5.4 Indefinite Integrals and the Net Change Theorem/1 - Revision history

Dvaezazizi@laalliance.org: Protected "5.4 Indefinite Integrals and the Net Change Theorem/1" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

2022-08-26T18:24:54Z

Protected "5.4 Indefinite Integrals and the Net Change Theorem/1" ([Edit=Allow only administrators] (indefinite) [Move=Allow only administrators] (indefinite))

← Older revision

Revision as of 18:24, 26 August 2022

2022-08-25T17:23:56Z

← Older revision		Revision as of 17:23, 25 August 2022
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	<math>\begin{align}		<math>\begin{align}
	a &= x^2+1 & b &= a^{1/2} \\[0.6ex]		a &= x^2+1 & b &= a^{1/2} \\[0.6ex]
	\frac{da}{dx} &=2x y & \frac{db}{da} &=\frac{1}{2}a^{-1/2}		\frac{da}{dx} &=2x & \frac{db}{da} &=\frac{1}{2}a^{-1/2}
	\end{align}</math>		\end{align}</math>
	<br><br>		<br><br>

	<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>		<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>

2022-08-25T17:23:45Z

← Older revision		Revision as of 17:23, 25 August 2022
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	\end{align}</math>		\end{align}</math>
	<br><br>		<br><br>

	~~<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>~~


	<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>		<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>

2022-08-25T17:23:38Z

← Older revision		Revision as of 17:23, 25 August 2022
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	<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>		<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>


	~~<math>\int\frac{x}{\sqrt{x^2+1}}dx=\sqrt{x^2+1}+c</math>~~

	~~<math>\frac{d}{dx}\left[(x^2+1)^\frac{1}{2}+c\right]= \frac{x}{\sqrt{x^2+1}}</math>~~

	~~<br>~~
	~~<math>\begin{align}~~
	~~a &= x^2+1 & b &= a^{1/2} \\[0.6ex]~~
	~~\frac{da}{dx} &=2x y & \frac{db}{da} &=\frac{1}{2}a^{-1/2}~~
	~~\end{align}</math>~~
	~~<br><br>~~

	<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>		<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>

2022-08-25T17:23:26Z

← Older revision		Revision as of 17:23, 25 August 2022
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	<br>		<br>
	<math>\begin{align}		<math>\begin{align}
	a &= x^2+1 b &= a^{1/2} \\[0.6ex]		a &= x^2+1 & b &= a^{1/2} \\[0.6ex]
	\frac{da}{dx} &=2x & \frac{db}{da} &=\frac{1}{2}a^{-1/2}		\frac{da}{dx} &=2x y & \frac{db}{da} &=\frac{1}{2}a^{-1/2}
			\end{align}</math>
			<br><br>

			<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>


			<math>\int\frac{x}{\sqrt{x^2+1}}dx=\sqrt{x^2+1}+c</math>

			<math>\frac{d}{dx}\left[(x^2+1)^\frac{1}{2}+c\right]= \frac{x}{\sqrt{x^2+1}}</math>

			<br>
			<math>\begin{align}
			a &= x^2+1 & b &= a^{1/2} \\[0.6ex]
			\frac{da}{dx} &=2x y & \frac{db}{da} &=\frac{1}{2}a^{-1/2}
	\end{align}</math>		\end{align}</math>
	<br><br>		<br><br>

	<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>		<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>

2022-08-25T17:22:29Z

← Older revision		Revision as of 17:22, 25 August 2022
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	<br>		<br>
	<math>\begin{align}		<math>\begin{align}
	a &= x^2+1 b & &= a^{1/2} \\[0.6ex]		a &= x^2+1 b &= a^{1/2} \\[0.6ex]
	\frac{da}{dx} & &=2x \frac{db}{da} &=\frac{1}{2}a^{-1/2}		\frac{da}{dx} &=2x & \frac{db}{da} &=\frac{1}{2}a^{-1/2}
	\end{align}</math>		\end{align}</math>
	<br><br>		<br><br>

	<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>		<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>

2022-08-25T17:22:19Z

← Older revision		Revision as of 17:22, 25 August 2022
Line 5:		Line 5:
	<br>		<br>
	<math>\begin{align}		<math>\begin{align}
	a &= x^2+1 b &= a^{1/2} \\[0.6ex]		a &= x^2+1 b & &= a^{1/2} \\[0.6ex]
	\frac{da}{dx} &=2x \frac{db}{da} &=\frac{1}{2}a^{-1/2}		\frac{da}{dx} & &=2x \frac{db}{da} &=\frac{1}{2}a^{-1/2}
	\end{align}</math>		\end{align}</math>
	<br><br>		<br><br>

	<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>		<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>

2022-08-25T17:22:07Z

← Older revision		Revision as of 17:22, 25 August 2022
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	Show that: <math>\frac{d}{dx}\left[(x^2+1)^\frac{1}{2}+c\right]= \frac{x}{\sqrt{x^2+1}}</math>		Show that: <math>\frac{d}{dx}\left[(x^2+1)^\frac{1}{2}+c\right]= \frac{x}{\sqrt{x^2+1}}</math>

	~~<br>~~<br>		<br>
	<math>\begin{align}		<math>\begin{align}
	a &= x^2+1 b &= a^{1/2} \\[0.6ex]		a &= x^2+1 b &= a^{1/2} \\[0.6ex]
	\frac{da}{dx} &=2x & \frac{db}{da} &=\frac{1}{2}a^{-1/2}		\frac{da}{dx} &=2x \frac{db}{da} &=\frac{1}{2}a^{-1/2}
	\end{align}</math>		\end{align}</math>
	<br><br>		<br><br>

	<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>		<math>\frac{da}{dx}\cdot\frac{db}{da} = \left(2x\right)\left(\frac{1}{2}a^{-1/2}\right) = xa^{-1/2} = x(x^2+1)^{-1/2} = \frac{x}{\sqrt{x^2+1}}</math>

2022-08-25T17:20:50Z

← Older revision		Revision as of 17:20, 25 August 2022
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	<math>\int\frac{x}{\sqrt{x^2+1}}dx=\sqrt{x^2+1}+c</math~~><br~~><br>		<math>\int\frac{x}{\sqrt{x^2+1}}dx=\sqrt{x^2+1}+c</math><br>

	Show that: <math>\frac{d}{dx}\left[(x^2+1)^\frac{1}{2}+c\right]= \frac{x}{\sqrt{x^2+1}}</math>		Show that: <math>\frac{d}{dx}\left[(x^2+1)^\frac{1}{2}+c\right]= \frac{x}{\sqrt{x^2+1}}</math>

	<br>		<br><br>
	<math>\begin{align}		<math>\begin{align}
	a &= x^2+1 b &= a^{1/2} \\[0.6ex]		a &= x^2+1 b &= a^{1/2} \\[0.6ex]

2022-08-25T17:18:27Z