5.3 The Fundamental Theorem of Calculus/31 - Revision history

Dvaezazizi@laalliance.org: Protected "5.3 The Fundamental Theorem of Calculus/31" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-06T21:25:03Z

Protected "5.3 The Fundamental Theorem of Calculus/31" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 21:25, 6 September 2022
(No difference)

2022-09-06T21:24:49Z

← Older revision		Revision as of 21:24, 6 September 2022
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	<math>		<math>
	\int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,dt = \tan(t)\bigg\|_{0}^{\frac{\pi}{4}=\tan\left(\frac{\pi}{4}\right)-\tan(0)=1-0=1		\int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,dt = \tan(t)\bigg\|_{0}^{\frac{\pi}{4}}=\tan\left(\frac{\pi}{4}\right)-\tan(0)=1-0=1
	</math>		</math>

2022-09-06T21:24:36Z

← Older revision		Revision as of 21:24, 6 September 2022
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	<math>		<math>
	\int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,dt = \tan\left(\frac{\pi}{4}\right)-\tan(0)=1-0=1		\int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,dt = \tan(t)\bigg\|_{0}^{\frac{\pi}{4}=\tan\left(\frac{\pi}{4}\right)-\tan(0)=1-0=1
	</math>		</math>

2022-09-06T21:23:48Z

← Older revision		Revision as of 21:23, 6 September 2022
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	<math> \int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,dt = \tan\left(\frac{\pi}{4}\right)-\tan(0)=1-0=1		<math>
			\int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,dt = \tan\left(\frac{\pi}{4}\right)-\tan(0)=1-0=1
	</math>		</math>

	~~Therefore, <math> \int_{0}^\frac{\pi}{4}sec^{2}(t)dt = 1 </math>~~

	~~(Use FTC #2,) <math> \int_{a}^{b}f(x)dt = F(b)-F(a) </math>~~

2022-09-06T21:23:37Z

← Older revision		Revision as of 21:23, 6 September 2022
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	<math> \int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,dt = tan\left(\frac{\pi}{4}\right)-tan(0)=1-0=1		<math> \int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,dt = \tan\left(\frac{\pi}{4}\right)-\tan(0)=1-0=1

	</math>		</math>

2022-09-06T21:23:26Z

← Older revision		Revision as of 21:23, 6 September 2022
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	<math> \int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,~~dt = \int_{0}^\frac{\pi}{4}\sec^{2}(t)~~dt= tan\left(\frac{\pi}{4}\right)-tan(0)=1-0=1		<math> \int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,dt = tan\left(\frac{\pi}{4}\right)-tan(0)=1-0=1

	</math>		</math>

2022-09-06T21:23:10Z

← Older revision		Revision as of 21:23, 6 September 2022
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	<math> \int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,dt = \int_{0}^\frac{\pi}{4}sec^{2}(t)dt= tan(\frac{\pi}{4})-tan(0)=1-0=1		<math> \int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,dt = \int_{0}^\frac{\pi}{4}\sec^{2}(t)dt= tan\left(\frac{\pi}{4}\right)-tan(0)=1-0=1

	</math>		</math>

2022-09-06T21:22:44Z

← Older revision		Revision as of 21:22, 6 September 2022
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	~~Evaluate the integral~~		<math> \int_{0}^{\frac{\pi}{4}}\sec^{2}(t)\,dt = \int_{0}^\frac{\pi}{4}sec^{2}(t)dt= tan(\frac{\pi}{4})-tan(0)=1-0=1

	<math> \int_{0}^\frac{\pi}{4}sec^{2}(t)dt ~~</math>~~

	~~<math>~~

	\int_{0}^\frac{\pi}{4}sec^{2}(t)dt= tan(\frac{\pi}{4})-tan(0)=1-0=1

	</math>		</math>

2022-08-26T19:16:46Z

← Older revision		Revision as of 19:16, 26 August 2022
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	Therefore, <math> \int_{0}^\frac{\pi}{4}sec^{2}(t)dt = 1 </math>		Therefore, <math> \int_{0}^\frac{\pi}{4}sec^{2}(t)dt = 1 </math>

	(Use FTC #2, <math> \int_{a}^{b}f(x)dt = F(b)-F(a) </math>)		(Use FTC #2,) <math> \int_{a}^{b}f(x)dt = F(b)-F(a) </math>

2022-08-25T19:34:31Z

← Older revision		Revision as of 19:34, 25 August 2022
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	Therefore, <math> \int_{0}^\frac{\pi}{4}sec^{2}(t)dt = 1 </math>		Therefore, <math> \int_{0}^\frac{\pi}{4}sec^{2}(t)dt = 1 </math>

	Use FTC #2 <math> \int_{a}^{b}f(x)dt = F(b)-F(a) </math>		(Use FTC #2, <math> \int_{a}^{b}f(x)dt = F(b)-F(a) </math>)