6.1 Areas Between Curves/27 - Revision history

Johnny C. at 19:21, 20 September 2022

2022-09-20T19:21:20Z

← Older revision		Revision as of 19:21, 20 September 2022
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	<math>\int_{0}^{1} \left(x-\frac{1}{4}x\right)dx + \int_{1}^{2} \left(\frac{1}{x}-\frac{1}{4}x\right)dx = \int_{0}^{1} \left(\frac{3}{4}x\right)dx + \int_{1}^{2} \left(\frac{1}{x}\right)dx - \int_{1}^{2} \left(\frac{1}{4}x\right)dx =		<math>\int_{0}^{1} \left(x-\frac{1}{4}x\right)dx + \int_{1}^{2} \left(\frac{1}{x}-\frac{1}{4}x\right)dx = \int_{0}^{1} \left(\frac{3}{4}x\right)dx + \int_{1}^{2} \left(\frac{1}{x}\right)dx - \int_{1}^{2} \left(\frac{1}{4}x\right)dx


	</math>		</math>

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	\begin{align}		\begin{align}

	&= \left[\frac{3x^2}{8}\right]\Bigg\|_{0}^{1} + \left[\ln\|x\|\right]\Bigg\|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg\|_{1}^{2}=

			&= \left[\frac{3x^2}{8}\right]\Bigg\|_{0}^{1} + \left[\ln\|x\|\right]\Bigg\|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg\|_{1}^{2} \\[2ex]
	&= \left[\frac{3(1)^2}{8}\right] + \left[ln\|2\|-ln\|1\|\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]		&= \left[\frac{3(1)^2}{8}\right] + \left[ln\|2\|-ln\|1\|\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]
	&= \frac{3}{8} + ln\|2\| - \frac{3}{8}		&= \frac{3}{8} + ln\|2\| - \frac{3}{8}

2022-09-20T19:16:35Z

← Older revision		Revision as of 19:16, 20 September 2022
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	</math>		</math>

	<math>\int_{0}^{1} \left(x-\frac{1}{4}x\right)\,dx + \int_{1}^{2} \left\|(\frac{1}{x}-\frac{1}{4}x)\right\|dx = \int_{0}^{1} \left\|(\frac{3}{4}x)\right\|dx + \int_{1}^{2} \left\|(\frac{1}{x})\right\|dx - \int_{1}^{2} \left\|(\frac{1}{4}x)\right\|dx =
			<math>\int_{0}^{1} \left(x-\frac{1}{4}x\right)dx + \int_{1}^{2} \left(\frac{1}{x}-\frac{1}{4}x\right)dx = \int_{0}^{1} \left(\frac{3}{4}x\right)dx + \int_{1}^{2} \left(\frac{1}{x}\right)dx - \int_{1}^{2} \left(\frac{1}{4}x\right)dx =
	</math>		</math>

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	\begin{align}		\begin{align}

	&= \left[\frac{3x^2}{8}\right]\Bigg\|_{0}^{1} + \left[\ln(x)\right]\Bigg\|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg\|_{1}^{2}=		&= \left[\frac{3x^2}{8}\right]\Bigg\|_{0}^{1} + \left[\ln\|x\|\right]\Bigg\|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg\|_{1}^{2}=
	&= \left[\frac{3(1)^2}{8}\right] + \left[ln(2)-ln(1)\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]		&= \left[\frac{3(1)^2}{8}\right] + \left[ln\|2\|-ln\|1\|\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]
	&= frac{3}{8} + ln(2)] - \frac{3}{8}		&= \frac{3}{8} + ln\|2\| - \frac{3}{8}
	&= ln(2)		&= ln(2)
	\end{align}		\end{align}
	</math>		</math>

2022-09-20T19:08:11Z

← Older revision		Revision as of 19:08, 20 September 2022
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	&= \left[\frac{3x^2}{8}\right]\Bigg\|_{0}^{1} + \left[\ln(x)\right]\Bigg\|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg\|_{1}^{2}=		&= \left[\frac{3x^2}{8}\right]\Bigg\|_{0}^{1} + \left[\ln(x)\right]\Bigg\|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg\|_{1}^{2}=
	&= \left[\frac{3(1)^2}{8}\right] + \left[ln(2)-ln(1)\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]		&= \left[\frac{3(1)^2}{8}\right] + \left[ln(2)-ln(1)\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]
	&= ~~\left[\~~frac{3}{8}~~\right]~~ + ~~\left[~~ln(2)~~\right~~] - ~~\left[~~\frac{3}{8}~~\right]~~		&= frac{3}{8} + ln(2)] - \frac{3}{8}
	&= ln(2)		&= ln(2)
	\end{align}		\end{align}
	</math>		</math>

2022-09-20T19:05:54Z

← Older revision		Revision as of 19:05, 20 September 2022
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	</math>		</math>

