6.2 Volumes/25: Difference between revisions
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R + f(y) = 1 | R + f(y) = 1 | ||
R = 1-f(y) | R = 1-f(y) | ||
r = 1 | r = 1 | ||
\usepackage{amsmath} | |||
\begin{document} | |||
\begin{align} | |||
2x^2 + 3(x-1)(x-2) & = 2x^2 + 3(x^2-3x+2)\\ \nonumber &= 2x^2 + 3x^2 - 9x + 6\\ &= 5x^2 - 9x + 6 | |||
\end{align} | |||
</math> | </math> | ||
Revision as of 03:02, 12 September 2022
Failed to parse (unknown function "\usepackage"): {\displaystyle R + f(y) = 1 R = 1-f(y) r = 1 \usepackage{amsmath} \begin{document} \begin{align} 2x^2 + 3(x-1)(x-2) & = 2x^2 + 3(x^2-3x+2)\\ \nonumber &= 2x^2 + 3x^2 - 9x + 6\\ &= 5x^2 - 9x + 6 \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_0^1\left[(1)^2-(1-y^2)^2\right]dy & = \pi\int_0^1\left[(1-(1-2y^2+y^4)\right]dy = \pi\int_0^1\left[(2y^2-y^4)\right]dy \\[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} }