6.2 Volumes/1: Difference between revisions

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<math>
<math>
y=2-\frac{1}{2}x, y=0, x=1, x=2  
y=2-\frac{1}{2}x, y=0, x=1, x=2; x-axis
</math>
</math>



Revision as of 03:40, 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 y=2-\frac{1}{2}x, y=0, x=1, x=2; x-axis }


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}-\left(4(1)-(1)^{2}+{\frac {1}{12}}(1)^{3}\right)\right]\\[2ex]&=\pi \left[8-4+{\frac {8}{12}}-\left(4-1+{\frac {1}{2}}\right)\right]\\[2ex]&=\pi \left[4+{\frac {8}{12}}-3-{\frac {1}{3}}\right]=\pi \left[1+{\frac {7}{12}}\right]\\[2ex]&=\pi \left[{\frac {12}{12}}+{\frac {7}{12}}\right]=\pi \left[{\frac {19}{12}}\right]\\[2ex]&={\frac {19\pi }{12}}\end{aligned}}}