← Older revision		Revision as of 18:49, 26 August 2022
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	<math>\int\limits_{2}^{-1}\left(4y^3+\frac{2}{y^3}\right)dy</math><br>		<math>\int\limits_{2}^{-1}\left(4y^3+\frac{2}{y^3}\right)dy</math><br>
	<math>= y^4-y^-2\bigg\|_{-2}^{-1}=(1-1)-\left(16-\frac{1}{4}\right)</math><br>		<math>= y^4-y^-2\bigg\|_{-2}^{-1}</math><br>
			<math>= (1-1)-\left(16-\frac{1}{4}\right)</math><br>
	<math>= \frac{-63}{4}		<math>= \frac{-63}{4}
	</math>		</math>

2022-08-26T18:48:31Z

← Older revision		Revision as of 18:48, 26 August 2022
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	<math>\int\limits_{2}^{-1}\left(4y^3+\frac{2}{y^3}\right)dy</math><br>		<math>\int\limits_{2}^{-1}\left(4y^3+\frac{2}{y^3}\right)dy</math><br>
	<math>y^4-y^-2\bigg\|_{-2}^{-1}=(1-1)-\left(16-\frac{1}{4}\right)=\frac{-63}{4}		<math>= y^4-y^-2\bigg\|_{-2}^{-1}=(1-1)-\left(16-\frac{1}{4}\right)</math><br>
			<math>= \frac{-63}{4}
	</math>		</math>

2022-08-26T16:16:47Z

← Older revision		Revision as of 16:16, 26 August 2022
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	<math>\int\limits_{2}^{-1}\left(4y^3+\frac{2}{y^3}\right)dy</math><br>		<math>\int\limits_{2}^{-1}\left(4y^3+\frac{2}{y^3}\right)dy</math><br>
	<math>y^4-y^-2\bigg\|_{-2}^{-1}=(1-1)-(16-1/4)=-63/4		<math>y^4-y^-2\bigg\|_{-2}^{-1}=(1-1)-\left(16-\frac{1}{4}\right)=\frac{-63}{4}
	</math>		</math>

2022-08-26T16:15:43Z

← Older revision		Revision as of 16:15, 26 August 2022
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	<math>\int\limits_{2}^{-1}\left(4y^3\frac{2}{y^3}\right)dy</math><br>		<math>\int\limits_{2}^{-1}\left(4y^3+\frac{2}{y^3}\right)dy</math><br>
	<math>y^4-y^-2\bigg\|_{-2}^{-1}=(1-1)-(16-1/4)=-63/4		<math>y^4-y^-2\bigg\|_{-2}^{-1}=(1-1)-(16-1/4)=-63/4
	</math>		</math>