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-\left(4(1)-(1)^2+\frac{1}{12}(1)^3\right)\right] \\[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}{ | &= \pi\left[8-4+\frac{8}{12}-\left(4-1+\frac{1}{12}\right)\right] \\[2ex] | ||
&= \pi\left[4+\frac{8}{12}-3-\frac{1}{3}\right]= \pi\left[1+\frac{7}{12}\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] | &= \pi\left[\frac{12}{12}+\frac{7}{12}\right]= \pi\left[\frac{19}{12}\right] \\[2ex] | ||
Revision as of 03:45, 24 November 2022
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[\left(2-{\frac {1}{2}}x\right)^{2}\right]dx&=\pi \int _{1}^{2}\left[\left(4-2x+{\frac {1}{4}}x^{2}\right)\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}{12}}\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}}}