6.2 Volumes/1: Difference between revisions
No edit summary |
No edit summary |
||
| Line 15: | Line 15: | ||
&= \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]= \pi\left[8-4+\frac{8}{12}-(4-1+\frac{1}{2})\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]= \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[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 03:36, 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} \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]= \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{align} }