5.3 The Fundamental Theorem of Calculus/11 - Revision history

Dvaezazizi@laalliance.org: Protected "5.3 The Fundamental Theorem of Calculus/11" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-06T20:15:24Z

Protected "5.3 The Fundamental Theorem of Calculus/11" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

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

Dvaezazizi@laalliance.org at 20:06, 6 September 2022

2022-09-06T20:06:27Z

← Older revision		Revision as of 20:06, 6 September 2022
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	<math> g(x)= \int_{x}^{\pi}\sqrt{1+sec(t)}\cdot dt </math> <br><br>		<math> g(x)= \int_{x}^{\pi}\sqrt{1+sec(t)}\cdot dt </math> <br><br>
	<math> \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\~~int_\x~~^{\pi}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}</math> <br><br>		<math> \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_x^{\pi}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}</math> <br><br>
	<math>\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}</math>		<math>\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}</math>

Dvaezazizi@laalliance.org at 20:06, 6 September 2022

2022-09-06T20:06:19Z

← Older revision		Revision as of 20:06, 6 September 2022
Line 1:		Line 1:
	<math> g(x)= \int_{x}^{\pi}\sqrt{1+sec(t)}\cdot dt </math> <br><br>		<math> g(x)= \int_{x}^{\pi}\sqrt{1+sec(t)}\cdot dt </math> <br><br>
	<math> \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\pi^{x}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}</math> <br><br>		<math> \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\x^{\pi}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}</math> <br><br>
	<math>\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}</math>		<math>\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}</math>

Dvaezazizi@laalliance.org at 20:05, 6 September 2022

2022-09-06T20:05:51Z

← Older revision		Revision as of 20:05, 6 September 2022
Line 1:		Line 1:
	<math> g(x)= \int_{~~\pi~~}^{x}\sqrt{1+sec(t)}\cdot dt </math> <br><br>		<math> g(x)= \int_{x}^{\pi}\sqrt{1+sec(t)}\cdot dt </math> <br><br>
	<math> \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\pi^{x}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}</math> <br><br>		<math> \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\pi^{x}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}</math> <br><br>
	<math>\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}</math>		<math>\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}</math>

Elizabethh58413: Created page with " $g(x)= \int_{\pi}^{x}\sqrt{1+sec(t)}\cdot dt$

$\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\pi^{x}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}$

$\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}$ "

2022-08-25T19:24:02Z

Created page with "<math> g(x)= \int_{\pi}^{x}\sqrt{1+sec(t)}\cdot dt </math> <math> \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\pi^{x}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}</math> <math>\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}</math>"

New page

<math> g(x)= \int_{\pi}^{x}\sqrt{1+sec(t)}\cdot dt </math> 
<math> \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\pi^{x}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}</math> 
<math>\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}</math>

5.3 The Fundamental Theorem of Calculus/11 - Revision history

Dvaezazizi@laalliance.org: Protected "5.3 The Fundamental Theorem of Calculus/11" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

Dvaezazizi@laalliance.org at 20:06, 6 September 2022

Dvaezazizi@laalliance.org at 20:06, 6 September 2022

Dvaezazizi@laalliance.org at 20:05, 6 September 2022

Elizabethh58413: Created page with " g(x)= \int_{\pi}^{x}\sqrt{1+sec(t)}\cdot dt \frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\pi^{x}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)} \text{Therefore, } g'(x)=-\sqrt{1+sec(x)}"

Elizabethh58413: Created page with " $g(x)= \int_{\pi}^{x}\sqrt{1+sec(t)}\cdot dt$

$\frac{d}{dx}\left[g(x)\right]=\frac{d}{dx}\left[\int_\pi^{x}\sqrt{1+sec(t)}\cdot dt\right]=0 \cdot \sqrt{1+sec(\pi)} - 1\cdot \sqrt{1+sec(x)} = -\sqrt{1+sec(x)}$

$\text{Therefore, } g'(x)=-\sqrt{1+sec(x)}$ "