6.1 Areas Between Curves/19: Difference between revisions

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=\int_{-2}^{2} [4-y^2]dy \\
=\int_{-2}^{2} [4-y^2]dy \\


=4y-\frac{y^3}{3} \int_{2}^{-2}
=4y-\frac{y^3}{3} |_{-2}^{2}\\




=4(2)-\frac{(2)^3}{3}-(4(-2)-\frac{(-2)^3}{3})\\
=8-\frac{8}{3}-(-8+\frac{8}{3})\\
=16-\frac{16}{3}\\
=\frac{48}{3}-\frac{16}{3}\\
=\frac{32}{3}\\
\end{align}
\end{align}
</math>
</math>

Revision as of 00:55, 18 September 2022

6.1.19 graph.png

19)

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} & \color{red} \mathbf{x=4+y^2} & \color{royalblue}\mathbf{x=2y^2} \\ \end{align} }

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}4+y^{2}&=2y^{2}\\4&=y^{2}\\y&={\sqrt {4}}\\y&=\pm 2\\\int _{-2}^{2}[(4+y^{2})-(2y^{2})]dy\\=\int _{-2}^{2}[4-y^{2}]dy\\=4y-{\frac {y^{3}}{3}}|_{-2}^{2}\\=4(2)-{\frac {(2)^{3}}{3}}-(4(-2)-{\frac {(-2)^{3}}{3}})\\=8-{\frac {8}{3}}-(-8+{\frac {8}{3}})\\=16-{\frac {16}{3}}\\={\frac {48}{3}}-{\frac {16}{3}}\\={\frac {32}{3}}\\\end{aligned}}}


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