6.2 Volumes/1: Difference between revisions

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<math>
<math>
\begin{align}
\begin{align}
\pi\int_1^2\left[(2-\frac{1}{2})^{2}\right]dx & = \pi\int_1^2\left[(4-2x+\frac{1}{4}x^2)\right]dx = \\[2ex]
\pi\int_1^2\left[(2-\frac{1}{2})^{2}\right]dx & = \pi\int_1^2\left[(4-2x+\frac{1}{4}x^2)\right]dx \\[2ex]


&= \pi\left[\frac{2y^3}{3}-\frac{y^5}{5}\right]\Bigg|_0^1 \\[2ex]
&= \pi\left[\frac{2y^3}{3}-\frac{y^5}{5}\right]\Bigg|_0^1 \\[2ex]

Revision as of 03:22, 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 (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})^{2}\right]dx & = \pi\int_1^2\left[(4-2x+\frac{1}{4}x^2)\right]dx \\[2ex] &= \pi\left[\frac{2y^3}{3}-\frac{y^5}{5}\right]\Bigg|_0^1 \\[2ex] &= \pi\left[\frac{2}{3}-\frac{1}{5}\right]= \pi\left[\frac{10}{15}-\frac{3}{15}\right] \\[2ex] &= \frac{7\pi}{15} \end{align} }