6.2 Volumes/1: Difference between revisions

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&= \pi\left[4x-x^2+\frac{1}{12}x^3\right]\Bigg|_1^2 \\[2ex]
&= \pi\left[4x-x^2+\frac{1}{12}x^3\right]\Bigg|_1^2 \\[2ex]
&= \pi\left[4(2)-(2)^2+\frac{1}{12}(2)^3-4(1)-(1)^2+\frac{1}{12}(1)^3\right]= \pi\left[\frac{10}{15}-\frac{3}{15}\right] \\[2ex]
&= \pi\left[4(2)-(2)^2+\frac{1}{12}(2)^3-4(1)-(1)^2+\frac{1}{12}(1)^3\right]= \pi\left[8-4+\frac{8}{12}-[4-1+\frac{1}{2}\right] \\[2ex]
&= \frac{7\pi}{15}
&= \frac{7\pi}{15}


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

Revision as of 03:28, 24 November 2022

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} y=2-\frac{1}{2}x, x-axis\\[1ex] y=0\\[1ex] x=1\\[1ex] x=2\\[1ex] \end{align} }


Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\pi \int _{1}^{2}\left[(2-{\frac {1}{2}}x)^{2}\right]dy&=\pi \int _{1}^{2}\left[(4-2x+{\frac {1}{4}}x^{2})\right]dx\\[2ex]&=\pi \left[4x-x^{2}+{\frac {1}{12}}x^{3}\right]{\Bigg |}_{1}^{2}\\[2ex]&=\pi \left[4(2)-(2)^{2}+{\frac {1}{12}}(2)^{3}-4(1)-(1)^{2}+{\frac {1}{12}}(1)^{3}\right]=\pi \left[8-4+{\frac {8}{12}}-[4-1+{\frac {1}{2}}\right]\\[2ex]&={\frac {7\pi }{15}}\end{aligned}}}