5.3 The Fundamental Theorem of Calculus/37 - Revision history

Dvaezazizi@laalliance.org: Protected "5.3 The Fundamental Theorem of Calculus/37" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-09-06T21:59:29Z

Protected "5.3 The Fundamental Theorem of Calculus/37" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision	Revision as of 21:59, 6 September 2022
(No difference)

2022-09-06T21:58:51Z

← Older revision		Revision as of 21:58, 6 September 2022
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	\begin{align}		\begin{align}

	~~g(x)~~=\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{6}{\sqrt{1-t^2}} dt \\[2ex]		\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{6}{\sqrt{1-t^2}}\,dt &= 6\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{1}{\sqrt{1-t^2}}\,dt\\[2ex]

	g^\~~prime~~(x)~~=\frac{d}{dx~~}\~~left(\int\limits_{1/2}^{\sqrt~~{~~3}/2}~~\frac{6}{~~\sqrt{1-t^~~2}} ~~dt\right)=6sin~~^{~~-1}(x)~~\~~bigg\|_{1/2}^~~{\sqrt{3}/2}		&=6\arcsin{(x)}\bigg\|_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \\[2ex]

	=~~6sin^~~{~~-1}(~~\sqrt{3})/2)-(~~6sin^~~{-1}~~(1/~~2))		&=\left[6\arcsin\left(\frac{\sqrt{3}}{2}\right)\right]-\left[6\arcsin{\left(\frac{1}{2}\right)}\right] \\[2ex]

	=\pi		&=\left[6\cdot\frac{\pi}{3}\right]-\left[6\cdot\frac{\pi}{6}\right] = 2\pi-\pi \\[2ex]

			&=\pi

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

2022-09-06T21:37:41Z

← Older revision		Revision as of 21:37, 6 September 2022
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	<math>		<math>
			\begin{align}

	g(x)=\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{6}{\sqrt{1-t^2}} dt		g(x)=\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{6}{\sqrt{1-t^2}} dt \\[2ex]

	g^\prime(x)=\frac{d}{dx}\left(\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt\right)=6sin^{-1}(x)\bigg\|_{1/2}^{\sqrt{3}/2}		g^\prime(x)=\frac{d}{dx}\left(\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt\right)=6sin^{-1}(x)\bigg\|_{1/2}^{\sqrt{3}/2}

2022-09-06T21:37:13Z

← Older revision		Revision as of 21:37, 6 September 2022
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	<math>		<math>

	g(x)=\int_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt		g(x)=\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}\frac{6}{\sqrt{1-t^2}} dt

	g^\prime(x)=\frac{d}{dx}\left(\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt\right)=6sin^{-1}(x)\bigg\|_{1/2}^{\sqrt{3}/2}		g^\prime(x)=\frac{d}{dx}\left(\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt\right)=6sin^{-1}(x)\bigg\|_{1/2}^{\sqrt{3}/2}

2022-09-06T21:36:54Z

← Older revision		Revision as of 21:36, 6 September 2022
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	<math>		<math>

	g(x)=\~~int\limits_~~{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt		g(x)=\int_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt

	g^\prime(x)=\frac{d}{dx}\left(\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt\right)=6sin^{-1}(x)\bigg\|_{1/2}^{\sqrt{3}/2}		g^\prime(x)=\frac{d}{dx}\left(\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt\right)=6sin^{-1}(x)\bigg\|_{1/2}^{\sqrt{3}/2}

2022-09-06T21:36:42Z

@@ Line 1: / Line 1: @@
-FTC # 2- the <math>\frac{d}{dx}\left(\int\limits_{a(x)}^{b(x)}F(x)dx\right)</math> is F(b)-F(a) where F is the antiderivitive of f such that <math>F^\prime=f</math>
+<math>
-) g(x)=<math>\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt </math>
+g(x)=\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt
-<math>g^\prime(x)=\frac{d}{dx}\left(\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt\right)=6sin^{-1}(x)\bigg|_{1/2}^{\sqrt{3}/2}</math>
+g^\prime(x)=\frac{d}{dx}\left(\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt\right)=6sin^{-1}(x)\bigg|_{1/2}^{\sqrt{3}/2}
-=<math>6sin^{-1}(\sqrt{3})/2)-(6sin^{-1}(1/2))</math>
+=6sin^{-1}(\sqrt{3})/2)-(6sin^{-1}(1/2))
-=<math>\pi</math>
+=\pi
-[[5.3 The Fundamental Theorem of Calculus/1|1]]
+</math>

