Juanr69976@students.laalliance.org at 18:05, 21 September 2022

2022-09-21T18:05:42Z

← Older revision		Revision as of 18:05, 21 September 2022
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	\int_{-2}^{2} [(x^3-x)-(3x)]dx \\		\int_{-2}^{2} [(x^3-x)-(3x)]dx=\int_{-2}^{2} [x^3-4x]dx \\[2ex]

	=\int_{-2}^{2} [x^3-4x]dx \\[2ex]

	=\left[\frac{x^4}{4}-2x^2\right]\Bigg\|_{-2}^{2}\\[2ex]		=\left[\frac{x^4}{4}-2x^2\right]\Bigg\|_{-2}^{2}\\[2ex]

Juanr69976@students.laalliance.org at 20:35, 18 September 2022

2022-09-18T20:35:16Z

← Older revision		Revision as of 20:35, 18 September 2022
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	x^2 &=4 \\		x^2 &=4 \\

			x &=\sqrt{4}\\

	x &= \pm2 \\[2ex]		x &= \pm2 \\[2ex]

Juanr69976@students.laalliance.org: Created page with "350px| 16) $\begin{align} & \color{red} \mathbf{y=x^3-x} & \color{royalblue}\mathbf{y=3x} \\ \end{align}$ $\begin{align} x^3-x &= 3x \\ x^3 &=4x \\ x^2(x) &=4(x) \\ x^2 &=4 \\ x &= \pm2 \\[2ex] \int_{-2}^{2} [(x^3-x)-(3x)]dx \\ =\int_{-2}^{2} [x^3-4x]dx \\[2ex] =\left[\frac{x^4}{4}-2x^2\right]\Bigg|_{-2}^{2}\\[2ex] =\frac{(2)^4}{4}-2(2)^2-\left(\frac{(-2)^4}{4}-2(-2)^2\right)\\[2ex] =\frac{16}{4}-2..."$

2022-09-18T20:31:51Z

Created page with "right|350px| 16) <math> \begin{align} & \color{red} \mathbf{y=x^3-x} & \color{royalblue}\mathbf{y=3x} \\ \end{align} </math> <math> \begin{align} x^3-x &= 3x \\ x^3 &=4x \\ x^2(x) &=4(x) \\ x^2 &=4 \\ x &= \pm2 \\[2ex] \int_{-2}^{2} [(x^3-x)-(3x)]dx \\ =\int_{-2}^{2} [x^3-4x]dx \\[2ex] =\left[\frac{x^4}{4}-2x^2\right]\Bigg|_{-2}^{2}\\[2ex] =\frac{(2)^4}{4}-2(2)^2-\left(\frac{(-2)^4}{4}-2(-2)^2\right)\\[2ex] =\frac{16}{4}-2..."

New page

[[File:6.1.16 graph.png|right|350px|]]
16)

<math>
\begin{align}

& \color{red} \mathbf{y=x^3-x}
& \color{royalblue}\mathbf{y=3x} \\

\end{align}
</math>

<math>
\begin{align}

x^3-x &= 3x \\

x^3 &=4x \\

x^2(x) &=4(x) \\

x^2 &=4 \\

x &= \pm2 \\[2ex]

\int_{-2}^{2} [(x^3-x)-(3x)]dx \\

=\int_{-2}^{2} [x^3-4x]dx \\[2ex]

=\left[\frac{x^4}{4}-2x^2\right]\Bigg|_{-2}^{2}\\[2ex]

=\frac{(2)^4}{4}-2(2)^2-\left(\frac{(-2)^4}{4}-2(-2)^2\right)\\[2ex]

=\frac{16}{4}-2(4)-(\frac{16}{4}-2(4))\\[2ex]

=4-8-(4-8)\\[2ex]
=-4-(-4)\\[2ex]
=-4+4\\[2ex]
=0

\end{align}
</math>

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6.1 Areas Between Curves/16 - Revision history

Juanr69976@students.laalliance.org at 18:05, 21 September 2022

Juanr69976@students.laalliance.org at 20:35, 18 September 2022