6.1 Areas Between Curves/23: Difference between revisions

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


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

Revision as of 02:03, 20 September 2022


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