5.3 The Fundamental Theorem of Calculus/7 - Revision history

Dvaezazizi@laalliance.org: Protected "5.3 The Fundamental Theorem of Calculus/7" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-06T20:14:26Z

Protected "5.3 The Fundamental Theorem of Calculus/7" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 20:14, 6 September 2022
(No difference)

2022-09-06T19:49:49Z

← Older revision		Revision as of 19:49, 6 September 2022
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	<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]=		<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]=
	(1)\left(\frac{1}{(x)^3+1}\right)-(0)\left(\frac{1}{(1)^3+1}\right)=\frac{1}{x^3+1}</math><Br>		(1)\left(\frac{1}{(x)^3+1}\right)-(0)\left(\frac{1}{(1)^3+1}\right)=\frac{1}{x^3+1}</math><Br>
			<br>
	<math>\text{Therefore, } g'(x)=\frac{1}{x^3+1}</math>		<math>\text{Therefore, } g'(x)=\frac{1}{x^3+1}</math>

2022-09-06T19:49:41Z

← Older revision		Revision as of 19:49, 6 September 2022
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	<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]=		<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]=
	(1)\left(\frac{1}{(x)^3+1}\right)-(0)\left(\frac{1}{(1)^3+1}\right)=\frac{1}{x^3+1}</math>		(1)\left(\frac{1}{(x)^3+1}\right)-(0)\left(\frac{1}{(1)^3+1}\right)=\frac{1}{x^3+1}</math><Br>
	<math>\text{Therefore, } g'(x)=\frac{1}{x^3+1}</math>		<math>\text{Therefore, } g'(x)=\frac{1}{x^3+1}</math>

2022-09-06T19:49:30Z

← Older revision		Revision as of 19:49, 6 September 2022
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	<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]=		<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]=
	(1)\left(\frac{1}{(x)^3+1}\right)-(0)\left(\frac{1}{(1)^3+1}\right)=\frac{1}{x^3+1}</math>		(1)\left(\frac{1}{(x)^3+1}\right)-(0)\left(\frac{1}{(1)^3+1}\right)=\frac{1}{x^3+1}</math>
			<math>\text{Therefore, } g'(x)=\frac{1}{x^3+1}</math>

2022-09-06T19:42:36Z

← Older revision		Revision as of 19:42, 6 September 2022
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	<math>g(x)=\int_{1}^{x}\frac{1}{t^3+1}dt</math><br>		<math>g(x)=\int_{1}^{x}\frac{1}{t^3+1}dt</math><br><br>

	<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]=		<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]=
	(1)\left(\frac{1}{(x)^3+1}\right)-(0)\left(\frac{1}{(1)^3+1}\right)=\frac{1}{x^3+1}</math>		(1)\left(\frac{1}{(x)^3+1}\right)-(0)\left(\frac{1}{(1)^3+1}\right)=\frac{1}{x^3+1}</math>

2022-09-06T19:42:22Z

← Older revision		Revision as of 19:42, 6 September 2022
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	<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]=		<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]=
	(1)\left(\frac{1}{1+x^3}\right)=\frac{1}{1+x^3}</math>		(1)\left(\frac{1}{(x)^3+1}\right)-(0)\left(\frac{1}{(1)^3+1}\right)=\frac{1}{x^3+1}</math>

2022-09-06T19:39:43Z

← Older revision		Revision as of 19:39, 6 September 2022
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	<math>g(x)=\int_{1}^{x}\frac{1}{t^3+1}dt</math><br>		<math>g(x)=\int_{1}^{x}\frac{1}{t^3+1}dt</math><br>

	<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]=		<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]=
	(1)\left(\frac{1}{1+x^3}\right)=\frac{1}{1+x^3}</math>		(1)\left(\frac{1}{1+x^3}\right)=\frac{1}{1+x^3}</math>

2022-08-25T19:43:46Z

← Older revision		Revision as of 19:43, 25 August 2022
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	<math>g(x)=\int_{1}^{x}\frac{1}{t^3+1}dt</math><br>		<math>g(x)=\int_{1}^{x}\frac{1}{t^3+1}dt</math><br>
	<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]~~</math>~~=		<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]=
	(1)\left(\frac{1}{1+x^3}\right)=\frac{1}{1+x^3}</math>		(1)\left(\frac{1}{1+x^3}\right)=\frac{1}{1+x^3}</math>

2022-08-25T19:41:56Z

← Older revision		Revision as of 19:41, 25 August 2022
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	<math>g(x)=\int_{1}^{x}\frac{1}{t^3+1}dt</math><br>		<math>g(x)=\int_{1}^{x}\frac{1}{t^3+1}dt</math><br>
	<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]</math>=(1)\left(\frac{1}{1+x^3}\right)		<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]</math>=
			(1)\left(\frac{1}{1+x^3}\right)=\frac{1}{1+x^3}</math>

2022-08-25T19:38:48Z

← Older revision		Revision as of 19:38, 25 August 2022
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	<math>g(x)=\int_{1}^{x}\frac{1}{t^3+1}dt</math><br>		<math>g(x)=\int_{1}^{x}\frac{1}{t^3+1}dt</math><br>
	<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]</math>=\(1)\left(\frac{1}{1+x^3}\right)~~</math>~~		<math>\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_{1}^{x}\frac{1}{t^3+1}dt\right]</math>=(1)\left(\frac{1}{1+x^3}\right)