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\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[\frac{10}{15}-\frac{3}{15}\right] \\[2ex] | ||
&= \frac{7\pi}{15} | &= \frac{7\pi}{15} | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 03:27, 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)^3-4(1)-(1)^2+\frac{1}{12}(1)^3\right]= \pi\left[\frac{10}{15}-\frac{3}{15}\right] \\[2ex] &= \frac{7\pi}{15} \end{align} }