6.1 Areas Between Curves/23: Difference between revisions

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<math>
<math>
\begin{align}
\begin{align}
\int_{0}^{\frac{\pi}{6}} \left(\cos(x) - \sin(2x) \right)dx  
\int_{0}^{\frac{\pi}{6}} \left(\cos(x) - \sin(2x) \right)dx &= \left[\sin(x)+\frac{1}{2}\cos(2x) \right]\Bigg|_{0}^{\frac{\pi}{6}} \\[2ex]


&= \left[\sin(x)+\frac{1}{2}\cos(2x) \right]\Bigg|_{0}^{\frac{\pi}{6}} \\[2ex]
&= \frac{1}{2}+\frac{1}{4}-(0+\frac{1}{2})


\end{align}
\end{align}
</math>
</math>

Revision as of 01:47, 20 September 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} \cos(x) &= \sin(2x) \\ x &= \frac{\pi}{2} \\ x &= \frac{\pi}{6} \\ \end{align} }


Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}\int _{0}^{\frac {\pi }{6}}\left(\cos(x)-\sin(2x)\right)dx&=\left[\sin(x)+{\frac {1}{2}}\cos(2x)\right]{\Bigg |}_{0}^{\frac {\pi }{6}}\\[2ex]&={\frac {1}{2}}+{\frac {1}{4}}-(0+{\frac {1}{2}})\end{aligned}}}