	<math>\int_{0}^{1} \left(x-\frac{1}{4}x\]right)\,dx + \int_{1}^{2} \left\|(\frac{1}{x}-\frac{1}{4}x)\right\|dx = \int_{0}^{1} \left\|(\frac{3}{4}x)\right\|dx + \int_{1}^{2} \left\|(\frac{1}{x})\right\|dx - \int_{1}^{2} \left\|(\frac{1}{4}x)\right\|dx = </math>		<math>\int_{0}^{1} \left(x-\frac{1}{4}x\right)\,dx + \int_{1}^{2} \left\|(\frac{1}{x}-\frac{1}{4}x)\right\|dx = \int_{0}^{1} \left\|(\frac{3}{4}x)\right\|dx + \int_{1}^{2} \left\|(\frac{1}{x})\right\|dx - \int_{1}^{2} \left\|(\frac{1}{4}x)\right\|dx =
			</math>

	<math>		<math>

2022-09-20T19:04:38Z

← Older revision		Revision as of 19:04, 20 September 2022
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	</math>		</math>

	<math>\int_{0}^{1} \left(x-\frac{1}{4}x)\~~right\|~~dx + \int_{1}^{2} \left\|(\frac{1}{x}-\frac{1}{4}x)\right\|dx = \int_{0}^{1} \left\|(\frac{3}{4}x)\right\|dx + \int_{1}^{2} \left\|(\frac{1}{x})\right\|dx - \int_{1}^{2} \left\|(\frac{1}{4}x)\right\|dx = </math>		<math>\int_{0}^{1} \left(x-\frac{1}{4}x\]right)\,dx + \int_{1}^{2} \left\|(\frac{1}{x}-\frac{1}{4}x)\right\|dx = \int_{0}^{1} \left\|(\frac{3}{4}x)\right\|dx + \int_{1}^{2} \left\|(\frac{1}{x})\right\|dx - \int_{1}^{2} \left\|(\frac{1}{4}x)\right\|dx = </math>

	<math>		<math>

2022-09-20T19:01:31Z

← Older revision		Revision as of 19:01, 20 September 2022
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	</math>		</math>

	<math>\int_{0}^{1} \left\|(x-\frac{1}{4}x)\right\|dx + \int_{1}^{2} \left\|(\frac{1}{x}-\frac{1}{4}x)\right\|d = \int_{0}^{1} \left\|(\frac{3}{4}x)\right\|dx + \int_{1}^{2} \left\|(\frac{1}{x})\right\|dx - \int_{1}^{2} \left\|(\frac{1}{4}x)\right\|dx = </math>		<math>\int_{0}^{1} \left(x-\frac{1}{4}x)\right\|dx + \int_{1}^{2} \left\|(\frac{1}{x}-\frac{1}{4}x)\right\|dx = \int_{0}^{1} \left\|(\frac{3}{4}x)\right\|dx + \int_{1}^{2} \left\|(\frac{1}{x})\right\|dx - \int_{1}^{2} \left\|(\frac{1}{4}x)\right\|dx = </math>

	<math>		<math>

2022-09-20T18:59:28Z

2022-09-20T18:58:28Z

← Older revision		Revision as of 18:58, 20 September 2022
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	<math>		<math>

2022-09-19T06:30:40Z

← Older revision		Revision as of 06:30, 19 September 2022
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	[[File:Desmos-graph.png\|right\|500px\|]]		[[File:Desmos-graph.png\|right\|500px\|]]

			<math>
			\begin{align}

			& \color{red}\mathbf{y= \frac{1}{x} }
			& \color{royalblue}\mathbf{y=x} \\
			& \color{green}\mathbf{y= \frac{1}{4} x } \\
			& x>0 \\

			\end{align}

			</math>

			<math>\int_{0}^{1} \left\|(x-\frac{1}{4}x)\right\|dx + \int_{1}^{2} \left\|(\frac{1}{x}-\frac{1}{4}x)\right\|d = \int_{0}^{1} \left\|(\frac{3}{4}x)\right\|dx + \int_{1}^{2} \left\|(\frac{1}{x})\right\|dx - \int_{1}^{2} \left\|(\frac{1}{4}x)\right\|dx = </math>

			<math>
			\begin{align}

			&= \left[\frac{3x^2}{8}\right]\Bigg\|_{0}^{1} + \left[\ln(x)\right]\Bigg\|_{1}^{2} - \left[\frac{1}{8} x^2\right]\Bigg\|_{1}^{2}=
			&= \left[\frac{3(1)^2}{8}\right] + \left[ln(2)-ln(1)\right] - \left[\frac{1}{8} (2)^2 - \frac{1}{8} (1)^2\right] \\[2ex]
			&= \left[\frac{3}{8}\right] + \left[ln(2)\right] - \left[\frac{3}{8}\right]
			&= ln(2)
			\end{align}
			</math>

2022-09-19T05:33:04Z

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