5.4 Indefinite Integrals and the Net Change Theorem/27 - Revision history

Dvaezazizi@laalliance.org: Protected "5.4 Indefinite Integrals and the Net Change Theorem/27" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-21T19:40:41Z

Protected "5.4 Indefinite Integrals and the Net Change Theorem/27" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

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

2022-09-21T15:14:36Z

← Older revision		Revision as of 15:14, 21 September 2022
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	&=\left(\frac{2t^{\frac{3}{2}}}{3}+\frac{2t^{\frac{5}{2}}}{5}\right)\Bigg\|_{1}^{4} \\[2ex]		&=\left(\frac{2t^{\frac{3}{2}}}{3}+\frac{2t^{\frac{5}{2}}}{5}\right)\Bigg\|_{1}^{4} \\[2ex]

	&=\left[\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}\right]-\left[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}\right] \\[2ex]		&=\left[\frac{2(4)^{\frac{3}{2}}}{3}+\frac{2(4)^{\frac{5}{2}}}{5}\right]-\left[\frac{2(1)^{\frac{3}{2}}}{3}+\frac{2(1)^{\frac{5}{2}}}{5}\right] \\[2ex]

	&=\left[\frac{16}{3}+\frac{64}{5}\right]-\left[\frac{2}{3}+\frac{2}{5} \right] \\[2ex]		&=\left[\frac{16}{3}+\frac{64}{5}\right]-\left[\frac{2}{3}+\frac{2}{5} \right] \\[2ex]

2022-09-21T15:13:56Z

← Older revision		Revision as of 15:13, 21 September 2022
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	\int_{1}^{4}\sqrt{t}(1+t)dt &=\int_{1}^{4}\left(t^{\frac{1}{2}}+t^{\frac{3}{2}}\right)dt \\[2ex]		\int_{1}^{4}\sqrt{t}(1+t)dt &=\int_{1}^{4}\left(t^{\frac{1}{2}}+t^{\frac{3}{2}}\right)dt \\[2ex]

	&=\left(\frac{~~2(t)~~^{3/2}}{3}+\frac{~~2(t)~~^{5/2}}{5}\right)\Bigg\|_{1}^{4} \\[2ex]		&=\left(\frac{2t^{\frac{3}{2}}}{3}+\frac{2t^{\frac{5}{2}}}{5}\right)\Bigg\|_{1}^{4} \\[2ex]

	&=\left[\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}\right]-\left[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}\right] \\[2ex]		&=\left[\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}\right]-\left[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}\right] \\[2ex]

2022-09-21T15:12:25Z

← Older revision		Revision as of 15:12, 21 September 2022
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	&=\left[\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}\right]-\left[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}\right] \\[2ex]		&=\left[\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}\right]-\left[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}\right] \\[2ex]

	&=\left[\frac{16}{3}+\frac{64}{5}]-[\frac{2}{3}+\frac{2}{5} \right]		&=\left[\frac{16}{3}+\frac{64}{5}\right]-\left[\frac{2}{3}+\frac{2}{5} \right] \\[2ex]

	&=\frac{256}{15}		&=\frac{256}{15}

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

2022-09-21T15:11:57Z

← Older revision		Revision as of 15:11, 21 September 2022
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	&=\left[\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}\right]-\left[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}\right] \\[2ex]		&=\left[\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}\right]-\left[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}\right] \\[2ex]

			&=\left[\frac{16}{3}+\frac{64}{5}]-[\frac{2}{3}+\frac{2}{5} \right]
	&=\frac{256}{15}		&=\frac{256}{15}

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

2022-09-21T15:10:49Z

← Older revision		Revision as of 15:10, 21 September 2022
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	&=\left(\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\right)\Bigg\|_{1}^{4} \\[2ex]		&=\left(\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\right)\Bigg\|_{1}^{4} \\[2ex]

	&=\left[\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}\right]-\left[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}\right]		&=\left[\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}\right]-\left[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}\right] \\[2ex]

	&=\frac{256}{15}		&=\frac{256}{15}

2022-09-21T15:10:30Z

← Older revision		Revision as of 15:10, 21 September 2022
Line 6:		Line 6:
	&=\left(\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\right)\Bigg\|_{1}^{4} \\[2ex]		&=\left(\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\right)\Bigg\|_{1}^{4} \\[2ex]

	&=left[\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}\right]-\left[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}\right]		&=\left[\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}\right]-\left[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}\right]

	&=\frac{256}{15}		&=\frac{256}{15}

2022-09-21T15:10:20Z

← Older revision		Revision as of 15:10, 21 September 2022
Line 6:		Line 6:
	&=\left(\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\right)\Bigg\|_{1}^{4} \\[2ex]		&=\left(\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\right)\Bigg\|_{1}^{4} \\[2ex]

	&=\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}-\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}		&=left[\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}\right]-\left[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}\right]

	&=\frac{256}{15}		&=\frac{256}{15}

2022-09-21T15:09:59Z

← Older revision		Revision as of 15:09, 21 September 2022
Line 6:		Line 6:
	&=\left(\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\right)\Bigg\|_{1}^{4} \\[2ex]		&=\left(\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\right)\Bigg\|_{1}^{4} \\[2ex]

	&=~~left[~~\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}~~\right]~~-~~\left[~~\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}~~\right]~~		&=\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}-\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}

	&=\frac{256}{15}		&=\frac{256}{15}

2022-09-21T15:09:56Z

← Older revision		Revision as of 15:09, 21 September 2022
Line 6:		Line 6:
	&=\left(\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\right)\Bigg\|_{1}^{4} \\[2ex]		&=\left(\frac{2(t)^{3/2}}{3}+\frac{2(t)^{5/2}}{5}\right)\Bigg\|_{1}^{4} \\[2ex]

	&=\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}-\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}		&=left[\frac{2(4)^{3/2}}{3}+\frac{2(4)^{5/2}}{5}\right]-\left[\frac{2(1)^{3/2}}{3}+\frac{2(1)^{5/2}}{5}\right]

	&=\frac{256}{15}		&=\frac{256}{15}