6.5 Average Value of a Function/2 - Revision history

Dvaezazizi@laalliance.org: Protected "6.5 Average Value of a Function/2" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-07T18:10:24Z

Protected "6.5 Average Value of a Function/2" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision

Revision as of 18:10, 7 September 2022

2022-09-07T18:05:55Z

2022-09-07T18:05:44Z

2022-09-07T18:05:27Z

2022-09-07T18:04:39Z

← Older revision		Revision as of 18:04, 7 September 2022
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	&= \frac{1}{2\pi}\int_{-4\pi}^{4\pi}\sin{(u)}\left(\frac{1}{4}\,du\right) = \frac{1}{8\pi}\int_{-4\pi}^{4\pi}\sin(u)\,du \\[2ex]		&= \frac{1}{2\pi}\int_{-4\pi}^{4\pi}\sin{(u)}\left(\frac{1}{4}\,du\right) = \frac{1}{8\pi}\int_{-4\pi}^{4\pi}\sin(u)\,du \\[2ex]

	&= -\frac{1}{8\pi}\cos(u)\bigg\|_{-4\pi}^{4\pi} = \\[2ex]		&= -\frac{1}{8\pi}\cos(u)\bigg\|_{-4\pi}^{4\pi} \\[2ex]

	&= \left[-\frac{1}{8\pi}\cos(4\pi)\right]-\left[-\frac{1}{8\pi}\cos(-4\pi)\right] = \left[-\frac{1}{8\pi}(1)\right]+\left[\frac{1}{8\pi}(1)\right] \\[2ex]		&= \left[-\frac{1}{8\pi}\cos(4\pi)\right]-\left[-\frac{1}{8\pi}\cos(-4\pi)\right] = \left[-\frac{1}{8\pi}(1)\right]+\left[\frac{1}{8\pi}(1)\right] \\[2ex]

2022-09-07T18:04:26Z

← Older revision		Revision as of 18:04, 7 September 2022
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	&= -\frac{1}{8\pi}\cos(u)\bigg\|_{-4\pi}^{4\pi} = \\[2ex]		&= -\frac{1}{8\pi}\cos(u)\bigg\|_{-4\pi}^{4\pi} = \\[2ex]

	\left[-\frac{1}{8\pi}\cos(4\pi)\right]-\left[-\frac{1}{8\pi}\cos(-4\pi)\right] = \left[-\frac{1}{8\pi}(1)\right]+\left[\frac{1}{8\pi}(1)\right] \\[2ex]		&= \left[-\frac{1}{8\pi}\cos(4\pi)\right]-\left[-\frac{1}{8\pi}\cos(-4\pi)\right] = \left[-\frac{1}{8\pi}(1)\right]+\left[\frac{1}{8\pi}(1)\right] \\[2ex]

	&= 0		&= 0

2022-09-07T18:04:12Z

← Older revision		Revision as of 18:04, 7 September 2022
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	&= \frac{1}{2\pi}\int_{-4\pi}^{4\pi}\sin{(u)}\left(\frac{1}{4}\,du\right) = \frac{1}{8\pi}\int_{-4\pi}^{4\pi}\sin(u)\,du \\[2ex]		&= \frac{1}{2\pi}\int_{-4\pi}^{4\pi}\sin{(u)}\left(\frac{1}{4}\,du\right) = \frac{1}{8\pi}\int_{-4\pi}^{4\pi}\sin(u)\,du \\[2ex]

	&= -\frac{1}{8\pi}\cos(u)\bigg\|_{-4\pi}^{4\pi} = ~~\left[-\frac{1}{8\pi}\cos(4\pi)\right]-\left[-\frac{1}{8\pi}\cos(-4\pi)\right]~~ \\[2ex]		&= -\frac{1}{8\pi}\cos(u)\bigg\|_{-4\pi}^{4\pi} = \\[2ex]

	&= \left[-\frac{1}{8\pi}(1)\right]+\left[\frac{1}{8\pi}(1)\right] \\[2ex]		\left[-\frac{1}{8\pi}\cos(4\pi)\right]-\left[-\frac{1}{8\pi}\cos(-4\pi)\right] = \left[-\frac{1}{8\pi}(1)\right]+\left[\frac{1}{8\pi}(1)\right] \\[2ex]

	&= 0		&= 0

2022-09-07T18:03:40Z

← Older revision		Revision as of 18:03, 7 September 2022
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	&= \frac{1}{2\pi}\int_{-4\pi}^{4\pi}\sin{(u)}\left(\frac{1}{4}\,du\right) = \frac{1}{8\pi}\int_{-4\pi}^{4\pi}\sin(u)\,du \\[2ex]		&= \frac{1}{2\pi}\int_{-4\pi}^{4\pi}\sin{(u)}\left(\frac{1}{4}\,du\right) = \frac{1}{8\pi}\int_{-4\pi}^{4\pi}\sin(u)\,du \\[2ex]

	&= -\frac{1}{8\pi}\cos(u)\bigg\|_{-4\pi}^{4\pi} ~~\\[2ex]~~		&= -\frac{1}{8\pi}\cos(u)\bigg\|_{-4\pi}^{4\pi} = \left[-\frac{1}{8\pi}\cos(4\pi)\right]-\left[-\frac{1}{8\pi}\cos(-4\pi)\right] \\[2ex]

	&= \left[-\frac{1}{8\pi}\cos(4\pi)\right]-\left[-\frac{1}{8\pi}\cos(-4\pi)\right] \\[2ex]

	&= \left[-\frac{1}{8\pi}(1)\right]+\left[\frac{1}{8\pi}(1)\right] \\[2ex]		&= \left[-\frac{1}{8\pi}(1)\right]+\left[\frac{1}{8\pi}(1)\right] \\[2ex]

2022-09-07T18:03:07Z

← Older revision		Revision as of 18:03, 7 September 2022
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	&= \left[-\frac{1}{8\pi}\cos(4\pi)\right]-\left[-\frac{1}{8\pi}\cos(-4\pi)\right] \\[2ex]		&= \left[-\frac{1}{8\pi}\cos(4\pi)\right]-\left[-\frac{1}{8\pi}\cos(-4\pi)\right] \\[2ex]

	&= \left[-\frac{1}{8\pi}(1)\right]-\left[-\frac{1}{8\pi}(1)\right] \\[2ex]		&= \left[-\frac{1}{8\pi}(1)\right]+\left[\frac{1}{8\pi}(1)\right] \\[2ex]

	&= 0		&= 0
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	\end{align}		\end{align}
	</math>		</math>


	<math>		<math>

2022-09-07T18:02:45Z

← Older revision		Revision as of 18:02, 7 September 2022
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	&= -\frac{1}{8\pi}\cos(u)\bigg\|_{-4\pi}^{4\pi} \\[2ex]		&= -\frac{1}{8\pi}\cos(u)\bigg\|_{-4\pi}^{4\pi} \\[2ex]

	&= \left[-\frac{1}{8\pi}\cos(4\pi)\right]-\left[-\frac{1}{8\pi}\cos(-4\pi)\right]		&= \left[-\frac{1}{8\pi}\cos(4\pi)\right]-\left[-\frac{1}{8\pi}\cos(-4\pi)\right] \\[2ex]

			&= \left[-\frac{1}{8\pi}(1)\right]-\left[-\frac{1}{8\pi}(1)\right] \\[2ex]

			&= 0

	\end{align}		\end{align}