2022-08-26T16:03:53Z

← Older revision		Revision as of 16:03, 26 August 2022
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	FTC # 2- the <math>\frac{d}{dx}\left[\int\limits_{a(x)}^{b(x)}F(x)dx\right]</math> is F(b)-F(a) where F is the antiderivitive of f such that <math>F^\prime=f</math>		FTC # 2- the <math>\frac{d}{dx}\left(\int\limits_{a(x)}^{b(x)}F(x)dx\right)</math> is F(b)-F(a) where F is the antiderivitive of f such that <math>F^\prime=f</math>

	37) g(x)=<math>\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt </math>		37) g(x)=<math>\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt </math>

2022-08-26T16:03:05Z

← Older revision		Revision as of 16:03, 26 August 2022
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	FTC # 2- the <math>\frac{d}{dx}[\int\limits_{a(x)}^{b(x)}F(x)dx]</math> is F(b)-F(a) where F is the antiderivitive of f such that <math>F^\prime=f</math>		FTC # 2- the <math>\frac{d}{dx}\left[\int\limits_{a(x)}^{b(x)}F(x)dx\right]</math> is F(b)-F(a) where F is the antiderivitive of f such that <math>F^\prime=f</math>

	37) g(x)=<math>\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt </math>		37) g(x)=<math>\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt </math>

2022-08-25T19:31:10Z

← Older revision		Revision as of 19:31, 25 August 2022
Line 1:		Line 1:
	FTC # 2- the <math>\frac{d}{dx}[\int\limits_{a(x)}^{b(x)}F(x)dx]</math> is F(b)-F(a) where F is the antiderivitive of f such that <math>F^\prime=f</math>		FTC # 2- the <math>\frac{d}{dx}[\int\limits_{a(x)}^{b(x)}F(x)dx]</math> is F(b)-F(a) where F is the antiderivitive of f such that <math>F^\prime=f</math>

	37) g(x)=<math>\int\limits_{1/2}^{\sqrt{3})/2}\frac{6}{\sqrt{1-t^2}} dt </math>		37) g(x)=<math>\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt </math>

	<math>g^\prime(x)=\frac{d}{dx}\left(\int\limits_{1/2}^{\sqrt{3})/2}\frac{6}{\sqrt{1-t^2}} dt\right)=6sin^{-1}(x)\bigg\|_{1/2}^{\sqrt{3}/2}</math>		<math>g^\prime(x)=\frac{d}{dx}\left(\int\limits_{1/2}^{\sqrt{3}/2}\frac{6}{\sqrt{1-t^2}} dt\right)=6sin^{-1}(x)\bigg\|_{1/2}^{\sqrt{3}/2}</math>

	=<math>6sin^{-1}(\sqrt{3})/2)-(6sin^{-1}(1/2))</math>		=<math>6sin^{-1}(\sqrt{3})/2)-(6sin^{-1}(1/2))</math>

2022-08-25T19:30:09Z

← Older revision		Revision as of 19:30, 25 August 2022
Line 1:		Line 1:
	FTC # 2- the <math>\frac{d}{dx}[\int\limits_{a(x)}^{b(x)}F(x)dx]</math> is F(b)-F(a) where F is the antiderivitive of f such that <math>F^\prime=f</math>		FTC # 2- the <math>\frac{d}{dx}[\int\limits_{a(x)}^{b(x)}F(x)dx]</math> is F(b)-F(a) where F is the antiderivitive of f such that <math>F^\prime=f</math>

	37) <math>\int\limits_{1/2}^{\sqrt{3})/2}\frac{6}{\sqrt{1-t^2}} dt </math>		37) g(x)=<math>\int\limits_{1/2}^{\sqrt{3})/2}\frac{6}{\sqrt{1-t^2}} dt </math>

	~~using the rule we get~~		<math>g^\prime(x)=\frac{d}{dx}\left(\int\limits_{1/2}^{\sqrt{3})/2}\frac{6}{\sqrt{1-t^2}} dt\right)=6sin^{-1}(x)\bigg\|_{1/2}^{\sqrt{3}/2}</math>
	<math>6sin^{-1}(x)\bigg\|_{1/2}^{\sqrt{3}/2}</math>

	=<math>6sin^{-1}(\sqrt{3})/2)-(6sin^{-1}(1/2))</math>		=<math>6sin^{-1}(\sqrt{3})/2)-(6sin^{-1}(1/2))</math>