5.3 The Fundamental Theorem of Calculus/13 - Revision history

Dvaezazizi@laalliance.org: Protected "5.3 The Fundamental Theorem of Calculus/13" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-06T20:15:33Z

Protected "5.3 The Fundamental Theorem of Calculus/13" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 20:15, 6 September 2022
(No difference)

2022-08-25T20:22:36Z

← Older revision		Revision as of 20:22, 25 August 2022
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	=\frac{-\arctan\frac{1}{x}}{x^2}</math> <br><br>		=\frac{-\arctan\frac{1}{x}}{x^2}</math> <br><br>

	<math>\text{Therefore, } g'(x)=</math>		<math>\text{Therefore, } h'(x)=\frac{-\arctan\frac{1}{x}}{x^2}</math>

2022-08-25T20:21:57Z

← Older revision		Revision as of 20:21, 25 August 2022
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	=\frac{-1}{x^2}\cdot\left(\arctan\left(\frac{1}{x}\right)\right)-0\cdot(\arctan\left(2)\right)		=\frac{-1}{x^2}\cdot\left(\arctan\left(\frac{1}{x}\right)\right)-0\cdot(\arctan\left(2)\right)

	=\frac{-\arctan\frac{1}{x}}{x}</math> <br><br>		=\frac{-\arctan\frac{1}{x}}{x^2}</math> <br><br>

	<math>\text{Therefore, } g'(x)=</math>		<math>\text{Therefore, } g'(x)=</math>

2022-08-25T20:21:24Z

← Older revision		Revision as of 20:21, 25 August 2022
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	=\frac{d}{dx}\left[\int_{2}^{1/x}\arctan(t)dt\right]		=\frac{d}{dx}\left[\int_{2}^{1/x}\arctan(t)dt\right]

	=\frac{-1}{x^2}\cdot\left(\arctan(\frac{1}{x})\right)-0\cdot(\arctan\left(2)\right)		=\frac{-1}{x^2}\cdot\left(\arctan\left(\frac{1}{x}\right)\right)-0\cdot(\arctan\left(2)\right)

	=\frac{-\arctan\frac{1}{x}}{x}</math> <br><br>		=\frac{-\arctan\frac{1}{x}}{x}</math> <br><br>

	<math>\text{Therefore, } g'(x)=</math>		<math>\text{Therefore, } g'(x)=</math>

2022-08-25T20:21:07Z

← Older revision		Revision as of 20:21, 25 August 2022
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	=\frac{d}{dx}\left[\int_{2}^{1/x}\arctan(t)dt\right]		=\frac{d}{dx}\left[\int_{2}^{1/x}\arctan(t)dt\right]

	=\frac{-1}{x^2}\cdot(\arctan~~\left~~(\frac{1}{x}\right))-0\cdot(\arctan\left(2)\right)		=\frac{-1}{x^2}\cdot\left(\arctan(\frac{1}{x})\right)-0\cdot(\arctan\left(2)\right)

	=\frac{-\arctan\frac{1}{x}}{x}</math> <br><br>		=\frac{-\arctan\frac{1}{x}}{x}</math> <br><br>

	<math>\text{Therefore, } g'(x)=</math>		<math>\text{Therefore, } g'(x)=</math>

2022-08-25T20:20:28Z

← Older revision		Revision as of 20:20, 25 August 2022
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	=\frac{d}{dx}\left[\int_{2}^{1/x}\arctan(t)dt\right]		=\frac{d}{dx}\left[\int_{2}^{1/x}\arctan(t)dt\right]


	=\frac{-1}{x^2}\cdot(\arctan\left(\frac{1}{x}\right))-0\cdot(\arctan\left(2)\right)		=\frac{-1}{x^2}\cdot(\arctan\left(\frac{1}{x}\right))-0\cdot(\arctan\left(2)\right)


	=\frac{-\arctan\frac{1}{x}}{x}</math> <br><br>		=\frac{-\arctan\frac{1}{x}}{x}</math> <br><br>

	<math>\text{Therefore, } g'(x)=</math>		<math>\text{Therefore, } g'(x)=</math>

2022-08-25T20:19:55Z

2022-08-25T20:19:18Z

← Older revision		Revision as of 20:19, 25 August 2022
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	=\frac{-1}{x^2}\cdot(\arctan\left(\frac{1}{x}\right))-0\cdot(\arctan\left(2)\right)		=\frac{-1}{x^2}\cdot(\arctan\left(\frac{1}{x}\right))-0\cdot(\arctan\left(2)\right)
	~~</math>~~

	=\frac{-\arctan{\frac{1}{x}}</math> <br><br>
			=\frac{-\arctan\frac{1}{x}</math> <br><br>

	<math>\text{Therefore, } g'(x)=</math>		<math>\text{Therefore, } g'(x)=</math>

2022-08-25T20:18:56Z

← Older revision		Revision as of 20:18, 25 August 2022
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	=\frac{d}{dx}\left[\int_{2}^{1/x}\arctan(t)dt\right]		=\frac{d}{dx}\left[\int_{2}^{1/x}\arctan(t)dt\right]


			=\frac{-1}{x^2}\cdot(\arctan\left(\frac{1}{x}\right))-0\cdot(\arctan\left(2)\right)
	</math>		</math>
	~~=\frac{-1}{x^2}\cdot(\arctan\left(\frac{1}{x}\right))-0\cdot(\arctan\left(2)\right)~~

	=\frac{-\arctan{\frac{1}{x}}</math> <br><br>		=\frac{-\arctan{\frac{1}{x}}</math> <br><br>

	<math>\text{Therefore, } g'(x)=</math>		<math>\text{Therefore, } g'(x)=</math>

2022-08-25T20:18:46Z

← Older revision		Revision as of 20:18, 25 August 2022
Line 5:		Line 5:
	=\frac{d}{dx}\left[\int_{2}^{1/x}\arctan(t)dt\right]		=\frac{d}{dx}\left[\int_{2}^{1/x}\arctan(t)dt\right]

			</math>
	=\frac{-1}{x^2}\cdot(\arctan\left(\frac{1}{x}\right))-0\cdot(\arctan\left(2)\right)		=\frac{-1}{x^2}\cdot(\arctan\left(\frac{1}{x}\right))-0\cdot(\arctan\left(2)\right)