5.4 Indefinite Integrals and the Net Change Theorem/37 - Revision history

Dvaezazizi@laalliance.org: Protected "5.4 Indefinite Integrals and the Net Change Theorem/37" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-21T19:41:26Z

Protected "5.4 Indefinite Integrals and the Net Change Theorem/37" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision

Revision as of 19:41, 21 September 2022

2022-09-21T16:07:34Z

← Older revision		Revision as of 16:07, 21 September 2022
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	= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]		= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]

	&= \tan({\theta}) + \theta \Bigg\|_{0}^{\frac{\pi}{4}}\\[2ex]		&= (\tan({\theta}) + \theta)\Bigg\|_{0}^{\frac{\pi}{4}}\\[2ex]

	&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - \left[\tan{0} + 0\right] \\[2ex]		&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - \left[\tan{0} + 0\right] \\[2ex]

2022-09-21T16:07:23Z

← Older revision		Revision as of 16:07, 21 September 2022
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	= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]		= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]

	&= ~~\left~~\tan({\theta}) + \theta ~~\right]~~\Bigg\|_{0}^{\frac{\pi}{4}}\\[2ex]		&= \tan({\theta}) + \theta \Bigg\|_{0}^{\frac{\pi}{4}}\\[2ex]

	&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - \left[\tan{0} + 0\right] \\[2ex]		&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - \left[\tan{0} + 0\right] \\[2ex]

2022-09-21T16:07:11Z

← Older revision		Revision as of 16:07, 21 September 2022
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	= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]		= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]

	&= \left\tan({\theta}) + \theta \right]\~~Bigg_~~{0}^{\frac{\pi}{4}}\\[2ex]		&= \left\tan({\theta}) + \theta \right]\Bigg\|_{0}^{\frac{\pi}{4}}\\[2ex]

	&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - \left[\tan{0} + 0\right] \\[2ex]		&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - \left[\tan{0} + 0\right] \\[2ex]

2022-09-21T16:06:54Z

← Older revision		Revision as of 16:06, 21 September 2022
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	= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]		= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]

	&= \left[\tan({\theta}) + \theta \right]_{0}^{\frac{\pi}{4}}\\[2ex]		&= \left\tan({\theta}) + \theta \right]\Bigg_{0}^{\frac{\pi}{4}}\\[2ex]

	&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - \left[\tan{0} + 0\right] \\[2ex]		&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - \left[\tan{0} + 0\right] \\[2ex]

2022-09-21T16:06:34Z

← Older revision		Revision as of 16:06, 21 September 2022
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	= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]		= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]

	&= \tan({\theta}) + \theta \ ~~\bigg\|~~_{0}^{\frac{\pi}{4}}\\[2ex]		&= \left[\tan({\theta}) + \theta \right]_{0}^{\frac{\pi}{4}}\\[2ex]

	&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - \left[\tan{0} + 0\right] \\[2ex]		&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - \left[\tan{0} + 0\right] \\[2ex]

2022-09-21T16:05:58Z

← Older revision		Revision as of 16:05, 21 September 2022
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	= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]		= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]

	&= ~~\left[~~\tan~~\left~~({\~~frac~~{~~\pi~~}{~~4}}\right) +~~ \frac{\pi}{4}\~~right~~]		&= \tan({\theta}) + \theta \ \bigg\|_{0}^{\frac{\pi}{4}}\\[2ex]

	&= \tan({\~~theta~~}) + ~~\theta \ \bigg\|_{0}^{~~\frac{\pi}{4}}\\[2ex]		&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - \left[\tan{0} + 0\right] \\[2ex]

	&= 1+\frac{\pi}{4}		&= 1+\frac{\pi}{4}
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	\end{align}		\end{align}
	</math>		</math>

	~~&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right]~~

	~~- left[\tan{0} + 0\right] \\[2ex]~~

2022-09-21T16:05:17Z

← Older revision		Revision as of 16:05, 21 September 2022
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	&= \int_{0}^{\frac{\pi}{4}}\left(\frac{1}{\cos^2(\theta)} + \frac{\cos^2(\theta)}{\cos^2(\theta)}\right)d\theta		&= \int_{0}^{\frac{\pi}{4}}\left(\frac{1}{\cos^2(\theta)} + \frac{\cos^2(\theta)}{\cos^2(\theta)}\right)d\theta
	= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]		= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]

			&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right]

	&= \tan({\theta}) + \theta \ \bigg\|_{0}^{\frac{\pi}{4}}\\[2ex]		&= \tan({\theta}) + \theta \ \bigg\|_{0}^{\frac{\pi}{4}}\\[2ex]
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	</math>		</math>

	&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - left[\tan{0} + 0\right] \\[2ex]		&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right]

			- left[\tan{0} + 0\right] \\[2ex]

2022-09-21T16:04:55Z

← Older revision		Revision as of 16:04, 21 September 2022
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	&= \tan({\theta}) + \theta \ \bigg\|_{0}^{\frac{\pi}{4}}\\[2ex]		&= \tan({\theta}) + \theta \ \bigg\|_{0}^{\frac{\pi}{4}}\\[2ex]

	~~&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - left[\tan{0} + 0\right] \\[2ex]~~
	&= 1+\frac{\pi}{4}		&= 1+\frac{\pi}{4}

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

			&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - left[\tan{0} + 0\right] \\[2ex]

2022-09-21T16:04:46Z

← Older revision		Revision as of 16:04, 21 September 2022
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	= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]		= \int_{0}^{\frac{\pi}{4}}\left(\sec^2(\theta) + 1\right)d\theta \\[2ex]

	& =\tan({\theta}) + \theta \ \bigg\|_{0}^{\frac{\pi}{4}}\\[2ex]		&= \tan({\theta}) + \theta \ \bigg\|_{0}^{\frac{\pi}{4}}\\[2ex]
	& =left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - left[\tan{0} + 0\right] \\[2ex]
	& =1+\frac{\pi}{4}		&= \left[\tan\left({\frac{\pi}{4}}\right) + \frac{\pi}{4}\right] - left[\tan{0} + 0\right] \\[2ex]
			&= 1+\frac{\pi}{4}

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