6.1 Areas Between Curves/23: Difference between revisions
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<math> | <math> | ||
\begin{align} | \begin{align} | ||
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<math> | <math> | ||
\int_{0}^{\frac{\pi}{6}} \left(\cos(x) - \sin(2x) \right)dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \left(\sin(2x)- \cos(x) \right)dx | \int_{0}^{\frac{\pi}{6}} \left(\cos(x) - \sin(2x) \right)dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \left(\sin(2x)- \cos(x) \right)dx = \frac{1}{4}+\frac{1}{4} = \frac{2}{4} = \frac{1}{2} | ||
</math> | </math> | ||
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&= \left[\sin(\frac{\pi}{6})+\frac{1}{2}\cos(\frac{2\pi}{6})\right]-\left[\sin(0)+\frac{1}{2}\cos(2(0))\right] \\[2ex] | &= \left[\sin(\frac{\pi}{6})+\frac{1}{2}\cos(\frac{2\pi}{6})\right]-\left[\sin(0)+\frac{1}{2}\cos(2(0))\right] \\[2ex] | ||
&= \frac{1}{2}+\frac{1}{2}(\frac{1}{2}) | &= \left[\frac{1}{2}+\frac{1}{2}\left(\frac{1}{2}\right)\right]-\left[0-\frac{1}{2} (1)\right] \\[2ex] | ||
&= \frac{1}{2}+\frac{1}{4}-( | &= \frac{1}{2}+\frac{1}{4}-\frac{1}{2} \\ | ||
&= \frac{1}{4} | |||
\end{align} | |||
</math> | |||
<math> | |||
\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}} \\ [2ex] | |||
&= \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] \\[2ex] | |||
&= \left[-\frac{1}{2}\left(-1\right)-1\right]-\left[-\frac{1}{2}\left(\frac{1}{2}\right)-\frac{1}{2}\right]\\ | |||
&= \frac{1}{2}-1+\frac{1}{4}+\frac{1}{2} \\ | |||
&= \frac{1}{4} | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Latest revision as of 02:24, 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} & \color{red} \mathbf{y=\cos(x)} & \color{royalblue}\mathbf{y=\sin(2x)} \\ & x=0 & x=\frac{\pi}{2}\\ \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} \cos(x) &= \sin(2x) \\ x &= \frac{\pi}{2} \\ x &= \frac{\pi}{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 \int_{0}^{\frac{\pi}{6}} \left(\cos(x) - \sin(2x) \right)dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \left(\sin(2x)- \cos(x) \right)dx = \frac{1}{4}+\frac{1}{4} = \frac{2}{4} = \frac{1}{2} }
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} \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(\frac{\pi}{6})+\frac{1}{2}\cos(\frac{2\pi}{6})\right]-\left[\sin(0)+\frac{1}{2}\cos(2(0))\right] \\[2ex] &= \left[\frac{1}{2}+\frac{1}{2}\left(\frac{1}{2}\right)\right]-\left[0-\frac{1}{2} (1)\right] \\[2ex] &= \frac{1}{2}+\frac{1}{4}-\frac{1}{2} \\ &= \frac{1}{4} \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} \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}} \\ [2ex] &= \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] \\[2ex] &= \left[-\frac{1}{2}\left(-1\right)-1\right]-\left[-\frac{1}{2}\left(\frac{1}{2}\right)-\frac{1}{2}\right]\\ &= \frac{1}{2}-1+\frac{1}{4}+\frac{1}{2} \\ &= \frac{1}{4} \end{align} }